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 DearCs Stereotype Edition. 
 
 ELEMENTS 
 
 OF 
 
 GEOMETRY: 
 
 CONTAINING 
 
 THE FIRST SIX BOOKS OF EUCLID, 
 
 SUPPLEMENT 
 
 QUADRATURE OF THE CIRCLE. AND THE 
 GEOMETRY OF SOLIDS : 
 
 TO WHICH ARE ADDED, 
 
 ELEMENTS OF PLANE AND SPHERICALE 
 TRIGONOMETRY. 
 
 BY 
 
 JOHN PLAYFAIR, F.R.S. Lond. & Edin. 
 
 PBOnftOB OF NATURAL PHILOSOPHY, PORMERLY OP MATHEMATICS, IX 
 UNIVERSITY OF EDINBURGH. 
 
 FROM THE LAST LONDON EDITION, ENLARGED. 
 
 PHILADELPHIA: 
 J. B. LIPPINCOTT & 0.
 
 Entered according to the Act of Congress, in the year One Thousand 
 Eight Hundred and Forty-Jive, by W. E. Dean, in tJte Clerk's Of- 
 fice of the Southern District of New-York.
 
 °Wl2Ml7 5& 
 
 t*R£F ACE. 
 
 Ir is a remarkable fact in the history ol science, that the oldest book ot 
 Elementary Geometry is still considered as the best, and that the writings 
 of Euclid, at the distance of two thousand years, continue to form the most 
 approved introduction to the mathematical sciences. This remarkable 
 distinction the Greek Geometer owes not only to the elegance and correct- 
 ness of his demonstrations, but to an arrangement most happily contrived 
 for the purpose of instruction, — advantages which, when they reach a cer- 
 tain eminence, secure the works of an author against the injuries of time 
 more effectually than even originality of invention. The Elements of Eu- 
 clid, however, in passing through the hands of the ancient editors during 
 the decline of science, had suffered some diminution of their excellence, and 
 much skill and learning have been employed by the modern mathemati- 
 cians to deliver them from blemishes which certainly did not enter into their 
 original composition. Of these mathematicians, Dr. Sim son, as he may 
 be accounted the last, has also been the most successful, and has left very 
 little room for the ingenuity of future editors to be exercised in, either by 
 amending the text of Euclid, or by improving the translations from it. 
 
 Such being the merits of Dr. Simson's edition, and the reception it has 
 met with having been every way suitable, the work now offered to the pub- 
 lic will perhaps appear unnecessary. And indeed, if the geometer just 
 named had written with a view of accommodating the Elements of Euclid 
 to the present state of the mathematical sciences, it is not likely that any 
 thing new in Elementary Geometry would have been soon attempted. But 
 his design was different; it was hisobject to restore the writings of Euclid 
 to their original perfectio.n, and to give them to Modern Europe as nearly 
 as possible in the state wherein they made their first appearance in Ancient 
 Greece. For this undertaking, nobody could be better qualified than Dr. 
 Simson ; who, to an accurate knowledge of the learned languages, and an 
 indefatigable spirit of research, added a profound skill in the ancient Geome- 
 try, and an admiration of it almost enthusiastic. Accordingly, he not only 
 restored the text of Euclid wherever it had been corrupted, but in some 
 cases removed imperfections that probably belonged to the original work : 
 though his extreme partiality for his author never permitted him to suppose 
 that such honour could fall to the share either of himself, or of any other of 
 the moderns. 
 
 But, after all this was accomplished, something still remained to be dour, 
 since, notwithstanding the acknowledged excellence of Euclid's Ele- 
 ments, it could not be doubted that some alterations might bo made that 
 would accommodate them better to a state of the mathematical sciences, so 
 much more improved and extended than at the period when they were 
 written. Accordingly, the object of the edition now offered to the public, is 
 not so much to give the writings of Euclid the form which thi-y originally 
 had, as that which may at present render them most useful.
 
 PREFACE. 
 
 One ol the alterations made with this view, respects the Doctrine of 
 Proportion, the method of treating which, as it is laid down in the fifth of 
 Euclid, has gieat advantages accompanied with considerable defects ; 01 
 which, however, it must be observed, that the advantages are essential, and 
 the defects only accidental. To explain the nature of the former requires 
 a more minute examination than is suited to this place, and must therefore 
 be reserved for the Notes ; but, in the mean time, it may be remarked, that 
 no definition, except that of Euclid, has ever been given, from which the 
 properties of proportionals can be deduced by reasonings, which, at the 
 same time that they are perfectly rigorous, are also simple and direct. As 
 to the defects, the prolixness and obscurity that have so often been com- 
 plained of in the fifth Book, they seem to arise chiefly from the nature of 
 the language employed, which being no other than that of ordinary dis- 
 course, cannot express, without much tediousness and circumlocution, the 
 relations of mathematical quantities, when taken in their utmost generality, 
 and when no assistance can be received from diagrams. As it is plain that 
 the concise language of Algebra is directly calculated to remedy this in- 
 convenience, I have endeavoured to introduce it here, in a very simple form 
 however, and without changing the nature of the reasoning, or departing 
 in any thing from the rigour of geometrical demonstration. By this means, 
 the steps of the reasoning which were before far separated, are brought 
 near to one another, and the force of the whole is so clearly and directly 
 perceived, that I am persuaded no more difficulty will be found in under- 
 standing the propositions of the fifth Book than those of any other of the 
 Elements. 
 
 In the second Book, also, some algebraic signs have been introduced, for 
 the sake of representing more readily the addition and subtraction of the 
 rectangles on which the demonstrations depend. The use of such sym- 
 bolical writing, in translating from an original, where no symbols are used, 
 cannot, I think, be regarded as an unwarrantable liberty : for, if by that 
 means the translation is not made into English, it is made into that univer- 
 sal language so much sought after in all the sciences, but destined, it would 
 seem, to be enjoyed only by the mathematical. ■ 
 
 The alterations above mentioned are the most material that have been 
 attempted on the books of Euclid. There are, however, a few others, 
 which, though less considerable, it is hoped may in some degree facilitate 
 the study of the Elements. Such are those made on the definitions in the 
 first Book, and particularly on that of a straight line. A new axiom- is also 
 introduced in the room of the 12th, for the purpose of demonstrating more 
 easily some of the properties of parallel lines. In the third Book, the re- 
 marks concerning the angles made by a straight line, and the circumference 
 of a circle, are left out, as tending to perplex one who has advanced no 
 farther than the elements of the science. Some propositions also have 
 been added ; but for a fuller detail concerning these changes, I must refer 
 to the Notes, in which several of the more difficult, or more interesting sub- 
 jects of Elementary Geometry are treated at considerable length. 
 
 College of Edinburgh, 
 Dec. 1, 1813
 
 ELEMENTS 
 
 OF 
 
 GEOMETRY 
 
 BOOK I. 
 
 THE PRINCIPLES. 
 
 EXPLANATION OF TERMS AND SIGNS. 
 
 1 Geometry is a science which has for its object the measurement of mag 
 nitudts. 
 
 Magnitudes may be considered under three dimensions, — length, breadth, 
 height or thickness. 
 
 2. In Geometry there are several general terms or principles ; such as, 
 Definitions, Propositions, Axioms, Theorems, Problems, Lemmas, Scho 
 hums, Corollaries, &c. 
 
 3. A Definition is the explication of any term or word in a science, show 
 ing the sense and meaning in which the term is employed. 
 
 Every delinition ought to be clear, and expressed in words that are 
 common and perfectly well understood. 
 
 4. An Axiom, or Maxim, is a self-evident proposition, requiring no formal 
 demonstration to prove the truth of it ; but is received and assented to as 
 soon as mentioned. 
 
 Such as, the whole of any thing is greater than a part of it ; or, the 
 whole is equal to all its parts taken together ; or, two quantities that 
 are each of them equal to a third quantity, are equal to each other. 
 
 f. A Theorem is a demonstrative proposition ; in which some property is 
 
 asserted, and the truth of it required to be proved. 
 
 Thus, when it is said that the sum of the three angles of any plane tri- 
 angle is equal to two right angles, this is called a Theorem ; and tho 
 method of collecting the several arguments ami proofs, and laying 
 them together in proper order, by menus of which the truth of tho 
 proposition becomes evident, is called a Dunntustrntion. 
 
 6 A Direct Demonstration is that which concludes with the direct and cc 
 tain proof of the proposition in hand. 
 
 It is also called Positive or Affirmative, and sometimes an Ostensive De 
 mrnstration, because it is most satisfactory to the mind
 
 d ELEMENTS 
 
 7 An Indirect or Negative Demonstration is that which shows a proposition 
 to be true, by proving that some absurdity would necessarily follow if 
 the proposition advanced were false. 
 
 This is sometimes called Reductio ad Absurdum ; because it shows the 
 absurdity and falsehood of all suppositions contrary to that contained 
 in the proposition. 
 
 8 A Problem is a proposition or a question proposed, which requires a so- 
 lution. 
 
 As, to draw one line perpendicular to another ; or to divide a line into 
 two equal parts. 
 
 9. Solution of a problem is the resolution or answer given to it. 
 
 A Numerical or Numeral solution, is the answer given in numbers. A 
 Geometrical solution, is the answer given by the principles of Geome- 
 try. And a Mechanical solution, is one obtained by trials. 
 
 10. A Lemma is a preparatory proposition, laid down in order to shorten 
 the demonstration of the main proposition which follows it. 
 
 1 1 . A Corollary, or Consectary, is a consequence drawn immediately from 
 some proposition or other premises. 
 
 i2. A Scholium is a remark or observation made on some foregoing propo- 
 sition or premises. 
 
 13. An Hypothesis is a supposition assumed to be true, in order to argue 
 from, or to found upon it the reasoning and demonstration of some pro- 
 position. 
 
 14. A Postulate, or Petition, is something required to be done, which is so 
 easy and evident that no person will hesitate to allow it. 
 
 15. Method is the art of disposing a train of arguments in a proper order, 
 to investigate, the truth or falsity of a proposition, or to demonstrate it to 
 others when it has been found out. This is either Analytical or Syn- 
 thetical. 
 
 16. Analysis, or the Analytic method, is the art or mode of finding out the 
 truth of a proposition, by first supposing the thing to be done, and then 
 reasoning step by step, till we arrive at some known truth. This is also 
 called the Method of Invention, or Resolution ; and is that which is com- 
 monly used in Algebra. 
 
 1 7. Synthesis, or the Synthetic Method, is the searching out truth, by first 
 laying down simple principles, and pursuing the consequences flowing 
 from them till we arrive at the conclusion. This is also called the Me- 
 thod of Composition ; and is that which is commonly used in Geometry. 
 
 (8. The sign = (or two parallel lines), is the sign of equality; thus, 
 A = B, implies that the quantity denoted by A is equal to the quantity 
 denoted by B, and is read A equal to B. 
 
 19. To signify that A is greater than B, the expression A 7 B is used. A.nd 
 to sigp'fv that A is less than B, the expression K/_ B is used. 
 
 \
 
 OF GEOMETRY. BOOK I. 7 
 
 20. The sign of Addition is an erect cross ; thus A+B implies the sum o 
 A and B, and is called A plus B. 
 
 21. Subtraction is denoted by a single line; as A — B, which is read A 
 minus B ; A — B represents their difference, or the part of A remainii.^, 
 when a part equal to B has been taken away from it. 
 
 In like manner, A — B+C, or A+C — B, signifies that A and C are to 
 be added together, and that B is to be subtracted from their sum. 
 
 22. Multiplication is expressed by an oblique cross, by a point, or by simple 
 apposition : thus, A X B, A . B, or AB, signifies that the quantity de- 
 noted by A is to be multiplied by the quantity denoted by B. The ex- 
 pression AB should not be employed when there is any danger of con- 
 founding it with that of the line AB, the distance between the points A 
 and B. The multiplication of numbers cannot be expressed by simple 
 apposition. 
 
 23. When any quantities are enclosed in a parenthesis, or have aline drawn 
 over them, they are considered as one quantity with respect to other 
 symbols: thus, the expression AX(B+C — D), or Ax B+C — D, re- 
 presents the product of A by the quantity B+C — D. In like manner, 
 (A+B)X(A — B+C), indicates the product of A+B by the quantity 
 A— B+C. 
 
 24. The Co-efficient of a quantity is the number prefixed to it: thus, 2AB 
 signifies that the line AB is to be taken 2 times ; £AB signifies the half 
 of the line AB. 
 
 25. Division, or the ratio of one quantity to another, is usually denoted by 
 placing one of the two quantities over the other, in the form of a fraction ■ 
 
 thus, — signifies the ratio or quotient arising from the division of the 
 
 quantity A by B. In fact, this is division indicated. 
 
 26. The Square, Cube, &c. of a quantity, are expressed by placing a small 
 figure at the right hand of the quantity : thus, the square of the line 
 AB is denoted by AB 2 , the cube of the line AB is designated by AB 3 ; 
 and so on. 
 
 27. The Roots of quantities are expressed by means of the radical sign -/, 
 with the proper index annexed ; thus, the square root of 5 is indicated 
 
 i/5 ; V( A X B) means the square root of the product of A and B, or the 
 mean proportional between them. The roots of quantities are some- 
 times expressed by means of fractional indices : thus, the cube root of 
 
 A x B x C may be expressed by VAxBxC, or (A x B x C)», and 
 so on. 
 
 28. Numbers in a parenthesis, such as (15. 1.), refers back to the numbei 
 of the proposition and the Book in which it has been announced or de- 
 monstrated. The expression (15. 1.) denotes the fifteenth proposition, 
 first book, and so on. Id like manner, (3. Ax.) designates the third 
 axiom; (2. Post.) the second postulate; (Def. 3.) the third lefinition, 
 and so on
 
 8 ELEMENTS 
 
 29. The word, therefore, or hence, frequently occurs. To express either of 
 these words, the sign .*. is generally used. 
 
 30. If the quotients of two pairs of numbers, or quantities, are equal, the 
 
 A C 
 quantities are said to be proportional : thus, if p =-; then, A. is to B 
 
 asCtoD. And the abbreviations of the proportion is, A : B : : C : D ; 
 it is sometimes written A : B=C : D. 
 
 DEFINITIONS. 
 
 1. "A Point is that which has position, but not magnitude*.'' (See 
 Notes.) 
 
 2. A line is length without breadth. 
 
 " Corollary. The extremities of a line are points ; and the intersections 
 " of one line with another are also points." 
 
 3. " If two lines are such that they cannot coincide in any two points, with- 
 " out coinciding altogether, each of them is called a straight line." 
 
 " Cor. Hence two straight lines cannot inclose a space. Neither can two 
 " straight lines have a common segment ; that is, they cannot coincide 
 " in part, without coinciding altogether." 
 
 <i. A superficies is that which has only length and breadth. 
 • Cor. The extremities of a superficies are lines ; and the intersections of 
 " one superficies with another are also lines." 
 
 5. A plane superficies is that in which any two points being taken, tho 
 straight line between them lies wholly in that superficies. 
 
 6. A plane rectilineal angle is the inclination of two straight lines to one 
 another, which meet together, but are not in the same straight line 
 
 E 
 
 N. B. 'When several angles are at one point B, any one of them is ex- 
 pressed by three letters, of which the letter that is at the vertex of the an- 
 gle, that is, at the point in which the straight lines that contain the angle 
 meet one another, is put between the other two letters, and one of these 
 two is somewhere upon one of those straight lines, and the other upon the 
 other line : Thus the angle which is contained by the straight lines, AB 
 OB, is named the angle ABC, or CBA ; that which is contained by AB, 
 
 * The definitions marked with inverted commas are different from those of Kaiiid.
 
 OF GEOMETRY. BOOK I. q 
 
 • BD, is named the angle ABD, or DBA ; and that which is contained by 
 BD, CB, is called the angle DBC, or CBD ; but, if there be only one an- 
 gle at a point, it may be expressed by a letter placed at that point ; as th« 
 
 • angle at E.' 
 
 When a straight line standing on another 
 straight line makes the adjacent angles equal 
 to one another, each of the angles is called 
 a right angle ; and the straight line which 
 stands on the other, is called a perpendicu- 
 lar to it. 
 
 8. An obtuse angle is that which is greater than a right angle. 
 
 9. An acute angle is that which is less than a right angle. 
 
 10. A figure is that which is enclosed by one or more boundaries. — The 
 word area denotes the quantity of space contained in a fgure, without any 
 reference to the nature of the line or lines which bound it. 
 
 11. A circle is a plane figure contained by one line, which is called the 
 circumference, and is such that all straight lines drawn from a certain 
 point within the figure to the circumference, are equal to one another 
 and are called radii 
 
 12. And this point is called the centre of the circle. 
 
 IS. A diameter of a circle is a straight line drawn through the cenire, and 
 terminated both ways by the circumference. 
 
 14. A semicircle is the figure contained by a diameter and the part o! 'Ur 
 circumference cut off by the diameter 
 
 2
 
 10 
 
 ELEMENTS 
 
 15. Rectilineal figures are those which are contained by straight line* 
 10. Trilateral figures, or triangles, by three straight lines. 
 
 17. Quadrilateral, by four straight lines. 
 
 18. Multilateral figures, or polygons, by more than four straight lines. 
 
 19. Of three sided figures, an equilateral triangle is that which has chree 
 equal sides. 
 
 20. An isosceles triangle is that which has only two sides equal. 
 
 21. A scalene triangle is that which has three unequal sides. 
 
 22 A right angled triangle is that which has a right angle. 
 
 23. An obtuse angled triangle is that which has an obtuse angle. 
 
 24 An acute angled triangle is that which has three acute angles. 
 
 25 Of four sided figures, a square is that which has all its sides equal 
 and all its angles right angles. 
 
 26. An oblong is that which has all its angles right angles, but has not all 
 'ts sides equal. 
 
 27 A rhombus is that which has all its sides equal, but its angles are not 
 right angles.
 
 OF GEOMETRY. BOOK I. 11 
 
 28. A rhomboid is that which has its opposite sides equal to one another, 
 but all its sides are not equal, nor its angles right angles. 
 
 29. All other four sided figures besides these, are called trapeziums. 
 
 30. Parallel straight lines are such as are in the same plane, and which 
 • being produced ever so far both ways, do not meet. 
 
 POSTULATES. 
 
 1 . Let it be granted that a straight line may be drawn from any one point 
 to any other point. 
 
 2. That a terminated straight line may be produced to any length in a 
 straight line. 
 
 3. And that a circle may be described from any centre at any distance 
 from that centre 
 
 AXIOMS. 
 
 1. Things which are equal to the same thing are equal to ono another. 
 
 2. If equals be added to equals, the wholes are equal. 
 
 3. If equals be taken from equals, the remainders are equal. 
 
 4. If equals be added to unequals, the wholes are unequal. 
 
 5. If equals be taken from unequals, the remainders are unequal. 
 
 8. Things which are doubles of the same thing, are equal to one another. 
 
 7. Things which are halves of the same thing, are equal to one another. 
 
 8. Magnitudes which coincide with one another, that is, which exactly 
 fill the same space, are equal to one another. 
 
 9. The whole is greater than its part. 
 
 10. All right angles are equal to one another. 
 
 11." Two straight lines which intersect one another, cannot be both pa- 
 " ralle to the same straight line."
 
 12 
 
 ELEMENTS 
 
 PROPOSITION I. PROBLEM. 
 
 To describe an equilateral triangle uv- a a given finite straight line. 
 
 Let AB be the given straight line ; it is required to describe an equi« 
 lateral triangle upon it. 
 
 From the centre A, at the dis- 
 tance AB, describe (3. Postulate) 
 the circle BCD, and from the cen- 
 tre B, at the distance BA, describe 
 the circle ACE ; and from the point 
 C, in which the circles cut one an- 
 other, draw the straight lines (1. 
 Post.) CA, CB to the points A, B ; 
 ABC is an equilateral triangle. 
 
 Because the point A is the cen- 
 tirt of the circle BCD, AC is equal 
 
 (11. Definition) to AB ; and because the point B is the centre of the cir- 
 cle ACE, BC is equal to AB : But it has been proved that CA is equal 
 to AB ; therefore CA, CB are each of them equal to AB ; now things 
 which are equal to the same are equal to one another, (1.- Axiom) ; there- 
 fore CA is equal to CB ; wherefore CA, AB, CB are equal to one another ; 
 and the triangle ABC is therefore equilateral, and it is described upon the 
 given straight line AB. 
 
 PROP. II. PROB. 
 
 From a given point to draw a straight line equal to a given straight line. 
 
 Let A be the given point, and BC the given straight line ; it is required 
 to draw, from the point A, a straight line equal to BC. 
 
 From the point A to B draw (1. Post.) 
 the straight line AB ; and upon it describe 
 (1. 1.) the equilateral triangle DAB, and 
 produce (2. Post.) the straight lines DA, 
 BD, to E and F ; from the centre B, at the 
 distance BC, describe (3. Post.) the circle 
 CGH, and from the centre D, at the dis- 
 tance DG, describe the circle GKL, AL is 
 equal to BC. 
 
 Because the point B is the centre of the 
 circle CGH, BC is equal (11. Def.) to BG ; 
 and because D is the centre of the circle 
 GKL, DL is equal to DG, and DA, DB, 
 parts of them, are equal ; therefore the re- 
 mainder AL is equal to the remainder (3. 
 
 Ax.) BG: But it has been shewn that BC is equal to BG ; wherefore 
 AL and BC are each of them equal to BG ; and things that are equal
 
 OF GEOMETRY. BOOK I. 
 
 13 
 
 to the same are equal to one another ; therefore the straight line AL is 
 equal to BC. Wherefore, from the given point A, a straight line AL hat 
 been drawn equal to the given straight line BC. 
 
 PROP. III. PROB. 
 
 From the greater of two given straight lines to et»J i£ a part equal to the 
 
 less. 
 
 Let AB and C be the two given straight 
 lines, whereof AB is the greater. It is 
 required to cut off from AB, the greater, 
 a part equal to C, the less. 
 
 From the point A draw (2. 1.) the 
 straight line AD equal to C ; and from 
 the centre A, and at the distance AD, 
 describe (3. Post.) the circle DEF; and 
 because A is the centre of the circle 
 DEF, AE is equal to AD; but the 
 straight line C is likewise equal to AD ; 
 
 whence AE and C are each of them equal to AD ; wherefore the straight 
 line AE is equal to (1. Ax.) C, and from AB the greater of two straight 
 lines, a part AE has been cut off equal to C the less. 
 
 PROP. IV. THEOREM. 
 
 If two triangles have two sides of the one equal to two sides of the other, each 
 to each ; and have likewise the angles contained by those sides equal to 
 one another, their bases, or third sides, shall be equal ; and the areas of 
 the triangles shall be equal ; and their other angles shall be equal, each to 
 each, viz. those to which the equal sides are oppos te* 
 
 Let ABC, DEF be two triangles which have the two sides AB, AC 
 equal to the two sides DE DF, each to each. viz. AB to DE. and AC to 
 DF , and let the angle 
 BAC be also equal to the 
 angle EDF: then shall 
 .lie base BC be equal to 
 the base EF ; and the tri- 
 angle ABC to the triangle 
 DEF; and the other an- 
 gles, to which the equnl 
 sides are opposite, shall 
 be equal, each to each, ** O E r 
 
 viz. the angle ABC to the angle DEF, and the angle ACB to DFE. 
 
 For, if the triangle ABC be applied to the triangle DEF, so that the 
 point A may be on D, and the straight line AB upon DE ; the point B 
 shall coincide with the point E, because AB is equal to DE ; and AB 
 
 * The three conclusions in this enunciation are more briefly expressed by say ng, that tk, 
 tnonfttt a-e every way equal.
 
 i4 ELEMENTS 
 
 coinciding with DE, AC shall coincide with DF, because the angle BAC 
 is equal to the angle EDF ; wherefore also the point C shall coincide with 
 ihe point F, because AC is equal to DF : But the point B coincides with 
 the point E ; wherefore the base BC shall coincide with the base EF 
 cor. def. 3.), and shall be equal to it. Therefore also the whole triangle 
 ABC shall coincide with the whole triangle DEF, so that the spaces whicr 
 they contain or their areas are equal ; and the remaining angles of the 
 one shall coincide with the remaining angles of the other, and be equal to 
 them, viz. the angle ABC to the angle DEF, and the angle ACB to the 
 angle DFE. Therefore, if two triangles have two sides of the one equal 
 to two sides of the other, each to each, and have likewise the angles con- 
 tained by those sides equal to one another ; their bases shall be equal, 
 and their areas shall be equal, and their other angles, to which the equal 
 sides are opposite, shall be equal, each to each. 
 
 PROP. V. THEOR. 
 
 The angles at the base of an Isosceles triangle are equal to one another ; and 
 if the equal sides be produced, the angles upon the other side of the base 
 shall be equal. 
 
 Let ABC be an isosceles triangle, of which the side AB is equal to AC 
 and let the straight lines AB, AC be produced to D and E, the angle ABC 
 shall be equal to the angle ACB, and the angle CBD to the angle BCE. 
 
 In BD take any point F, and from AE the greater cut off AG equal 
 (3. 1.) to AF, the less, and join FC, GB. 
 
 Because AF is equal to AG, and A B to AC, the two sides FA, AC are equal 
 to the two GA, AB, each to each ; and they contain the angle FAG com- 
 mon to the two triangles, AFC, AGB ; 
 therefore the base FC is equal (4. 1.) to 
 the base GB, and the triangle AFC to 
 the triangle AGB ; and the remaining 
 angles of the one are equal (4. 1.) to 
 the remaining angles of the other, each to 
 each, to which the equal sides are oppo- 
 site, viz. the angle ACF to the angle 
 ABG, and the angle AFC to the angle 
 AGB : And because the whole AF is 
 equal to the whole AG, and the part AB 
 to the part AC ; the remainder BF shall 
 be equal (3. Ax.) to the remainder CG ; 
 and FC was proved to be equal to GB, "' \E 
 
 therefore the two sides BF, FC are equal to the two CG, GB, each to 
 each ; but the angle BFC is equal to the angle CGB ; wherefore the tri- 
 angles BFC, CGB are equal (4. 1.), and their remaining angles are equal- 
 to which the equal sides are opposite ; therefore the angle FBC is equal 
 to the angle GCB, and the angle BCF to the angle CBG. Now, since 
 it has been demonstrated, that the whole angle ABG is equal to the whole 
 ACF, and the part CBG to the part BCF, the remaining angle ABC is 
 therefore equal to the remaining angle ACB, which are the angles at the
 
 OF GEOMETRY. BOOK I. 15 
 
 base of the triangle ABC : And it has also been proved that the angle 
 FBC is equal to the angle GCB, which are the angles upon the other side 
 of the base. 
 
 Corollary. Henco every equilateral triangle is also equiangular 
 
 PROP. VI. THEOR. 
 
 If twg angles of a triangle be equal to one another, the sides which subtend 
 or are opposite to them, are also equal to one another. 
 
 Let ABC be a triangle having the angle ABC equal to the angle ACB ; 
 the side AB is also equal to the side AC. 
 
 For, if AB be not equal to AC, one of them is 
 greater than the other : Let AB be the greater, 
 and from it cut (3. 1.) off DB equal to AC the 
 less, and join DC ; therefore, because in the tri- 
 angles DBC, ACB, DB is equal to AC, and BC 
 common to both, the two sides DB, BC are equal 
 to the two AC, CB, each to each ; but the angle 
 DBC is also equal to the angle ACB ; therefore 
 the base DC is equal to the base AB, and the area 
 of the triangle DBC is equal to that of the triangle 
 (4. 1.) ACB, the less to the greater ; which is ab- 
 surd. Therefore, AB is not unequal to AC, that 
 is, it is equal to it. 
 
 Cor. Hence every equiangular triangle is also equilateral. 
 
 PROP. VII. THEOR. 
 
 Upon the same base, and on the same side of it, there cannot be ttoo triangles, 
 that have their sides which are terminated in one extremity of the base 
 equal to one another, and likewise those which are terminated in the other 
 extremity, equal to one another. 
 
 Let there be two triangles ACB, ADB, upon the same base AB, and 
 upon the same side of it, which have their sides CA, DA, terminated in A 
 equal to one another ; then their sides CB, DB, terminated in B. cannot 
 be equal to one another. 
 
 Join CD, and if possible let CB be 
 equal to DB ; then, in the case in which 
 the vertex of eich of the triangles is with- 
 out the other triangle, because AC is 
 equal to AD, the angle ACD is equal (5. 
 1.) to the angle ADC : But the angle 
 ACD is greater than the angle BCD ; 
 therefore the angle ADC is greater also 
 than BCD ; much more then is the angle 
 BDC greater than the angle BCD. Again, 
 because CB is equal to DB, the angle 
 BDC is equal (5. 1.) to the angle BCD ; A^
 
 16 
 
 ELEMENTS 
 
 bui it has been demonstrated to be greater than It ; which is imposs* 
 ble. 
 
 But if one of the vertices, as D, 
 be within the other triangle ACB ; 
 produce AC, AD to E, F ; therefore, 
 because AC is equal to AD in the 
 triangle ACD, the angles ECD, FDC 
 upon the other side of the base CD 
 are equal (5. 1.) to one another, but 
 the angle ECD is greater than the 
 angle BCD ; wherefore the angle 
 FDC is likewise greater than BCD ; 
 much more then is the angle BDC greater than the angle BCD. Again, 
 because CB is equal to DB, the angle BDC is equal (5. 1.) to the angle 
 BCD ; but BDC has been proved to be greater than the same BCD ; 
 which is impossible. The case in which the vertex of one triangle is 
 upon a side of the other, needs no demonstration. 
 
 Therefore, upon the same base, and on the same side of it, there cannot 
 be two triangles that have their sides which are terminated in one extrem 
 ity of the base equal to one another, and likewise those which are termina 
 ted in the other extremity equal to one another. 
 
 PROP. VIII. THEOR. 
 
 If two triangles have two sides of the one equal to two sides of the othet 
 each to each, and have likewise their bases equal ; the angle which is contain 
 ed by the two sides of the one shall be equal to the angle contained by the twt 
 sides of the other. 
 
 Let ABC, DEF be two triangles having the two sides AB, AC, equal 
 to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF ; 
 
 The angle BAC is equal to 
 
 and also the base BC equal to the base EF. 
 •he angle EDF. 
 
 For, if the triangle ABC be applied to the triangle DEF, so that the 
 point B be on E, and the straight line BC upon EF ; the point C shall also 
 coincide with the point F, because BC is equal to EF : therefore BC coin- 
 ciding with EF, BA and AC shall coincide with ED and DF ; for, if 
 BA and CA do not coincide with ED and FD, but have a different situa-
 
 OF GEOMETRY. BOOK I. 
 
 A* 
 
 oon, as EG and FG ; then, upon the same base EF, and upon the same 
 side of it, there can be two triangles EDF, EGF,that have their sides which 
 are terminated in one extremity of the base equal to one another, and like- 
 wise their sides terminated in the other extremity ; but this is impossible 
 (7. 1.); therefore, if the base BC coincides with the base EF, the sides 
 BA, AC cannot but coincide with the sides ED, DF ; wherefore likewise 
 the angle BAC coincides with the angle EDF, and is equal (8. Ax.) to it. 
 
 PROP. IX. PROB. 
 
 To bisect a given rectilineal angle, that is, to divide it into two equal angles 
 
 Let BAC be the given rectilineal angle, it is required to bisect it. 
 
 Take any point D in AB, and from AC cut 
 (3. 1.) off AE equal to AD ; join DE, and upon 
 it describe (1. 1.) an equilateral triangle DEF ; 
 then join AF ; the straight line AF bisects 
 the angle BAC. 
 
 Because AD is equal to AE, and AF is com- 
 mon to the two triangles DAF, E AF ; the two 
 sides DA, AF, are equal to the two sides EA, 
 AF, each to each ; but the base DF is also 
 equal to the base EF ; therefore the angle 
 DAF is equal (8. 1.) to the angle EAF : where- 
 fore the given rectilineal angle BAC is bisect- 
 ed by the straight line AF. 
 
 SCHOLIUM. 
 
 By the same construction, each of the halves BAF, CAF, may be divi. 
 ded into two equal parts ; and thus, by successive subdivisions, a given an- 
 gle may be divided into four equal parts, into eight, into sixteen, and so on. 
 
 PROP. X. PROB. 
 
 To bisect a given finite straight line, that is, to divide it into two equal parts. 
 
 Let AB be the given straight line ; it is required to divide it into two equal 
 parts. 
 
 Describe (1. 1.) upon it an equilateral triangle ABC, and bisect (9. 1.) 
 the angle ACB by the straight line CD. AB is 
 cut into two equal parts in the point D. 
 
 Because AC is equal to CB, and CD common 
 to the two triangles ACD, BCD : the two sides 
 AC, CD, are equal to the two BC, CD, each to 
 each ; but the angle ACD is also equal to the an- 
 gle BCD ; therefore the base AD is equal to the 
 base (4. 1.) DB, and the straight line AB is divi- 
 ded into two equal parts in the point D.
 
 18 
 
 ELEMENTS 
 
 PROP. XL PROB. 
 
 To draw a straight line at right angles to a given straight line, from a given 
 
 point in that line. 
 
 Let AB be a given straight line, and C a point given in it ; it is requi- 
 red to draw a straight line from the point C at right angles to AB 
 
 Take any point D in AC, and (3. 1.) make CE equal to CD, and upon 
 DE describe (1.1.) the equilateral -p 
 
 triangle DFE, and join FC; the 
 straight line FC, drawn from the giv- 
 en point C, is at right angles to the 
 given straight line AB. 
 
 Because DC is equal to CE, and 
 FC common to the two triangles 
 DCF, EOF, the two sides DC, CF 
 are equal to the two EC, CF, each ADC E B 
 
 to each ; but the base DF is also equal to the base EF ; therefore the 
 angle DCF is equal (8. 1.) to the angle ECF ; and they are adjacent an- 
 gles. But, when the adjacent angles which one straight line makes with 
 another straight line are equal to one another, each of them is called a 
 right (7. def.) angle ; therefore each of the angles DCF, ECF, is a right 
 angle. Wherefore, from the given point C, in the given straight line AB, 
 FC has been drawn at right angles to AB. 
 
 PROP. XII. PROB. 
 
 To draw a straight line perpendicular to a given straight line, of an unlimited 
 length, from a given point without it. 
 
 Let AB be a given straight line, which may be produced to any length 
 both ways, and let C be a point without it. It is required to draw a straight 
 line perpendicular to AB from the 
 point C. 
 
 Take any point D upon the other 
 side of AB,and from the centre C, at 
 the distance CD, describe (3. Post.) 
 the circle EGF meeting AB in F, G : 
 and bisect (10. 1.) FG in H, and join 
 CF, CH, CG ; the straight line CH, 
 drawn from the given point C, is per- 
 pendicular to the given straight line AB. 
 
 Because FH is equal to HG, and HC common to the two triangles FHC, 
 GHC, the two sides FH, HC are equal to the two GH, HC, each to each , 
 but the base CF is also equal (11. Def. 1.) to the base CG ; therefore the 
 angle CHF is equal (8. 1.) to the angle CHG ; and they are adjacent an- 
 gles ; now when a straight line standing on a straight line makes the ad- 
 jacent angles equal to one another, each of them is a right angle, and 
 the straight line which stands upon the other is called a perpendicular to 
 it ; therefore from the given point C a perpendicular CH has been drawn 
 to the given sfc-aight line AB.
 
 OF GEOMETRY. BOOK i. 
 
 19 
 
 PROP. XIII. THEOR. 
 
 The angles which one straight line makes with another upon one side oj it, are 
 either two right angles, or are together equal to two right angles. 
 
 Let the straight line AB make with CD, upon one side of it the angles 
 CBA, ABD ; these are either two right angles, or are together equal to two 
 right angles. 
 
 For, if the angle CBA be equal to ABD, each of them is a right angle 
 (Def. 7.) ; but, if not, from the point B draw BE at right angles (11. 1.) 
 
 E A 
 
 D B D B C 
 
 to CD ; therefore the angles CBE, EBD are two right angles. Now, the 
 angle CBE is equal to the two angles CBA, ABE together ; add the an- 
 gle EBD to each of these equals, and the two angles CBE, EBD will be 
 equal (2. Ax.) to the three CBA, ABE, EBD. Again, the angle DBA is 
 equal to the two angles DBE, EBA ; add to each of these equals the angle 
 ABC ; then will the two angles DBA, ABC be equal to the three angles 
 DBE, EBA, ABC; but the angles CBE, EBD have been demonstrated 
 to be equal to the same three angles ; and things that are equal to the same 
 are equal (1. Ax.) to one another; therefore the angles CBE, EBD are 
 equal to the angles DBA, ABC ; but CBE, EBD, are two right angles ; 
 therefore DBA, ABC ; are together equal to two right angles. 
 
 Cor. The sum of all the angles, formed on the same side of a straight 
 line DC, is equal to two right angles ; because their sum is equal to that 
 of the two adjacent angles DBA, ABC. 
 
 PROP. XIV. THEOR. 
 
 If, at a point in a straight line, two other straight lines, upon the opposite 
 sides of it, make the adjacent angles together equal to two right angles, 
 these two straight lines are in one and the same straight line. 
 
 At the point B in the straight line AB, 
 let the two straight lines BC, BD upon 
 the opposite sides of A B. make the adja- 
 cent angles ABC, ABD equal together 
 to two right angles. BD is in the same 
 straight line with CB. 
 
 For if BD be not in the same straight 
 Une with CB, let BE !><: in the s:mio 
 •trtiiuht line with it; therefore, because 
 he straigtit lino AB mrikes angles with 
 jxq straight line CBE, upon one side of
 
 90 ELEMENTS 
 
 it, tV argles ABC, ABE are together equal (13. 1.) to two right angles ; 
 but the angles ABC, ABD are likewise together equal to two right angles : 
 therefore the angles CBA, ABE are equal to the angles CBA, ABD. 
 Take away the common angle ABC, and the remaining angle ABE is equal 
 (3. Ax.) to the remaining angle ABD, the less to the greater, which is im- 
 possible ; therefore BE is not in the same straight line with BC. And in 
 like manner, it may be demonstrated, that no other can be in the same 
 straight line with it but BD, which therefore is in the same straight line 
 with CB. 
 
 PROP. XV. THEOR. 
 
 If two straight lines cut one another, the vertical, or opposite angles are equal 
 
 Let the two straight lines AB, CD, cut one another in the point E : the 
 angle AEC shall be equal to the angle DEB, and CEB to AED. 
 
 For the angles CEA, AED, which the straight line AE makes with the 
 straight line CD, are together equal (13. 1.) to two right angles : and the 
 angles AED, DEB, which the 
 straight line DE makes with the 
 straight line AB, are also together 
 equal (13. 1.) to two right angles ; 
 therefore the two angles CEA, 
 AED are equal to the two AED, 
 DEB. Take away the common 
 angle AED, and the remaining 
 angle CEA is equal (3. Ax.) to the 
 remaining angle DEB. In the 
 same manner it may be demonstrated that the angles CEB, AED are 
 equal. 
 
 Cor. 1. From this it is manifest, that if two straight lines cut one an- 
 other, the angles which they make at the point of their intersection, are 
 together equal to four right angles. 
 
 Cor. 2. And hence, all the angles made by any number of straight line? 
 meeting in one point, are together equal to four right angles. 
 
 PROP. XVI. THEOR. 
 
 If one side of a triangle be produced, the exterior angle is greater than 
 either of the interior, and opposite angles. 
 
 Let ABC be a triangle, and let its side BC be produced to D, the ex- 
 terior angle ACD is greater than either of the interior opposite angles 
 CBA, BAC. 
 
 Bisect (10. 1.) AC in E, join BE and produce it to F, and make E> 
 equal to BE ; join also FC, and produce AC to G. 
 
 Because AE is equal to EC, and BE to EF ; AE, EB are equal to 
 CE, EF, each to each; and the angle AEB is equal (15. 1.) to the 
 »ngle CEF, because they are vertical angles ; therefore the base AB
 
 OF GEOMETRY. BOOK 
 
 21 
 
 is equal (4. 1 ) to the base CF, and 
 the triangle AEB to the triangle 
 CEF, and the remaining angles to 
 the remaining angles each to each, 
 to which the equal sides are oppo- 
 site ; wherefore the angle BAE is 
 equal to the angle ECF ; but the 
 angle ECD is greater than the an- 
 gle ECF ; therefore the angle ECD, 
 that is A CD, is greater than BAE : 
 In the same manner, if the side BC 
 be bisected, it may be demonstrated 
 that the angle BCG, that is (15. 1.), 
 the angle ACD, is greater than the 
 angle ABC. 
 
 PROP. XVII. THEOR. 
 
 Any two angles of a triangle are together less than two right angles. 
 
 Let ABC be any triangle ; any 
 two of its angles together are less 
 than two right angles. 
 
 Produce BC to D ; and because 
 ACD is the exterior angle of the tri- 
 angle ABC, ACD is greater (16. 1.) 
 than the interior and opposite angle 
 ABC ; to each of these add the angle 
 ACB ; therefore ihe angles ACD, 
 ACB are greater than the angles 
 ABC, ACB j but ACD, ACB are to- 
 gether equal (13. 1.) to two right an- 
 gles : therefore the angles ABC, BCA are less than two right angles. In 
 like manner, it may be demonstrated, that BAC, ACB as also CAB, ABC, 
 are less than two right angles. 
 
 PROP. XVIII. THEOR. 
 
 The greater side of every triangle has the greater angle opposite to it 
 
 Let ABC be a triangle of which the 
 side AC is greater than the side AB ; the 
 angle ABC is also greater than the angle 
 BCA. 
 
 From AC, which is greater than AB, 
 cut off (3. 1.) AD equal to AB, and join 
 BD : and because A I) B is the exterior 
 angle of the triangle BDC, it is greater 
 (16. 1.) than the interior and opposite
 
 29 ELEMENTS 
 
 angle DCB , but ADB is equal (5. 1.) to ABD, because the sid* At* » 
 equal to the side AD ; therefore the angle ABD is likewise greater than 
 the angle ACB; wherefore much more is the angle ABC greater than 
 ACB 
 
 PROP. XIX. THEOR. 
 
 The greater angle of every triangle is subtended by the greater side, cr hai 
 the greater side opposite to it. ■ 
 
 Let ABC be a triangle, of which the angle ABC is greater than the 
 angle BCA ; the side AC is likewise greater than the side AB. 
 
 For, If it be not greater, AC must either 
 be equal to AB, or less than it; it is not 
 equal, because then the angle ABC would 
 be equal (5. 1.) to the angle ACB ; but it is 
 not ; therefore AC is not equal to AB ; nei- 
 ther is it less ; because then the angle ABC 
 would be less (18. l.)than the angle ACB ; 
 but it is not ; therefore the side AC is not -r» 
 less than AB ; and it has been shewn that 
 it is not equal to AB ; therefore AC is greater than AB. 
 
 PROP. XX. THEOR. 
 Any two sides of a triangle are together greater than the third side. 
 
 Let ABC be a triangle ; any two sides of it together arc greater than 
 the third side, viz. the sides BA, AC greater than the side BC ; and AB, 
 BC greater than AC ; and BC, CA greater than AB. 
 
 Produce BA to the point D, and make _. 
 
 (3. 1.) AD equal to AC ; and join DC. 
 
 Because DA is equal to AC, the an- . 
 
 gle ADC is likewise equal (5. 1.) to •"■- 
 
 ACD ; but the angle BCD is greater 
 than the angle ACD ; therefore the an- 
 gle BCD is greater than the angle 
 ADC ; and because the angle BCD of 
 the triangle DCB is greater than its an- B C 
 
 gle BDC, and that the greater (19. 1.) side is opposite to the greater an- 
 gle ; therefore the side DB is greater than the side BC ; but DB is equal 
 to BA and AC together; therefore BA and AC together are greater than 
 BC. In the same manner it may be demonstrated, that the sides AB, 
 BC are greater than CA, and BC, CA greater than AB. 
 
 SCHOLIUM. 
 
 This may be demonstrated without producing any of the sides : thus, 
 •he line BC, for example, is the shortest distance from B to C ; therefore 
 BC is less than BA + AC or BA + AOBC.
 
 OF GEOMETRY. BOOK I. 
 
 21 
 
 PROP. XXI. THEOR. 
 
 If from the ends of one side of a triangle, there be drawn two straight 
 lines to a point within the triangle, these two lines shall be less than iht 
 other two sides of the triangle, but shall contain a greater angle. 
 
 Let the two straight lines BD, CD be drawn from B, C, the ends tf 
 the side BC of the triangle ABC, to the point D within it; BD and DC 
 are less than the other two sides BA, AC of the triangle, but contain an 
 angle BDC greater than the angle BAC. 
 
 Produce BD to E ; and because two sides of a triangle (20. 1.) are 
 greater than the third side, the two sides BA, 
 A.E of the triangle ABE are greater than BE. 
 To each of these add EC ; therefore the 
 sides BA, AC are greater than BE, EC : 
 Again, because the two sides CE, ED, of 
 the triangle CED are greater than CD, if 
 DB be added to each, the sides CE, EB, 
 will be greater than CD, DB ; but it has 
 been shewn that BA, AC are greater than 
 BE, EC ; much more then are BA, AC great- 
 er than BD, DC. 
 
 Again, because the exterior angle of a 
 triangle (16. 1.) is greater than the interior and opposite angle, the exte- 
 rior angle BDC of the triangle CDE is greater than CED ; for the same 
 reason, the exterior angle CEB of the triangle ABE is greater than BAC ; 
 and it has been demonstrated that the angle BDC is greater than the 
 angle CEB ; much more then is the angle BDC greater than the angle 
 BAC. 
 
 PROP. XXII. PROB. 
 
 To construct a triangle of which the sides shall be equal to three given 
 straight lines ; but any two whatever of these lines must be greater than 
 the third (20. 1.). 
 
 Let A, B, C be the three given 
 straight lines, of which any two 
 whatever are greater than the 
 third, viz. A and B greater than 
 C ; A and C greater than B ; and 
 B and C than A. It is required 
 to make a triangle of which the 
 sides shall be equal to A, B, C, 
 each to each. 
 
 Take a straight line DE, ter- 
 minated at the point D, but un- 
 limited towards E, and make 
 (a 1.) DP equal to A, FG to B, 
 and Kill equai to C ; and from
 
 24 ELEMENTS 
 
 tne cer.tre F, -xt the distance FD, describe (3. Post.) the circle DKL , 
 and from the centre G, at the distance GH, describe (3. Post.) another 
 circle HLK ; and join KF, KG ; the triangle KFG has its sides equal to 
 the three straight lines, A, B, C. 
 
 Because the point F is the centre of the circle DKL, FD is equal (11 
 Def.) to FK ; but FD is equal to the straight line A ; therefore FK is 
 equal to A: Again, because G is the centre of the circle LKH, GH is 
 equal (11. Def.) to GK ; but GH is equal to C; therefore, also, GK is 
 equal to C ; and FG is equal to B ; therefore the three straight lines KF, 
 FG, GK, are equal to the three A, B, C : And therefore the triangle 
 KFG has its three sides KF, FG, GK equal to the three given straight 
 lines, A, B C. 
 
 SCHOLIUM. 
 
 If one of the sides were greater than the sum of the other two, the arcs 
 would not intersect each other : but the solution will always be possible, 
 when the sum of two sides, any how taken (20. 1.) is greater than the 
 Lhird. 
 
 PROP. XXIII. PROB. 
 
 At a given point in a given straight line, to make a rectilineal angle equal 
 to a given rectilineal angle. 
 
 Let AB be the given straight line, and A the given point in it, and DOE 
 the given rectilineal angle ; it is required to make an angle at the given 
 point A in the given straight line 
 AB, that shall be equal to the 
 given rectilineal angle DCE. 
 
 Take in CD, CE any points D, 
 E, and join DE ; and make (22. 
 1.) the triangle AF®, the sides 
 of which shall be equal to the 
 three straight lines, CD, DE, CE, 
 so that CD be equal to AF, CE to 
 AG, and DE to FG ; and because 
 DC, CE are equal to FA, AG, 
 each to each, and the base DE to 
 the base FG ; the angle DCE is 
 equal (8. 1.) to the angle FAG. 
 Therefore, at the given point A in the given straight line AB, the angle 
 FAG is made equal to the given rectilineal angle DCE. 
 
 PROP. XXIV. THEOR. 
 
 If two triangles have two sides of the one equal to two sides of the other, each 
 to each, but the angle contained by the two sides of the one greater than 
 the angle contained by the two sides of the other ; the base of that which 
 has the greater angle shall be greater than the base of the other. 
 
 Let ABC, DEF be two triangles which have the two sides AB, AC 
 equal to the two DE, DF each to each, viz. AB equal to DE, and AC to
 
 OF GEOMETRY. BOOK I. 34 
 
 DF; but the angle BAC greater than the angle EDF ; the Dase BC i* 
 also greater than the base EF. 
 
 Of the two sides DE, DF, let DE be the side which is not greater than 
 the other, and at the point D, in the straight line DE, make (23. 1.) the 
 angle EDG equal to the angle BAC : and make DG equal (3. 1.) to AC 
 or DF, and join EG, GF. 
 
 Because AB is equal to DE, and AC to DG, the two sides BA, AC are 
 equal to the two ED, DG, each to each, and the angle BAC is equal to 
 the angle EDG, therefore 
 
 the base BC is equal (4. 1 .) A J> 
 
 to the base EG ; and be- 
 cause DG is equal to DF, 
 the angle DFG is equal 
 (5. 1.1 to the angle DGF; 
 but the angle DGF is 
 greater than the angle 
 EGF ; therefore the angle 
 DFG is greater than EGF; 
 and much more is the angle 
 EFG greater than the 
 angle EGF ; and because 
 the angle EFG of the triangle EFG is greater than its angle EGF, and 
 because the greater ( 1 9. 1 .) side is opposite to the greater angle, the side 
 EG is greater than the side EF ; but EG is equal to BC ; and therefore 
 also BC is greater than EF. 
 
 PROP. XXV. THEOR. 
 
 If two triangles have two sides of the one equal to two sides of the other, each 
 to each, but the base of the one greater than the base of the other ; the angle 
 contained by the sides of that which has the greater base, shall be greater 
 than the angle contained by the sides of the other. 
 
 Let ABC, DEF be two triangles which have the two sides, AB, AC, 
 equal to the two sides DE, DF, each to each, viz. AB equal to DE, and 
 AC to DF : but let the base CB be greater than the base EF, the angle 
 BAC is likewise greater than the angle EDF. 
 
 For, if it be not greater, it must either be equal to it, or less ; but the 
 angle BAC is not equal to the angle 
 EDF, because then the base BC 
 would be equal (4. 1 .) to E F ; but it is 
 not ; therefore the angle BAC is not 
 equal to the angle EDF ; neither is 
 it less ; because then the base BC 
 would be less (24. 1.) than the bas„ 
 EF ; but it is not ; therefore the an- 
 gle BAC is not less than the angle 
 EDF: and it was shewn that it is 
 ool equal to it : therefore the angle 
 BAC is greater than the angle EDF. 
 
 ■s
 
 26 
 
 ELEMENTS 
 
 PROP. XXVI. THEOR. 
 
 If two ttiangles have two angles of the one equal to two angles of the other 
 each to each ; and one side equal to one side, viz. either the sides adjacent 
 to the equal angles, or the sides opposite to the equal angles in each ; then 
 shall the other side be equal, each to each ; and also the third angle of the 
 one to the third angle of the other 
 
 Let ABC, DEF be two trian- 
 gles which have the angles 
 ABC, BCA equal to the angles 
 DEF, EFD, viz. ABC to DEF, 
 and BCA to EFD, also one side 
 equal to one side ; and first, let 
 those sides be equal which are 
 adjacent to the angles that are 
 equal in the two triangles, viz. 
 BC to EF ; the other sides 
 
 shall be equal, each to each, viz. ^ 
 
 ABtoDE.andAC toDF; and B ^ JS * 
 
 the third angle BAC to the third angle EDF. 
 
 For, if AB be not equal to DE, one of them must be the greater. Let 
 AB be the greater of the two, and make BG equal to DE, and join GC ; 
 therefore, because BG is equal to DE, and BC to EF, the two sides GB, 
 BC are equal to the two, DE, EF, each to each ; and the angle GBC is 
 equal to the angle DEF; therefore the base GC is equal (4. 1.) to the 
 base DF, and the triangle GBC to the triangle DEF, and the other angles 
 to the other angles, each to each, to which the equal sides are opposite ; 
 therefore the angle GCB is equal to the angle DFE, but DFE is, by the 
 hypothesis, equal to the angle BCA ; wherefore also the angle BCG is 
 equal to the angle BCA, the less to the greater, which is impossible ; 
 therefore AB is not unequal to DE, that is, it is equal to it; and BC is 
 equal to EF ; therefore the two AB, BC are equal to the two DE, EF, 
 each to each ; and the angle ABC is equal to the angle DEF ; therefore 
 the base AC is equal (4. 1.) to the base DF, and the angle BAC to the 
 angle EDF. 
 
 Next, let the sides which are 
 opposite to equal angles in each 
 triangle be equal to one another, 
 viz. AB to DE ; likewise in this 
 case, the other sides shall be 
 equal, AC to DF, and BC to EF ; 
 and also the third angle BAC to 
 the third EDF. 
 
 For, if BC be not equal to EF, 
 let BC be the greater of them, 
 and make BH equal to EF, and 
 ioin AH ; and because BH is 
 equal to EF, and AB to DE ; the two AB, BH are equal to the two 
 DE, EF each to each; and they contain equal angles; therefore (4. 1.)
 
 OF GEOMETRY. BOOK I. 
 
 27 
 
 the base AH is equal to the base DF, and the triangle ABII to the trian 
 gle DEF, and the other angles are equal, each to each, to which the equai 
 sides are opposite ; therefore the angle BHA is equal to the angle EFD 
 but EFD is equal to the angle BCA ; therefore also the angle BHA is equai 
 to the angle BCA, that is, the exterior angle BHA of the triangle AHC is 
 equal to its interior and opposite angle BCA, which is impossible (16. I.); 
 wherefore BC is not unequal to EF, that is, it is equal to it; and AB is 
 equal to DE ; therefore the two, AB, BC are equal to the two DE, EF, each 
 to each ; and they contain equal angles ; wherefore the base AC is equa. 
 to the base DF, and the third angle BAC to the third angle EDF. 
 
 PROP. XXVII. THEOR. 
 
 lj a straight line falling upon two other straight lines makes the alternate 
 ingles equal to one another, these two straight lines are parallel. 
 
 Let the straight line EF, which falls upon the two straight lines AB, 
 CD make the alternate angles AEF, EFD equal to one another ; AB is 
 parallel to CD. 
 
 For, if it be not parallel, AB and CD being produced shall meet either 
 towards B, D, or towards A, C ; let them be produced and meet towards 
 B, 1) in the point G ; therefore GEF is a triangle, and its exterior angle 
 AEF is greater (16- 1.) than the interior and opposite angle EFG ; but it 
 is also equal to it, which is im- 
 possible : therefore, AB and CD 
 being produced, do not meet to- 
 wards B, D. In. like manner it 
 may be demonstrated that they 
 do not meet towards A, C ; but 
 those straight lines which meet 
 neither way, though produced 
 ever so far, are parallel (30. Def.) 
 to one another. AB therefore is parallel to CD. 
 
 PROP. XXVIII. THEOR. 
 
 If a straight line falling upon two other straight lines makes the exterior an 
 gle equal to the interior and opposite upon the same side of the line ; or 
 makes the interior angles upon the same side together equal to two right 
 angles ; the two straight lines are parallel to one another. 
 
 Let the straight line EF, which 
 falls upon the two straight lines AB, 
 CD, make the exterior angle EGB 
 equal to Gill), the interior and oppo- 
 site ancle npon the same side ; or let it 
 make the interior angles on the same 
 side BGH, GHD together equal to two 
 right angles; AB is parallel to CD. 
 
 Because the angle EGB is equal to 
 the angle GHD, and also (15. 1.) to the \jf 
 
 II>
 
 25 ELEMENTS 
 
 angle AGH, the angle AGH is equal to the angle GHD ; and they are th 
 alternate angles ; therefore AB is parallel (27. 1.) to CD. Again, becauat* 
 the angles BGH, GHD are equal (hyp. )to two right angles, and AGH, BGH, 
 are also equal (13. 1.) to two right angles, the angles AGH, BGH are equal 
 to the angles BGH, GHD : Take away the common angle BGH ; therefore 
 the remaining angle AGH is equal to the remaining angle GHD ; and they 
 are alternate angles ; therefore AB is parallel to CD. 
 
 Cor. Hence, when two straight lines are perpendicular to a third line, 
 they will be parallel to each other. 
 
 PROP. XXIX. THEOR. 
 
 If a straight line fall upon two parallel straight lines, it makes the alternate 
 angles equal to one another ; and the exterior angle equal to the interior 
 and opposite upon the same side ; and likewise the two interior angles upon 
 the same side together equal to two right angles. 
 
 Let the straight line EF fall upon the parallel straight lines AB, CD t 
 the alternate angles AGH, GHD are equal to one another ; and the exte- 
 rior angle EGB is equal to the interior and opposite, upon the same side, 
 GHD ; and the two interior angles BGH, GHD upon the same side are 
 together equal to two right angles. 
 
 For if AGH be not equal to GHD, let KG be drawn making the angle 
 KGH equal to GHD, and produce KG to L ,• then KL will be parallel to 
 CD (27. 1.) ; but AB is also paral- 
 lel to CD ; therefore two straight 
 lines are drawn through the same 
 point G, parallel to CD, and yet 
 not coinciding with one another, 
 which is impossible (11. Ax.) The 
 angles AGH, GHD therefore are 
 not unequal, that is, they are equal 
 to one another. Now, the angle 
 EGB is equal to AGH (15. 1.) ; 
 and AGH is proved to be equal 
 to GHD ; therefore EGB is like- 
 wise equal to GHD ; add to each of these the angle BGH ; therefore the 
 angles EGB, BGH are equal to the angles BGH, GHD ; but EGB, BGH 
 are equal (13. 1.) to two right angles; therefore also BGH, GHD are 
 equal to two right angles. 
 
 Cor. 1. If two lines KL and CD make, with EF, the two angles KGH, 
 GHC together less than two right angles, KG and CH will meet on the side 
 of EF on which the two angles are that are less than two right angles. 
 
 For, if not, KL and CD are either parallel, or they meet on the other 
 side of EF ; but they are not parallel ; for the angles KGH, GHC would 
 then be equal to two right angles. Neither do they meet on the other 
 side of EF; for the angles LGH, GHD would then be two angles of a 
 triangle, and less than two right angles ; but this is impossible ; for the 
 four angles KGH, HGL, CHG, GHD are together equal to four right 
 angles (13 1 .) of which the t\* o, KGH, CH G, are by supposition leas thah
 
 OF GEOMETRY. BOOK I. 2ft 
 
 ;wo right angles ; therefore the other two, HGL, GHD aro grea:er thai 
 two right angles. Therefore, since KL and CD are not parallel, and since 
 they do not meet towards L and D, they must meet if produced towards 
 K and C. 
 
 Cor. 2. If BGH is a right angle, GHD will be a right angle also; 
 therefore every line perpendicular to one of two parallels, is perpendiculai 
 to the other. 
 
 Cor. 3. Since AGE=BGH, and DHF=CHG; hence the four acute 
 angles BGH, AGE, GHC, DHF, are equal to each other. The same is 
 the case with the four obtuse angles EGB, AGH, GHD, CHF. It may 
 be also observed, that, in adding one of the acute angles to one of the ob- 
 tuse, the sum will always be equal to two right angles. 
 • 
 
 SCHOLIUM. 
 
 The angles just spoken of, when compared with each other, assume 
 different names. BGH, GHD, we have already named interior angles on 
 the same side ; AGH, GHC, have the same name ; AGH, GHD, are called 
 alternate interior angles, or simply alternate; so also, are BGH, GHC : 
 and lastly, EGB, GHD, or EGA, GHC, are called, respectively, the op- 
 posite exterior and interior angles ; and EGB, CHF, or AGE, DHF, the 
 alternate exterior angles. 
 
 PROP. XXX. THEOR. 
 
 Straight lines which are parallel to the same straight line are parallel to on* 
 
 another. 
 
 Let AB, CD, be each of them parallel to EF ; AB is also parallel to 
 CD. 
 
 Let the straight line GHK cut AB. EF, CD ; and because GHK cuts 
 the parallel straight lines AB, EF, the 
 angle AGH is equal (29. 1.) to the an- 
 gle GHF. Again, because the straight 
 line GK cuts the parallel straight lines 
 EF, CD, the angle GHF is equal (29. 
 1.) to the angle GKD : and it was 
 shewn that the angle AGK is equal to 
 the angle GHF; therefore also AGK 
 is equal to GKD ; and they are alter- 
 nate angles ; therefore AB is parallel 
 (27. 1.) to CD. 
 
 PROP. XXXI. PROB. 
 
 Tc draw a straight line through a given point parallel to a given straight 
 
 line. 
 
 Let A be the given point, and BC the given straight line, it is required 
 io draw a straight line through the point A, parallel to the straight line 
 BC. 
 
 A 
 
 \G 
 
 B 
 
 E 
 
 \h 
 
 IT 
 
 
 \ 
 
 J' 
 
 
 AX 

 
 30 ELEMENTS 
 
 In 13 (J take any point D, and join T ^- 
 
 AD , and at the point A, in the ■"-* 
 
 straight line AD, make (23. 1.) the 
 angle DAE equal to the angle ADC ; 
 
 and produce the straight line E A to F. D JJ C 
 
 Because the straight line AD, which meets the two straight lines BC, 
 EF, makes the alternate angles EAD, ADC equal to one another, EF i« 
 parallel (27. 1.) to BC. Therefore the straight line EAF is drawn 
 through the given point A parallel to the given straight line BC. 
 
 PROP. XXXII. THEOR. 
 
 If a side of any triangle be produced, the exterior angle is equal to the two 
 interior and opposite angles ; and the three interior angles of every triangle 
 are equal to two right angles. 
 
 Let ABC be a triangle, and let one of its sides BC be produced to D , 
 the exterior angle ACD is equal to the two interior and opposite angles 
 CAB, ABC ; and the three interior angles of the triangle, viz. ABC, BCA, 
 CAB, are together equal to two right angles. 
 
 Through the point C draw 
 CE parallel (31. 1.) to the 
 straight line AB ; and because 
 AB is parallel to CE, and AC 
 meets them, the alternate an- 
 gles BAC, ACE are equal (29. 
 • .) Again, because AB is pa- 
 rallel to CE, and BD falls upon 
 
 them, the exterior angle ECD is equal to the interior and opposite angle 
 ABC, but the angle ACE was shewn to be equal to the angle BAC ; 
 therefore the whole exterior angle ACD is equal to the two interior and 
 opposite angles CAB, ABC ; to these angles add the angle ACB, and 
 the angles ACD, ACB are equal to the three angles CBA, BAC, ACB ; 
 but the angles ACD, ACB are equal (13. 1.) to two right angles; there- 
 fore also the angles CBA, BAC, ACB are equal to two right angles. 
 
 Cor. 1. All the interior angles of any rectilineal figure are equal to 
 twice as many right angles as the figure has sides, wanting four right angles. 
 
 For any rectilineal figure ABCDE can be divided into as many trian- 
 gles as the figure has sides, by drawing straight lines from a point F 
 within the figure to each of its angles. And, by the preceding proposition, 
 all the angles of these triangles are equal 
 to twice as many right angles as there 
 are triangles, that is, as there are sides 
 of the figure ; and the same angles are 
 equal to the angles of the figure, together 
 with the angles at the point F, which 
 is the common vertex of the triangles ; 
 that is, (2 Cor. 15. 1.) together with four 
 right angles. Therefore, twice as many 
 right angles as the figure has sides, are 
 equal to all the angles of the figure, to-
 
 OF GEOMETRY. BOOK I. 31 
 
 gether with four right angles that is, the angles of the figure are equal 
 to twice as many right angles as the figure has sides, wanting four. 
 
 Cor. 2. All the exterior angles of any rectilineal figure are together 
 equal to four right angles. 
 
 Because every interior angle 
 
 ABC, with its adjacent exterior 
 
 ABD, is equal (13. I.) to two 
 right angles ; therefore all the 
 interior, together with all the 
 exterior angles of the figure, 
 are equal to twice as many 
 right angles as there are sides 
 of the figure ; that is, by the 
 foregoing corollary, they are 
 equal to all the interior angles 
 of the figure, together with 
 four right angles ; therefore all 
 the exterior angles are equal to four right angles. 
 
 Cor. 3. Two angles of a triangle being given, or merely their sum, the 
 third will be found by subtracting that sum from two right angles. 
 
 Cor. 4. If two angles of one triangle are respectively equal to two an- 
 gles of another, the third angles will also be equal, and the two triangles 
 will be mutually equiangular. 
 
 Cor. 5. In any triangle there can be but one right angle ; for if there 
 were two, the third angle must be nothing. Still less can a triangle have 
 more than one obtuse angle. 
 
 Cor. 6. In every right-angled triangle, the sum of the two acute an- 
 gles is equal to one right angle. 
 
 Cor. 7. Since every equilateral triangle (Cor. 5. 1.) is also equian- 
 gular, each of its angles will be equal to the third part of two right angles ; 
 so that if the right angle is expressed by unity, the angle of an equilateral 
 triangle will be expressed by % of one right angle. 
 
 Cor. 8. The sum of the angles in a quadrilateral is equal to two righ* 
 angles multiplied by 4 — 2, which amounts to four right angles ; hence, if 
 all the angles of a quadrilateral are equal, each of them will be a right an- 
 gle ; a conclusion which sanctions the Definitions 25 and 26. where the 
 four angles of a quadrilateral are said to be right, in the case of the rectan- 
 gle and the square. 
 
 Cor. 9. The sum of the angles of a pentagon is equal to two right an- 
 gles multiplied by 5 — 2, which amounts to six right angles ; hence, when 
 a pentagon is equiangular, each angle is equal to the fifth part of six right 
 angles, or f of one right angle. 
 
 Cor. 10. The sum of the angles of a hexagon is equal to 2 x (6 — 2), 
 or eight right angles ; hence, in the equiangular hexagon, each angle is 
 the sixth Dart of eight right angles, or £ of one right angle. 
 
 SCHOLIUM. 
 
 When (Cor. 1.) is applied to polygons, which have re-entrant angles. 
 as ABC each re-entrant angle must be regarded as greater than two rigbJ 
 angles.
 
 32 
 
 ELEMENTS 
 
 And, by joining BD, BE, BF, the 
 figure is divided into four triangles, 
 which contain eight right angles ; 
 that is, as many times two right an- 
 gles as there are units in the number 
 of sides diminished by two. 
 
 But to avoid all ambiguity, we shall 
 henceforth limit our reasoning to 
 polygons with salient angles, which 
 might otherwise be named convex 
 polygons. Every convex polygon is 
 such that a straight line, drawn at 
 pleasure, cannot meet the contour of 
 the polygon in more than two points. 
 
 PROP. XXXIII. THEOR. 
 
 The straight lines which join the extremities of two equal and parallel straight 
 lines, towards the same parts, are also themselves equal and parallel. 
 
 Let AB, CD, be equal and parallel straight lines, and joined towards 
 the same parts by the straight lines AC, BD; AC, BD are also equal and 
 parallel. 
 
 Join BC ; and because AB is parallel 
 to CD, and BC meets them, the alternate 
 angles ABC, BCD are equal (29. 1.); and 
 because AB is equal to CD, and BC com- 
 mon to the two triangles ABC, DCB, the 
 two sides AB, BC are equal to the two 
 DC, CB ; and the angle ABC is equal to C T> 
 
 the angle BCD ; therefore the base AC is equal (4. 1.) to the base BD, 
 and the triangle ABC to the triangle BCD, and the other angles to the 
 other angles (4. 1.) each to each, to which the equal sides are opposite ; 
 therefore the angle ACB is equal to the angle CBD ; and because the 
 straight line BC meets the two straight lines AC, BD, and makes the al- 
 ternate angles ACB, CBD equal to one another, AC is parallel (27. 1.) to 
 BD ; and it was shewn to be equal to it. 
 
 Cor. 1. Hence, if two opposite sides of a quadrilateral are equal and 
 parallel, the remaining sides will also be equal and parallel, and the figure 
 will be a parallelogram. 
 
 Cor. 2. And every quadrilateral, whose opposite sides are equal, is a 
 parallelogram, or has its opposite sides parallel. 
 
 For, having drawn the diagonal BC ; then, the triangles ABC, CBD, 
 being mutually equilateral (hyp.), they are also mutually equiangular 
 (Th. 8.), or have their corresponding angles equal ; consequently, the op 
 posite sides are parallel ; namely, the side AB parallel to CD, and BD pa 
 rallel to AC ; and, therefore, the figure is a parallelogram. 
 
 Cor. 3. Hence, also, if the opposite angles of a quadrilateral be equal 
 the opposite sides will likewise be equal and parallel. 
 
 For all the angles of the figure being equal to four right angles (Cor. 8
 
 OF GEOMETRY. BOOK I. 33 
 
 Th. 32.), and the opposite angles being mutually equal, each pair of adja 
 cent angles must be equal to two right angles ; therefore, the opposite sides 
 must be equal and parallel. 
 
 PROP. XXXIV. THEOR. 
 
 The opposite sides and angles of a parallelogram are equal to one another, and 
 the diagonal bisects it ; that is, divides it into two equal parts. 
 
 N. B. A Parallelogram is a four-sided figure, of which the opposite sides are parallel ; and 
 the diameter is a straight line joining two of its opposite angles. 
 
 Let ACDB be a parallelogram, of which BC is a diameter ; the oppo- 
 site sides and angles of the figure are equal to one another ; and the diam- 
 eter BC bisects it. 
 
 Because AB is parallel to CD, and BC 
 meets them, the alternate angles ABC, 
 BCD are equal (29. 1.) to one another ; and 
 because AC is parallel to BD, and BC meets 
 "them, the alternate angles ACB, CBD are 
 equal (29. 1.) to one another; wherefore 
 the two triangles ABC, CBD have two an- 
 gles ABC, BCA in one, equal to two angles 
 
 BCD, CBD in the other, each to each, and the side BC, which is adja- 
 cent to these equal angles, common to the two triangles ; therefore their 
 other sides are equal, each to each, and the third angle of the one to the 
 third angle of the other (26. 1.) ; viz. the side AB to the side CD, and 
 A.C to BD, and the angle BAC equal to the angle BDC. And because 
 the angle ABC is equal to the angle BCD, and the angle CBD to the 
 angle ACB, the whole angle ABD is equal to the whole angle ACD : 
 And the angle BAC has been shewn to be equal to the angle BDC : there- 
 fore the opposite sides and angles of a parallelogram are equal to one an- 
 other ; also, it3 >diameter bisects it ; for AB being equal to CD, and BC 
 common, the two AB, BC are equal to the two DC, CB, each to each ; 
 now the angle ABC is equal to the angle BCD ; therefore the triangle 
 ABC is equal (4. 1.) to the triangle BCD, and the diameter BC divides 
 the parallelogram ACDB into two equal parts. 
 
 Cor. 1. Two parallel lines, included between two other parallels, are 
 equal. 
 
 Cor. 2. Hence, two parallels are every where equally distant. 
 
 Cor. 3. Hence, also, the sum of any two adjacent angles of a paral 
 lelogram is equal to two right angles. 
 1 
 
 PROP. XXXV. THEOR. 
 
 Parallelograms upon the same base and between the same parallels, are eqva* 
 
 to one another. 
 
 (see the 2d and 3d figures.) 
 
 Let the parallelograms ABCD, EBCF be upon the same base BC, and 
 between the same parallels AF, BC ; the parallelogram ABCD is equal to 
 the parallelogram EBCF.
 
 94 
 
 ELEMENTS 
 
 If the sides AD, DF of the parallelo- 
 grams ABCD, DBCF opposite to the base 
 BC be terminated in the same point D ; 
 it is plain that each of the parallelograms 
 is double (34. 1.) of the triangle BDC ; 
 and they are therefore equal to one an- 
 other. 
 
 But, if the sides AD, EF, opposite to the base BC of the parallelograms 
 ABCD,EBCF, be not terminated in the same point ; then, because ABCD 
 is a parallelogram, AD is equal (34. l.)to BC ; for the same reason EF 
 is equal to BC ; wherefore AD is equal (1. Ax.) to EF ; and DE is com- 
 mon ; therefore the whole, or the remainder, AE is equal (2. or 3. Ax.) to 
 the whole, or the remainder DE ; now AB is also equal to DC ; therefore 
 the two EA, AB are equal to the two FD, DC, each to each ; but the ex- 
 
 terior angle FDC is equal (29. 1.) to the interior EAB, wherefore the base 
 EB is equal to the base FC, and the triangle EAB (4. 1.) to the triangle 
 FDC. Take the triangle FDC from the trapezium ABCF, and from the 
 same trapezium take the triangle EAB ; the remainders will then be equal 
 (3. Ax.) that is, the parallelogram ABCD is equal to the parallelogram EBCF. 
 
 PROP. XXXVI. THEOR. 
 
 Parallelograms upon equal bases, and between the same paWllels, are equal U 
 
 one another. " 
 
 Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and 
 between the same parallels AH, . _ __ __ 
 
 BG; the parallelogram ABCD -A. 
 is equal to EFGH. 
 
 Join BE, CH ; and because 
 BC is equal to FG, and FG to 
 (34. 1.) EH, BC is equal to EH ; 
 and they are parallels, and join- 
 ed towards the same parts by the 
 straight lines BE, CH : But 
 straight lines which join equal and parallel straight lines towards the same 
 parts, are themselves equal and parallel (33. 1.) ; therefore EB, CH are 
 both equal and parallel, and EBCH is a parallelogram ; and it is equal 
 (35 l.j to ABCD, because it is upon the same base BC, and between the 
 same parallels BC, AH : For the like reason, the parallelogram EFGH 
 is equal to the same EBCH : Therefore also the parallelogiam ABCD it 
 equal to EFGH.
 
 OF GEOMETRY. BOOK 1. 
 
 PROP. XXXVII. THEOR. 
 
 Trim gles upon the same base, and between the same parallels, are equal to ok* 
 
 another. 
 
 Let the triangles ABC, DBC be upon the same base BC, and between 
 the same parallels, AD, BC : The 
 triangle ABC is equal to the trian- 
 gle OBC. 
 
 Produce AD both ways to the 
 points E, F, and through B draw (31. 
 1.) BE parallel to CA ; and through 
 C draw CF parallel to BD : There- 
 fore, each of the figures EBCA, 
 DBCF is a parallelogram; and EBCA 
 is equal (35. 1.) to DBCF, because they are upon the same base BC, and 
 between the same parallels BC, EF ; but the triangle ABC is the half ol 
 •the parallelogram EBCA, because the diameter AB bisects (34. 1.) it; 
 and the triangle DBC is the half of the parallelogram DBCF, because 
 the diameter DC bisects it ; and the halves of equal things are equal (7. 
 Ax.) ; therefore the triangle ABC is equal to the triangle DBC. 
 
 PROP. XXXVIII. THEOR. 
 
 Triangles upon equal bases, and between the same parallels, are equal to one 
 
 another. 
 
 Let the triangles ABC, DEF be upon equal bases BC, EF, and between 
 the same parallels BF, AD : The triangle ABC is equal to the triangle DEF 
 
 Produce AD both ways to the points G, H, and through B draw BG 
 parallel (31. 1.) to CA, and through F draw FH parallel to ED : Then 
 each of the figures GBCA, 
 DEFH is a parallelogram ; 
 and they are equal to (36. 1.) 
 one another, because they aie 
 upon equal bases BC, EF, and 
 between the same parallels 
 BF, GH ; and the triangle 
 ABC is the half (34. 1.) of the 
 parallelogram GBCA, because " L> t* r 
 
 •he diameter AB bisects it; and the triangle DEF is the half (34. 1.) of 
 the parallelogram DEFH, because the diameter DF bisects it: But the 
 halves of equal things are equal (7. Ax.) ; therefore the triangle ABC is 
 equal to the triangle DEF. 
 
 PROP. XXXIX. THEOR. 
 
 Equal triangles upon the same base, and upon the same side of it, are between 
 
 the same parallels. 
 
 Let the equal triangles ABC, DBC be upon the same base BC, and udob 
 the same side of it ; tncy art* between the same parallels.
 
 36 
 
 ELEMENTS 
 
 Join AD ; AD is parallel to BC ; for, if it is no«, through the point A 
 draw (31. 1.) AE paiallel to BC, and join EC : * yv 
 
 The triangle ABC, is equal (37. 1.) to the tri- 
 angle EBC, because it is upon the same base 
 BC, and between the same parallels BC, AE : 
 But the triangle ABC is equal to the triangle 
 BDC ; therefore also the triangle BDC is equal 
 to the triangle EBC, the greater to the less, 
 which is impossible : Therefore AE is not par- 
 allel to BC. In the same manner, it may be 
 demonstrated that no other line but AD is parallel to BC ; AD is there* 
 fore parallel to it. 
 
 PROP. XL. THEOR. 
 
 Equal triangles on the same side of bases which are equal and in the same 
 straight line, are between the same parallels. 
 
 Let the equal triangles ABC, DEF be upon equal bases BC, EF, in 
 the same straight line BF, and to- 
 wards the same parts ; they are be- 
 tween the same parallels. 
 
 Join AD ; AD is parallel to BC ; 
 for, if it is not,, through A draw (31. 
 1.) AG parallel to BF, and join GF. 
 The triangle ABC is equal (38. 1.) 
 to the triangle GEF, because they 
 are upon equal bases BC, EF, and 
 between the same parallels BF, 
 AG : But the triangle ABC is equal to the triangle DEF ; therefore also 
 the triangle DEF is equal to the triangle GEF, the greater to the less, 
 which is impossible ; therefore AG is not parallel to BF ; and in the same 
 manner it may be demonstrated that there is no other parallel to it but 
 AD ; AD is therefore parallel to BF. 
 
 PROP. XLI. THEOR. 
 
 If a parallelogram and a triangle be upon the same base, and between the 
 same parallel ; the parallelogram is double of the triangle. 
 
 Let the parallelogram ABCD and the tri- 
 angle EBC be upon the same base BC and 
 between the same parallels BC, AE ; the 
 parallelogram ABCD is double of the trian- 
 gle EBC. 
 
 Join AC ; then the triangle ABC is equal 
 (37. 1.) to the triangle EBC, because they 
 are upon the same base BC, and between the 
 same parallels BC, AE. But the parallelo- 
 gram ABCD is double (34. 1.) of the triangle 
 ABC, because the diameter AC divides it 
 into two equal parts ; wherefore ABCD is also double of the triangle EBC
 
 OF GEOMETRY. BOOK 1. 
 
 PROP. XLII. PROB. 
 
 To describe a parallelogram that shall be equal to a given triangle, and haw 
 one of its angles equal to a given rectilineal angle. 
 
 Let ABC be the given triangle, and D the given rectilineal angle. I 
 is required to describe a parallelogram that shall be equal to the given tri 
 angle ABC, and have one of its angles equal to D. 
 
 Bisect (10. 1.) BC in E, join AE, and at the point E in the straight line 
 EC make (23. I.) the angle CEF equal to D ; and through A draw (31. 
 1.) AG parallel to BC, and through C draw CG (31. 1.) parallel to EF; 
 Therefore FECG is a parallelogram : a t^ 
 
 And because BE is equal to EC, the 
 triangle ABE is likewise equal (38. 
 1.) to the triangle AEC, since they 
 are upon equal bases BE, EC, and 
 between the same parallels BC, AG ; 
 therefore the triangle ABC is double 
 of the triangle AEC. And the paral- 
 lelogram FECG is likewise double \ \l \J D 
 (41. 1.) of the triangle AEC, because 
 it is upon the same base, and between 
 the same parallels : Therefore the parallelogram FECG is equal to the 
 triangle ABC, and it has one of its angles CEF equal to the given angle 
 D : Wherefore there has been described a parallelogram FECG equal to 
 a given triangle ABC, having one of its angles CEF equal to the given 
 angle D. 
 
 Cor. Hence, if the angle D be a right angle, the parallelogram EFGC 
 will be a rectangle, equivalent to the triangle ABC ; and therefore the 
 same construction will apply to the problem : to make a triangle equivalent 
 to a given rectangle. 
 
 PROP. XLIII. THEOR. 
 
 The complements of the parallelograms which are about the diameter of any 
 parallelogram, are equal to one another. 
 
 Let ABCD be a parallelogram of which the diameter is AC ; let EH, 
 FG be the parallelograms about AC, that is, through which AC passes, and 
 let BK, KD be the other parallelograms, 
 which make up the whole figure ABCD, 
 and are therefore called the complements ; 
 The complement BK is equal to the com- 
 plement KD. 
 
 Because ABCD is a parallelogram and 
 AC its diameter, the triangle ABC is 
 equal (31. 1.) to the triangle ADC: And 
 because EKHA is a parallelogram, and 
 \K its diameter, the triangle AEK is 
 equal to the triangle A UK: l'or the same 
 reason, the triangle KGC is equal to the
 
 38 
 
 ELEMENTS 
 
 triangle KFC. Then because the triangle AER is equal to the triangle 
 A.HK, and the triangle KGC to the triangle KFC ; the triangle AEK, to- 
 gether with tho triangle KGC, is equal to the triangle AHK, together with 
 the triangle K FC : But the whole triangle ABC is equal to the whole 
 ADC ; therefore the remaining complement BK is equal to the remaining 
 complement KD. 
 
 PROP. XL1V. PROB. 
 
 To a given straight line to apply a parallelogram, which shall be equal to a given 
 triangle, and have one of its angles ejual to a given rectilineal angle. 
 
 Let AB be the given straight line, and C the given triangle, and D the 
 given rectilineal angle. It is required to apply to the straight line AB a 
 parallelogram equal to the triangle C, and having an angle equal to D. 
 Make (42. 1.) the parallelogram BEFG equal to the triangle C, having the 
 
 H A. 
 
 angle EBG equal to the angle D, and the side BE in the same straight 
 line with AB : produce FG to H, and through A draw (31. 1.) AH parallel 
 to BG or EF, and join HB. Then because the straight line HF falls upon 
 the parallels AH, EF, the angles AHF, HFE, are together equal (29. 1.) 
 to two right angles ; wherefore the angles BHF, HFE are less than two 
 right angles ; But straight lines which with another straight line make the 
 interior angles, upon the same side less than two right angles, do meet if pro- 
 duced (1 Cor. 29. 1.) : Therefore HB, FE will meet, if produced ; let them 
 meet in K, and through K draw KL parallel to E A or FH, and produce HA, 
 G B to the points L, M : Then HLKF is a parallelogram, of which the diam- 
 eter is HK,and AG, ME are the parallelograms about HK; and LB, BF are 
 the complements ; therefore LB is equal (43. 1.) to BF : but BF is equal 
 to the triangle C ; wherefore LB is equal to the triangle C ; and because 
 the angle GBE is equal (15. 1.) to the angle ABM, and likewise to the an- 
 gle D ; the angle ABM is equal to the angle D : Therefore the parallelo- 
 gram LB, which is applied to the straight line AB, is equal to the triangle 
 C, and has the angle ABxM equal to the angle D. 
 
 Cor. Hence, a triangle may be converted into an equivalent rectangle, 
 having a side of a given length : for, if the angle D be a right angle, and 
 AB the given side, the parallelogram ABML will be a rectangle eqviva 
 lent to the triangle C
 
 OF GEOMETRY. BOOK I 
 
 PROP. XLV. PROB. 
 
 To describe a parallelogram equal to a given rectilineal figure, and. having 
 an angle equal to a given rectilineal angle. 
 
 Let ABCD be the given rectilineal figure, and E the given rectilineaj 
 angle. It is required to describe a parallelogram equal to ABCD, and hav- 
 ing an angle equal to E. 
 
 Join DB.and describe (42. 1.) the parallelogram FH equal to the tri- 
 angle ADB, and having the angle HKF equal to the angle E ; and to the 
 straight line GH (44. 1.) apply the parallelogram GM equal to the triangle 
 DBC, having the angle GHM equal to the angle E. And because the an- 
 gle E is equal to each of the angles FKH, GHM, the angle FKH is equal 
 to GHM ; add to each of these the angle KHG ; therefore the angles 
 FKH, KHG are equal to the angles KHG, GHM ; but FKH, KHG are 
 equal (29. 1.) to two right angles ; therefore also KHG, GHM are equal 
 to two right, angles : and because at the ooint H in the straight lines GH. 
 
 the two straight lines KH, HM, upon the opposite sides of GH, make the 
 adjacent angles equal to two right angles, KH is in the same straight line 
 (14. 1.) with HM. And because the straight line HG meets the parallels 
 KM, FG.the alternate angles MHG, HGF are equal (29. 1.) ; add to each 
 of these the angle HGL: therefore the angles MHG, HGL, are equal to 
 the angles HGF, HGL : But the angles MHG, HGL, are equal (29. 1.) to 
 two right angles ; wherefore also the angles HGF, HGL, are equal to two 
 right angles, and FG is therefore in the same straight line with GL. And 
 because KF is parallel to HG, and HG to ML, KF is parallel (30. 1.) to 
 ML ; but KM, FL are parallels : wherefore KFLM is a parallelogram. 
 And because the triangle ABD is equal to the parallelogram HF, and the 
 triangle DBC to the parallelogram GM, the whole rectilineal figure ABCD 
 is equal to the whole parallelogram KFLM ; therefore the parallelogram 
 KFLM lias been inscribed equal to the given rectilineal figure ABCD, hav- 
 ing the angle FKM equal to the given angle E. 
 
 Cor. From this it is manifest how to a given straight line to apply a 
 parallelogram, which shall have an angle aqua] to a given rectilineal angle, 
 and shall be equal to a given rectilineal figure, viz. by applying (4 1 1.) 
 to the given straight line a parallelogram equal to the first fiangle ABD. 
 and having an angle equal to the given angle.
 
 4? ELEMENTS 
 
 PROP. XLVI. PROB. 
 To describe a square upon a given straight line. 
 
 Let A B be the given straight line : it is required to describe a square 
 upon AB. 
 
 From the point A draw (11. 1.) AC at right angles to AB ; and make 
 (3. 1.) AD equal to AB, and through the point D draw DE parallel (31. 1.) 
 to AB, and through B draw BE parallel to AD ; therefore ADEB is a par- 
 allelogram ; whence AB is equal (34. 1.) to DE, and AD to BE ; but BA 
 is equal to AD : therefore the four straight _< 
 lines BA, AD, DE, EB are equal to one an- ** | 
 other, and the parallelogram ADEB is equi- 
 lateral ; it is likewise rectangular ; for the 
 straight line AD meeting the parallels, AB, DE, D 
 makes the angles BAD, ADE equal (29. 1.) to 
 two right angles ; but BAD is a right angle ; 
 therefore also ADE is a right angle now the 
 opposite angles of parallelograms are equal (34. 
 1 .) ; therefore each of the opposite angles ABE, 
 BED is a right angle ; wherefore the figure 
 
 ADEB is rectangular, and it has been demon- . I 
 
 strated that it is equilateral ; it is therefore a ■*• 
 square, and it is described upon the given straight line AB. 
 
 Cor. Hence every parallelogram that has one right angle has all its an 
 gles right angles. 
 
 E 
 
 B 
 
 PROP. XLVII. THEOR. 
 
 In any right angled triangle, the square which is described upon the side 
 subtending the right angle, is equal to the squares described upon the sides 
 which contain the right angle. 
 
 Let ABC be a right angled triangle having the right angle BAC ; the 
 6quare described upon the side BC is equal to the squares described upon 
 BA, AC. 
 
 On BC describe (46. 1.) the square BDEC, and on BA, AC the squares 
 GB, HC ; and through A draw (31. 1.) AL parallel to BD or CE, and join 
 AD, FC ; then, because each of the angles BAC, BAG is a right angle 
 (25. def.), the two straight lines AC, AG upon the opposite sides of AB, 
 make with it at the point A the adjacent angles equal to two right an- 
 gles ; therefore CA is in the same straight line (14. 1.) with AG; for 
 the same reason, AB and AH are in the same straight line. Now be- 
 cause the angle DBC is equal to the angle FBA, each of them being a 
 right angle, adding to each the angle ABC, the whole angle DBA will be 
 equal (2. Ax.) to the whole FBC ; and because the two sides AB, BD 
 are equal to the two FB, BC each to each, and the angle DBA equal to 
 the angle FBC, therefore the base AD is equal (4. 1.) to the bast FC, 
 and the triangle ABD to the triangle FBC. But the paiallelogram BL 
 is double (41. 1.) of the triangle ABD, because they are upon the same 
 base, BD, and between the same parallels, BD, AL ; and the square QB
 
 OF GEOMETRY. BOOK I. 
 
 is double of the triangle BFC be- 
 cause these also are upon the same 
 base FB, and between the same par- 
 allels FB, GC. Now the doubles 
 of equals are equal (6. Ax.) to one an- 
 other; therefore the parallelogram 
 BL is equal to the square GB : And 
 in the same manner, by joining AE, 
 BK, it is demonstrated that the par- 
 allelogram CL is equal to the square 
 HC. Therefore, the whole square 
 BDEC is equal to the two squares 
 GB, HC ; and the square BDEC is 
 described upon the straight line BC, 
 and the squares GB, HC upon BA, 
 AC : wherefore the square upon the 
 side BC is equal to the squares upon 
 the sides BA, AC. 
 
 Cor. 1. Hence, the square of one of the sides of a right angled triangle 
 is equivalent to the square of the hypotenuse diminished by the square of 
 the other side ; which is thus expressed : AB 2 = BC a — AC 2 . 
 
 Cor. 2. If AB=AC ; that is, if the triangle ABC be right angled and 
 isosceles; BC 2 =2AB 2 =2AC 2 ; therefore, BC = AB/ 2. 
 
 Cor. 3. Hence, also, if two right angled triangles have two sides of 
 the one, equal to two corresponding sides of the other ; their third sidei 
 will also be equal, and the triangles will be identical. 
 
 PROP. XLVIII. THisOR. 
 
 If the square described upon one of the sides of a triangle, be equal to th* 
 squares described upon the other ttoo sides of it ; the angle contained by 
 tfiese two sides is a right angle. 
 
 If the square described upon BC, one of the sides of the triangle ABC, 
 be equal to the squares upon the other sides BA, AC, the angle BAC ia 
 a right angle. 
 
 From the point A draw (11. 1.) AD at right angles to AC, and make 
 AD equal to BA, and join DC. Then because DA is equal to AB, the 
 square of DA is equal to the square of AB ; To 
 each of these add the square of AC ; therefore the 
 squares of DA, AC are equal to the squares of BA, 
 AC. But the square of DC is equal (47. 1.) to 
 the squares of DA, AC, because DAC is a right 
 angle ; and the square of BC, by hypothesis, is 
 equal to the squares of BA, AC ; therefore, the 
 square of DC is equal to the square of BC ; and 
 therefore also the side DC is equal to the side BC. 
 And because the side DA is equal to AB, and AC 
 common to the two triangles DAC, BAC, and the base DC likewise equa* 
 to the base BC, the angle DAC is equal (8. 1.) to the angle BAC ; Bui 
 DAC is a right ingle ; therefore <uso BAC is a right angle. 
 
 6
 
 42 
 
 ELEMENTS 
 
 ADDITIONAL PROPOSITIONS. 
 
 PROP. A. THEOR. 
 
 A perpendicular is the shortest line that can be drawn from a point, situated 
 without a straight line, to that line : any two oblique lines drawn from the 
 same point on different sides of the perpendicular , cutting off equal distances 
 on the other line, will be equal ; and any two other oblique lines, cutting off 
 unequal distances, the one which lies farther from the perpendicular will 
 be the longer. 
 
 If AB, AC, AD, &c. be lines drawn from the given point A, to the in- 
 definite straight line DE, of which AB is perpendicular; then shall the 
 perpendicular AB be less than AC, and AC less than AD, and so on. 
 
 For, the angle ABC being a right one, 
 the angle ACB is acute, (17. 1.) or less 
 than the angle ABC. But the less angle 
 of a triangle is subtended by the less side 
 (19. 1.) therefore, the side AB is less than 
 the side AC. 
 
 Again, if BC=BE; then the two ob- 
 lique lines AC, AE, are equal. For the 
 sido AB is common to the two triangles 
 ABC, ABE, and the contained angles ABC 
 and ABE equal ; the two triangles must 
 be equal (4. 1.) ; hence AE, AC are equal. 
 
 Finally, the angle ACB being acute, as before, the adjacent angle ACD 
 will be obtuse ; since (13. 1.) these two angles are together equal to two 
 right angles; and the angle ADC is acute, because the angle ABD is 
 right ; consequently, the angle ACD is greater than the angle ADC ; and, 
 since the greater side is opposite to the greater angle (19. 1.) ; therefore 
 the side AD is greater than the side AC. 
 
 Cor. 1. The perpendicular measures the true distance of a point from 
 a line, because it is shorter than any other distance. 
 
 Cor. 2. Hence, also, every point in a perpendicular at the middle point 
 of a given straight line, is equally distant from the extremities of that line. 
 
 Cor. 3. From the same point, three eq\ial straight lines cannot be 
 drawn to the same straight line ; for if there could, we should have two 
 equal oblique lines on the same side of the perpendicular, which is impos- 
 sible. 
 
 PROP. B. THEOR. 
 
 When the hypotenuse and one side of a right angled triangle, are respective' 
 ly equal to the hypotenuse and one side of another ; the two right angled 
 triangles are equal. 
 
 Suppose the hypotenuse AC = DF, and the side AB = DE ; the righ, 
 angled triangle ABC will be equal to the right angled triangle DEF
 
 OF GEOMETRY. BOOK I. 
 
 Their equality would be manifest, if the third sides BC and EF wer» 
 equal. If possible, suppose that those sides are not equal, and that BC is the 
 greater. Take BH = EF (3. 1.); andjoin AH. The triangle ABH = DEF; 
 for the right angles B and E are 
 equal, the side AB = DE, and BH 
 = EF ; hence, these triangles are 
 equal (4. 1.), and consequently 
 AH = DF. Now (by hyp.), we 
 have DF=AC; and therefore, 
 AH=AC. But by the last prop- 
 osition, the oblique line AC can- 
 not be equal to the oblique line 
 AH, which lies nearer to the per- 
 pendicular AB ; therefore it is 
 impossible that BC can differ 
 from EF ; hence, then, the trian- 
 gles ABC and DEF are equal. 
 
 PROP. C. THEOR. 
 
 Two angles are equal if their sides be parauei, each to each, and lying in tht 
 
 same direction. 
 
 If the straight lines AB, AC be parallel 
 to DF, DE ; the angle BAC is equal to 
 EDF. 
 
 For, draw GAD through the vertices. 
 And since AB is parallel to DF, the ex- 
 terior angle GAB is (29. 1.) equal to GDF ; 
 and, for the same reason, GAC is equal to 
 GDE ; there consequently remains the an- 
 gle BAC = EDF. 
 
 Cor. If BA, AC be produced to I and H, the angle BAC = HAI , 
 hence, the angle HAI is also equal to EDF. 
 
 SCHOLIUM. 
 
 The restriction of this proposition to the case where the side AB lies 
 in the same direction with DF, and AC in the same direction with DE, 
 is necessary ; because the angle CAI would have its sides parallel to those 
 of the angle EDF, but would not be equal to it. In that case, CAI and 
 EDF would be together equal to two right angles.
 
 44 
 
 ELEMENTS 
 
 PROP. D. PROB. 
 Two angles of a triangle being given, to find the third. 
 
 Draw any straight line CD ; at a 
 point therein, as B, make the angle 
 CBA equal to one of the given an- 
 gles, and the angle ABE equal to 
 the other: the remaining angle EBD 
 will be the third angle required ; be- 
 cause those three angles (Cor. 13. 1.) 
 are together equal to two right angles. 
 
 PROP. E. PROB. 
 Two angles of a triangle and a side being given, to construct the triangle 
 
 The two angles will either be both adjacent to the given side, or tiic 
 one adjacent and the other opposite : in the latter case, find the third angle 
 (Prop. D.) ; and the two adjacent angles will thus be known. 
 
 Draw the straight line BC equal to the 
 given side ; at the point B, make an angle 
 CBA equal to one of the adjacent angles, 
 and at C, an angle BCA equal to the other ; 
 the two lines BA, CA, will intersect each 
 other, and form with BC the triangle re- 
 quired ; for if they were parallel, the an- 
 gles B, C, would be together equal to two 
 right angles, and therefore could not be- 
 long to a triangle : hence, BAC will be the triangle required. 
 
 PROP. F. PROB. 
 
 Two sides and an angle opposite to one of them being given, to construct M 
 
 triangle. 
 
 This Problem admits of two cases. 
 
 First. When the given angle 
 is obtuse, make the angle BCA 
 equal to the given angle ; and take 
 CA equal to that side which is 
 adjacent to the given angle, the 
 arc described from A as a centre, 
 with a radius equal to AB, the 
 other given side, would cut BC on 
 opposite sides of C ; so that only -r> g^, 
 
 one obtuse angled triangle could be 
 formed ; that is, the triangle BCA will be the triangle required
 
 OF GEOMETRY. BOOK I. 45 
 
 And, if the given angle were right, although two triangles wouiJ be 
 formed, yet, as the hypotenuse would meet BC at equal distances from the 
 common perpendicular, these triangles would be equal. 
 
 Secondly. If the given angle be acute, and the side opposite to it greater 
 than the adjacent side, the same mode of construction will apply : for, mak 
 tng BCA equal to the given angle, and AC equal to the adjacent side 
 then, from A as centre, with a radius equal to the other given side, describe 
 an arc, cutting CB in B ; draw AB, and CAB will be the triangle requi- 
 red. 
 
 Bui if the given angle is acute, and the side opposite to it less than the 
 other given side ; make the angle CBA equal to the given angle, and take 
 BA equal to the adjacent side ; then, the arc described from the centre A, 
 with the radius AC equal to the opposite side, will cut the straight line 
 BC in two points C and C, lying on the same side of B : hence, there will 
 be two triangles BAC, BAC, either of which will satisfy the conditions 
 of the problem. 
 
 SCHOLIUM. 
 
 In the last case, if the opposite side was equal to the perpendicular from 
 the point A on the line BC, a right angled triangle would be formed. And 
 the problem would be impossible in all cases, if the opposite side was less 
 than the perpendicular let fall from the point A on the straight line BC. 
 
 PROP. G. PROB. 
 
 To find a triangle that shall he equivalent to any given rectilineal figure. 
 
 Let ABCDE be the given rectilineal figure. 
 
 Draw the diagonal CE, cutting off the triangle CDE ; draw DF paral* 
 lei to CE, meeting AE produced, and join CF: the polygon ABCDE 
 will be equivalent to the polygon 
 ABCF, which has one side less 
 than the original polygon. 
 
 For the triangles CDE, CFE, 
 have tho base CE common, and 
 they are between the same paral- 
 lels ; since their vertices D, F, are 
 situated in a line DF parallel to the 
 base : these triangles are therefore 
 equivalent (37. 1.) Draw, now, 
 the diagonal CA and BG parallel 
 to it, meeting EA produced : join 
 CG, and the polygon ABCF will be 
 reduced to an equivalent triangle ; 
 and thus the pentagon ABC DE 
 will bo reduced to an equivalent triangle GCF.
 
 46 ELEMENTS 
 
 The <jame process may be applied to every other polygon ; {or, by suc- 
 cessively diminishing the number of its sides, one being retrenched at each 
 step of the process, the equivalent triangle will at length be found. 
 
 Cor. Since a triangle may be converted into an equivalent rectangle 
 it follows that any polygon may be reduced to an equivalent rectangle. 
 
 PROP. H. PROB. 
 To find the side of a square that shall be equivalent to the sum of two squares 
 
 Draw the two indefinite lines AB, AC, per- 
 pendicular to each other. Take AB equal to 
 the side of one of the given squares, and AC 
 equal to the other ; join BC : this will be the 
 side of the square required. 
 
 For the triangle BAC being right angled, 
 the square constructed upon BC (47. 1.) is 
 equal to the sum of the squares described upon 
 AB and AC. 
 
 SCHOLIUM. 
 
 A square may be thus formed that shall be equivalent to the sum of any 
 number of squares ; for a similar construction which reduces two of them 
 to one, will reduce three of them to two, and these two to one, and so of 
 others. 
 
 PROP. I. PROB. 
 
 To find the side of a square equivalent to the difference of two given squares. 
 
 Draw, as in the last problem, (see the fig.) the lines AC, AD, at right angles 
 to each other, making AC equal to the side of the less square ; then, from 
 C as centre, with a radius equal to the side of the o*her square, describe 
 an arc cutting AD in D : the square described upon AD will be equivalent 
 to the difference of the squares constructed upon AC and CD. 
 
 For the triangle DAC is right angled ; therefore, the square described 
 upon DC is equivalent to the squares constructed upon AD and AC : hence 
 (Cor. 1. 47. :.j, AD 2 =CD 2 -AC 2 . 
 
 PROP. K. PROB. 
 
 A rectangle being given, to construct an equivalent one, having a side oj a 
 
 given length. 
 
 Let AEFH be the given rectangle, and produce one of its sides, as AH. till
 
 OF GEOMETRY. BOOK I 
 
 47 
 
 HB be the given length, and draw BFD 
 meeting the prolongation of AE in D ; 
 then produce EF till FG is equal to HB : 
 draw BGC, HFK, parallel to AED, and 
 through the point D draw DKC parallel 
 to AB or EG; then, the rectangle 
 GFKC, having the side FG of a given 
 length, is equal to the given rectangle 
 AEPH(43. 1.) 
 
 Cor. A polygon may be converted into an equivalent rectangle, having one 
 «l/ it s rides of a given length. 
 
 & 
 
 srlf.' I u 
 
 
 /: i 
 
 JLS 
 

 
 ELEMENTS 
 
 OF 
 
 GEOMETRY. 
 
 BOOK II. 
 
 DEFINITIONS. 
 
 1 Every right angled parallelogram, or rectangle, is said to be contained 
 by any two of the straight lines which are about one of the right an- 
 gles. 
 
 H Thus the right angled parallelogram AC is called the rectangle contain- 
 " ed by AD and DC, or by AD and AB, &c. For the sake of brevity, 
 " instead of the rectangle contained by AD and DC, we shall simply say 
 M the rectangle AD . DC, placing a Doint between the two sides of the 
 " rectangle." 
 
 A. In Geometry, the product of two lines means the same thing as their 
 rectangle, and this expression has passed into Arithmetic and Algebra, 
 where it serves to designate the product of two unequal numbers or 
 quantities, the expression square being employed to designate the pro 
 duct of a quantity multiplied by itself. 
 The arithmetical squares of 
 
 1, 2, 3, &c. are 1, 4, 9, &c. 
 
 So likewise the square de- 
 scribed on the double of 
 
 a* line is evidently four 
 
 times the square described 
 
 on a single one ; on a triple 
 
 line nine times that on a 
 
 single one, &c. 
 
 2 In every parallelogram, any of the 
 parallelograms about a diameter, to- 
 gether with the two complements, is 
 called a Gnomon. " Thus the paral- 
 " lelogram HG, together with the 
 u complements AF, FC, is the gno- 
 " mon of the parallelogram AC. This 
 "" gnomon may also, for the sake of 
 k brevity, be called the gnomon AGK 
 •or EHC." 
 
 ,4 
 
 A E
 
 OF GEOMETRY. BOOK II. 
 
 49 
 
 PROP. I. THEOR. 
 
 (J there be two straight lines, one of which is divided into any number of 
 parts ; the rectangle contained by the two straight lines is equal to the 
 rectangles contained by the undivided line, and the several parts of the 
 divided line. 
 
 Let A and BC be two straight lines ; and let BC be divided into any 
 parts in the points D, E ; the rectangle A.BC is equal to the several rect- 
 ingles A.BD, A.DE, A.EC. 
 
 From the point B draw (Prop. 11.1.) 
 BF at right angles to BC, and make BG 
 equal (Prop. 3. 1.) to A; and through 
 G draw (Prop. 31. 1.) GH parallel to 
 BC ; and through D, E, C, draw DK, 
 EL, CH parallel to BG ; then BH, BK, 
 DL, and EH are rectangles, and BH = 
 BK+DL+EH. 
 
 But BH = BG.BC= A.BC, because 
 BG=A: Also BK = BG.BD=A.BD, 
 because BG=A ; and DL=DK.DE = 
 A.DE, because (34. 1.) DK=BG=A. 
 In like manner, EH=A.EC. Therefore A.BC=A.BD+A.DE+A.EC ; 
 that is, the rectangle A.BC is equal to the several rectangles A.BD, A.DE, 
 A.EC. 
 
 SCHOLIUM. 
 
 The properties of the sections of lines, demonstrated in this Book, are 
 easily derived from Algebra. In this proposition, for instance, let the seg- 
 ments of BC be denoted by b, c, and d; then, k(b-\-c-\-d)-=kb-\- Ac+Ai. 
 
 B 
 
 ~« 
 
 
 
 
 
 K I 
 
 !i H 
 
 F 
 
 A 
 
 PROP. II. THEOR. 
 
 If a straight line be divided into any two parts, the rectangles contained by the 
 whole and each of the parts, are together equal to the square of the whol^line. 
 
 C B 
 
 Let the straight line AB be divided into any 
 two parts in the point C ; the rectangle AB.BC, 
 together with the rectangle AB.AC, is equal to 
 the square of AB ; or AB.AC+AB.BC=AB 3 . 
 
 On AB describe (Prop. 46. 1.) the square 
 ADEB, and through C draw CF (Prop. 31. 1.) 
 parallel to AD or BE ; then AF+CE=AE. 
 But AF=AD.AC=AB.AC, because AD=AB : 
 CE=BE.BC=AB.BC; and AE=AB 3 . There- 
 fore AB.AC+AB.BC=AB 2 . 
 
 SCHOLIUM. 
 
 This property is evident from Algebra : let AB be denoted by a, and th« 
 segments AC, CB, by b and d, respectively; then, a=b-\-d\ therefore, 
 nultiplying both members of this equality by a, we shall have a 2 =ab+ad
 
 50 
 
 ELEMENTS 
 
 PROP. III. THEOR. 
 
 If a straight line be divided into any two parts, the rectangle contained by the 
 whole and one of the parts, is equal to the rectangle contained by the tin* 
 parts, together with the square of the aforesaid part. 
 
 Let the straight line AB be divided into two parts, in the point C ; the 
 rectangle AB.BC is equal to the rect- 
 angle AC.BC, together with BC 2 . 
 
 Upon BC describe (Prop. 46. 1.) the 
 square CDEB, and produce ED to F, 
 and through A draw (Prop. 31. 1.) AF 
 parallel to CD or BE ; then AE=AD 
 + CE. 
 
 But AE = AB.BE = AB.BC, be- 
 cause BE=BC. So also AD=AC. 
 CD=AC.CB; and CE=BC 3 ; there- 
 fore AB.BC=AC.CB-f-BC 2 . 
 
 SCHOLIUM. 
 
 In this proposition let AB be denoted by a, and the segments AC and 
 CB, by b and c ; then a=b-{-c : therefore, multiplying both members of 
 this equality by c, we shall have ac=bc-\-c 2 . 
 
 PROP. IV. THEOR. 
 
 [J a straight line be divided into any two parts, the square of the whole line is 
 equal to the squares of the two parts, together with twice the rectangle con- 
 tained by the parts. 
 
 Let the straight line AB be divided into any two parts in C ; the square 
 of AB is equal to the squares of AC, CB, and to twice the rectangle con- 
 tained by AC, CB, that is, AB 2 =AC 2 +CB 2 +2AC.CB. 
 
 Upon AB describe (Prop. 46. 1.) the square ADEB, and join BD, and 
 through C draw (Prop. 31. 1.) CGF parallel to AD or BE, and through G 
 draw HK parallel to AB or DE. And because CF is parallel to AD, and 
 BD falls upon them, the exterior angle BGC 
 is equal (29. 1.) to the interior and opposite -^ 
 angle ADB ; but ADB is equal (5. 1.) to the 
 angle ABD, because BA is equal to AD, be- 
 ing sides of a square ; wherefore the angle H 
 CGB is equal to the angle GBC ; and there- 
 fore the side BC is equal (6. 1.) to the side 
 CG ; but CB is equal (34. 1.) also to GK and 
 CG to BK ; wherefore the figure CGKB is 
 equilateral. It is likewise rectangular ; for 
 the angle CBK being a right angle, the other 
 angles of the parallelogram CGKB are also right angles (Cor. 46. \.) 
 Wherefore CGKB is a square, and it is upon the side CB. For the same 
 
 B 
 
 F 
 
 J£
 
 OF GEOMETRY. BOOK II. 
 
 51 
 
 reason HF also is a square, and it is upon the side HG, whien is equal to 
 AC: therefore HF, CK are the squares of AC, CB. And because th« 
 complement AG is equal (43. l.)to the complement GE ; and because 
 AG = AC.CG=AC.CB, therefore also GE=AC.CB, and AG + GE = 
 2AC.CB. Now, HF=AC 2 and CK=CB 2 ; therefore, HF+CK+AG 
 + GE = AC 2 +CB 2 +2AC.CB. 
 
 But IIF-f-CK + AG + GE=the figure AE, or AB 2 ; therefore AB 2 » 
 AC 2 +CB 2 +2AC.CB. 
 
 Cor. From the demonstration, it is manifest that the parallelogram* 
 about the diameter of a square are likewise squares. 
 
 SCHOLIUM. 
 
 This property is derived from the square of a binomial. For, let the two 
 parts into which this line is divided be denoted by a and b\ then, (a-f £)* 
 =a 2 +2ab+b 2 . 
 
 PROP. V. THEOR. 
 
 If a straight line be divided into tico equal parts, and also into ttco unequal parts; 
 the rectangle contained by the unequal parts, together with the square oj the 
 line between the points of section, is equal to the square of half the line. 
 
 Let the straight line AB be divided into two equal parts in the point C, 
 and into two unequal parts in the point D ; the rectangle AD.DB, together 
 with the square of CD, is equal to the square of CB, or AD.DB-f-CD 2 — 
 CB 2 . 
 
 Upon CB describe (Prop. 46. 1.) the square CEFB, join BE, and through 
 D draw (Prop. 31. 1.) DHG parallel to CE or BF ; and through H draw 
 KLM parallel to CB or EF ; and 
 also through A draw AK parallel to 
 CLor BM : And because CH = HF, 
 if DM be added to both, CM=DF. 
 But AL=(36. 1.) CM, therefore AL 
 = DF, and adding CH to both, AH 
 =gnomon CMG. But AH = AD. 
 DH=AD.DB, because DH = DB 
 (Cot. 4. 2.) ; therefore gnomon CMG 
 =AD.DB. To each add LG=CD 2 , then, gnomon CMG + LG = AD.DB 
 + CD 2 . But CMG+LG = BC 2 ; therefore AD.DB + CD 2 =BC 3 . 
 
 '* Cor. From this proposition it is manifest, that the difference of the 
 "squares of two unequal lines, AC, CI), is equal to the rectangle contain- 
 ■' ed by their sum and difference, or that AC 2 — CD 2 =(AC + CD) (AC— 
 
 ' cd)." ; 
 
 scholium. 
 
 In this proposition, let AC be denoted by a, and CD by b ; then. AD=» 
 a+b, and DB=a — b; therefore, by Algebra, (u + b) X(fl — l>) = a 7 — b 2 ; 
 \hat is, the product of the sum and difference of two quantities, ts equivalent 
 »o the difference of their squares
 
 52 
 
 ELEMENTS 
 
 PROP. VI THEOR. 
 
 If a straight hnebe bisected, and produced to any point , the rectangle contained 
 by the whole line thus produced, and the part of it produced, together with the 
 square of half the line bisected, is equal to the square of the straight line which 
 is made up of the half and the part produced. 
 
 Let the straight line AB be bisected in C, and produced to the point D ; 
 the rectangle AD.DB together with the square of CB, is equal to the 
 square of CD. 
 
 Upon CD describe (Prop. 46.1.) the square CEFD, join DE, and 
 through B draw (Prop. 31.1.) BHG parallel to CE or DF, and through H 
 draw KLM parallel to AD or EF, and also through A draw AK parallel 
 to CL or DM. And because AC is 
 equal to CB, the rectangle AL is 
 equal (36.1.) to CH ; but CH is 
 equal (43. 1. ) to HF ; therefore also 
 AL is equal to HF : To each of these 
 add CM ; therefore the whole AM is 
 equal to the gnomon CMG. Now 
 AM=AD.DM = AD.DB, because 
 DM=DB. Therefore gnomon CMG 
 =AD.DB, and CMG+LG=AD. 
 DB + CB 2 . But CMG + LG=CF 
 =CD 2 , therefore AD.DB + CB 2 =CD 2 . 
 
 SCHOLIUM. 
 This property is evinced algebraically ; thus, let AB be denoted by 2a, 
 and BD by b ; then, AD=2a-\-b, and CD=a+b. Now by multiplication, 
 
 b(2a+b)=2ab+b 2 ; therefore, 
 by adding a 2 to each member of the equality, we shall have 
 b{2a+b) + a 2 =a 2 +2ab+b 2 ; 
 \ b(2o+ b) + a 2 =(a+b) 2 . 
 
 PROP. VII. THEOR. 
 
 If a straight line be divided into two parts, the squares of the whole line, and 
 of one of the parts, are equal to twice the rectangle contained by the whole and 
 that part, together with the square of the other part. 
 
 L et the straight line AB be divided into any 
 two parts in the point C ; the squares of AB, 
 BC, are equal to twice the rectangle AB.BC, 
 together with the square of AC, or AB 2 +BC 2 
 =2AB BC+AC 2 . 
 
 Upon AB describe (Prop. 46. 1.) the square 
 ADEB, and construct the figure as in the pre- 
 ceding propositions : Because AG=GE, AG 
 + CK = GE + CK, that is, AK = CE, and 
 rfierefore AK+CE=2AK. But AK+CE 
 =rgnomon AKF+CK ; and therefore AKF 
 
 E E
 
 OF GEOMETRY. BOOK II. 
 
 53 
 
 + CK=2AR = 2AB.BK=2AB.BC, because BK = (Cor. 4. 2.) BC. 
 Since then, AKF+CK=2AB.BC, AKF+CK+HF=2AB.BC+HF; 
 and because AKF-f HF=AE=AB 2 , AB 2 +CK=2AB.BC + HF, that 
 is, (since CK=CB 2 , and HF=AC 2 ,) AB 2 +CB 2 =2AB.BC+AC 2 . 
 
 " Cor. Hence, the sum of the squares of any two lines is equal U» 
 M twice the rectangle contained by the lines together with the square ol 
 " the difference of the lines." 
 
 SCHOLIUM. 
 
 In this proposition, let AB be denoted by a, and the segments AC and 
 CB by b and c ; 
 
 then a 2 =b 2 +2bc+c 2 ; 
 adding c 2 to each member of this equality, we shall have, 
 a 2 +c 2 =A 2 +26c+2c 2 ; 
 .-. a 2 +c 2 =b 2 +2c(b+c), 
 or a 2 -\-c 2 =2ac-{-b 2 . 
 
 Cor. From this proposition it is evident, that the square described on 
 the difference of two lines is equivalent to the sum of the squares described on 
 the lines respectively, minus twice the rectangle contained by the lines. For 
 a — c=b ; therefore, by involution, c 2 — 2ac-\-c 2 =b 2 . This may be also 
 derived from the above algebraical equality, by transposition. 
 
 PROP. VIII. THEOR. 
 
 [fa straight line be divided into any two parts, four times the rectangle con- 
 tained by the whole line, and one of the parts, together with the square of 
 the other part, is equal to the square of the straight line which is made up 
 of the whole and the first-mentioned part. 
 
 Let the straight line AB be divided into any two parts in the point C ; 
 four times the rectangle AB.BC, together with the square of AC, is equal 
 to the square of the straight line made up of AB and BC together. 
 
 Produce AB to D, so that BD be equal to CB, and upon AD describe 
 the square AEFD ; and construct two figures such as in the preceding. 
 Because GK is equal (34. 1.) to CB, and CB to BD, and BD to KN, GK 
 is equal to KN. For the same reason, PR 
 is equal to RO ; and because CB is equal 
 to BD, and GK to KN, the rectangles CK 
 and BN are equal, as also the rectangles 
 GR and RN : But CK is equal (43. 1.) 
 lo RN, because they are the complements 
 cf the parallelogram CO: therefore also 
 BN is equal to GR ; and the four rect- 
 angles BN, CK, GR, RN are there- 
 fore equal to one another, and so CK-f- 
 BN + GR + RN = 4CK. Again, be- 
 cause CB is equal to BD, and BD equal
 
 84 ELEMENTS 
 
 (On . 4. 2 I U BK, tl at is, to CG ; and CB equal to GK, that is, to GP - 
 therelore CG s equal to GP ; and because CG is equal to GP, and PR to 
 RO, the rectangle AG is equal to MP, and PL to RF : but MP is equal 
 (43. 1.) to PL, because they are the complements of the parallelogram 
 ML ; wherefore AG is equal also to RF. Therefore the four rectangles 
 AG, MP, PL, RF,are equal to one another, and so AG+MP + PL+RF 
 =4AG. And it was demonstrated, that CK+BN+GR+RN=4CK ; 
 wherefore, adding equals to equals, the whole gnomon AOH=4AK. 
 Now AK=AB.BK=AB.BC, and 4AK=4AB.BC ; therefore, gnomon 
 AOH=4AB.BC; and adding XH, or (Cor. 4. 2.) AC 2 , to both, gnomon 
 AOH + XH=4AB.BC+AC 2 . But AOH+XH=AF = AD 2 ; therefore 
 AD 2 =4AB.BC+AC 2 . 
 
 " Cor. 1. Hence, because AD is the sum, and AC the difference of 
 " the lines AB and BC, four times the rectangle contained by any two 
 " lines, together with the square of their difference, is equal to the square 
 " of the sum of the lines." 
 
 " Cor. 2. From the demonstration it is manifest, that since the square 
 " of CD is quadruple of the square of CB, the square of any line is qua- 
 " druple of the square of half that line." 
 
 SCHCLIUM. 
 
 In this proposition, let the line AB be denoted by a, and the parts AC 
 and CB by c and b ; then AD—c-\-2b. Now, since a=b-\-c, multiplying 
 both members by Ab, we shall have 
 
 4ai=46 2 +4&c; 
 and adding c 2 to each member of this equality, we shall have, 
 4a&+c 2 =c 2 +46c+4& 2 , 
 or 4ai+c 2 =(c4-24) 2 . 
 
 PROP. IX. THEOR. 
 
 If a straight line be divided into two equal, and also into two unequal parts , 
 the squares of the two unequal parts are together double of the square of halj 
 the line, and of the square of the line between the points of section. 
 
 Let the straight line AB be divided at the point C into two equal, and 
 at D into two unequal parts ; The squares of AD, DB are together double 
 of the squares AC, CD. 
 
 From the point C draw (Prop. 11.1.) CE at right angles to AB and 
 make it equal to AC or CB, and join EA, EB ; through D draw (Prop 31. 
 1.) DF parallel to CE, and through F draw FG parallel to AB ; and join 
 AF. Then, because AC is equal to CE, 
 the angle EAC is equal (5. 1.) to the 
 angle AEC ; and because the angle ACE 
 is a right angle, the two others AEC, 
 EAC together make one right angle (Cor. 
 4. 32. 1 .) ; and they are equal to one ano- 
 tner ; each of them therefore is half of a 
 right angle. For the same reason each
 
 OF GEOMETRY. BOOK II. 55 
 
 of the angles CEB, EBC i« half a right angle ; and therefore tue whoie 
 AEB is a right angle ; And because the angle GEF is half a right angle 
 and EGF a right angle, for it is equal (29. 1.) to the interior and opposite 
 angle ECB, the remaining angle EFG is half a right angle ; therefore the. 
 angle GEF is equal to the angle EFG, and the side EG equal (6. 1.) to the 
 side GF ; Again, because the angle at B is half a right angle, and FDB ? 
 right angle, for it is equal (29. 1.) to the interior and opposite angle ECB, 
 the remaining angle BFD is half a right angle ; therefore the angle at B it 
 equal to the angle BFD, and the side DF to (6. 1.) the side DB. Now, be- 
 cause AC=CE, AC 2 =CE 2 , and AC 2 +CE 2 =2AC 2 . But (47. 1 .) AE 2 =x 
 AC 2 +CE 2 ; therefore AE 2 =2AC 2 . Again, because EG=GF, EG 2 =GF 2 
 and EG 2 4-GF 2 =2GF 2 . But EF 2 =EG 2 +GF 2 ; therefore, EF 2 =2GF* 
 =2CD 2 , because (34. 1.) CD=GF. And it was shown that AE 2 =2AC 2 , 
 therefore AE 2 +EF 2 =2AC 2 +2CD 2 . But (47. 1.) AF 2 =AE 2 +EF* 
 and AD 2 +DF 2 =AF 2 , or AD 2 +DB 2 =AF 2 ; therefore, also, AD 2 +DB 2 =* 
 2AC 2 +2CD 2 . 
 
 SCHOLIUM. 
 This property is evident from the algebraical expression, 
 (a+&) 2 +(a— b) 2 =2a 2 +2b 2 ; 
 where a denotes AC, and b denotes CD ; hence, a-\-b =AD, a — 4=DB. 
 
 PROP. X. THEOR. 
 
 If a straight line bebisected, and produced to any point, the square of the whole 
 line thus produced, and the square of the part of it produced, are together 
 double of the square of half the line bisected, and of the square of the lint 
 made up of the half and the part produced. 
 
 Let the straight line AB be bisected in C, and produced to the point D ; 
 the squares of AD, DB are double of the squares of AC, CD. 
 
 From the point C draw (Prop. 11.1.) CE at right angles to AB, and make 
 it equal to AC or CB ; join AE, EB ; through E draw (Prop. 31. 1.) EF 
 parallel to AB, and through D draw DF parallel to CE. And because 
 the straight line EF meets the parallels EC, FD, the angles CEF, EFD 
 are equal (29. 1.) to two right angles ; and therefore the angles BEF, EFD 
 are less than two right angles ; But straight lines, which with another 
 straight line make the interior angles upon the same side less than two 
 right angles, do meet (29. 1.), if produced far enough; therefore EB, FD 
 will meet, if produced, towards B, D : let them meet in G, and join AG. 
 Then because AC is equal to CE, 
 the angle CEA is equal (5. 1.) to 
 the angle EAC ; and the angle 
 ACE is a right angle ; therefore 
 each of the angles CEA, EAC is 
 half a right angle (Cor. 4.32. 1.); 
 For the same reason, each of the 
 angles CEB, EBC is half a right 
 angle; therefore AEB is a right an- 
 gle ; And because EBC is half a
 
 66 
 
 ELEMENTS 
 
 right angie, DBG is also (15. 1.) half a right angle, for they are vertically 
 opposite : but BDG is a right angle, because it is equal (29. 1.) to the al- 
 ternate angle DCE ; therefore the remaining angle DGB is half a right 
 angle, and is therefore equal to the angle DBG; wherefore also the side 
 DB is equal (6. 1.) to the side DG. Again, because EGF is halt a right 
 angle, and the angle at F aright angle, being equal (34. 1.) to the 
 opposite angle ECD, the remaining angle FEG is half a right angle, 
 and equal to the angle EGF ; wherefore also the side GF is equal 
 (6. 1.) to the side FE. And because EC=CA, EC 2 + CA 2 = 2CA 2 . 
 Now AE 2 = (47. 1.) AC 2 4-CE2; therefore, AE 2 = 2AC 2 . Again, be- 
 cause EF = FG, EF 2 =FG 2 , and EF 2 +FG 2 =2EF 2 . ButEG 2 = (47. 1.) 
 EF 2 +FG 2 ; therefore EG 2 =2EF 2 ; and since EF=CD, EG 2 =2CD 2 . 
 And it was demonstrated, that AE 2 =2AC 2 ; therefore, AE 2 +EG 2 =2AC 2 
 +2CD 2 . Now, AG 2 =AE 2 -f-EG 2 , wherefore AG 2 =2AC 2 +2CD i . But 
 AG 2 (47. l.) = AD 2 +DG 2 = AD 2 +DB 2 , because DG=DB: Therefore 
 AD 2 +DB 2 =2AC 2 +2CD 2 . 
 
 SCHOLIUM. 
 
 Let AC be denoted by a, and BD, the part produced, by b ; then AD^z 
 2a-\-b, and CD=a+b. 
 
 Now, (2a + b) 2 +b 2 = 4a 2 +4ab+2b i ; but 4a 2 + 4ab+2b 2 =2a 2 +2 (a + 
 b) 2 ; hence, (2a+^) 2 +Z> 2 =2a 2 +2(a-f-5) 2 , and the proposition is evident 
 from this algebraical equality. 
 
 PROP. XL PROB. 
 
 To divide a given straight line into two parts, so that the rectangle contained 
 by the whole, and one of the parts, may be equal to the square of the other 
 part. 
 
 Let AB be the given straight line ; it is required to divide it into two 
 
 parts, so that the rectangle contained by 
 the whole, and one of the parts, shall be 
 equal to the square of the other part. 
 
 Upon AB describe (46. 1.) the square 
 ABDC ; bisect (10. 1.) AC in E, and join 
 BE ; produce CA to F, and make (3. 1.) 
 EF equal to EB, and upon AF describe 
 (46. 1.) the square FGHA, AB is divided 
 in H, so that the rectangle AB, BH is equal 
 to the square of AH. 
 
 Produce GH to K : Because the straight 
 line AC is bisected in E, and produced to 
 the point F, the rectangle CF.FA, to- 
 gether with the square of AE, is equal 
 (6. 2.) to the square of EF : But EF in 
 equal to EB ; therefore the rectangle CF 
 FA, together with the souare of AE. is
 
 OF GEOMETRY. BOOK II. 5? 
 
 equal to the square of EB ; And the squares of BA, AE are equai 
 (17. l.)to the square of EB, because the angle EAB is a right angle; 
 therefore the rectangle CF.FA, together with the square of AE, is equa. 
 to the squares of BA, AE : take away the square of AE, which is com 
 mon to both, therefore the remaining rectangle CF.FA is equal to the 
 square of AB. Now the figure FK is the rectangle CF.FA, for AF is 
 equal to FG ; and AD is the square of AB ; therefore FK is equal to AD : 
 take away the common part AK, and the remainder FH is equal to tho 
 remainder HD. But HD is the rectangle AB.BH for AB is equal to 
 BD ; and FH is the square of AH ; therefore the rectangle AB.BU is 
 equal to the square of AH : Wherefore the straight line AB is divided in 
 H, so that the rectangle AB.BH is equal to the square of AH. 
 
 PROP. XII. THEOR. 
 
 In obtuse angled triangles, if a perpendicular be drawn from any of the acute 
 angles to the opposite side produced, the square of the side subtending the 
 obtuse angle is greater than the squares of the sides containing the obtuse 
 angle, by twice the rectangle contained by the side upon which, wJien produced, 
 the perpendicular falls, and the straight line intercepted between the perpen- 
 dicular and the obtuse angle. 
 
 Let ABC be an obtuse angled triangle, having the obtuse angle ACB, 
 and from the point A let AD be drawn (12. 1.) perpendicular to BC pro- 
 duced : The square of AB is greater than the squares of AC, CB, by twice 
 the rectangle BC.CD. 
 
 Because the straight line BD is divided _A. 
 
 into two parts in the point C, BD 2 =(4. 2.) 
 BC 2 +CD 2 +2BC.CD; add AD 2 to both: 
 Then. BD 2 +AD 2 = BC 2 + CD 2 -f- AD 2 + 
 2BC.CD. But AB 2 =BD 2 +AD 2 (47. 1.), 
 and AC 2 = CD 2 + AD 2 (47. 1.); therefore, 
 AB 2 =BC 2 +AC 2 +2BC.CD; that is, AB 2 
 is greater than BC 2 +AC 2 by 2BC.CD. 
 
 PROP. XIII. THEOR. 
 
 fn every triangle the square of the sidt subtending any of the acute angles, ts 
 less than the squares of the sides containing that angle, by twice the rectan- 
 gle contained by either of these sides, and the straight line intercepted be- 
 tween the perpendicular, let fall upon it from the opposite angle, and the acute 
 angle. 
 
 Let ABC be any triangle, and the angle at B one of its acute angles, and 
 upon BC, one of the sides containing it, let fall the perpendicular (12. 1.) 
 AD from the opposite angle : The square of AC, opposite to tne angle B, 
 is lev than the squares of CB, BA bv twice the rectangle CB.BD. 
 
 8*
 
 ELEMENTS 
 
 P list, let AD fall within the triangle ABC ; 
 and because the straight line CB is divided 
 into two parts in the point D (7. 2), BC 2 + 
 BD 2 =2BC.BD + CD 2 . Addtoeach AD 2 ; 
 thenBC 2 +BD 2 +AD 2 =2BC.BD + CD 2 + 
 AD 2 . But BD 2 +AD 2 =AB 2 , and CD 2 + 
 DA 2 ^r AC 2 (47. 1.) ; therefore BC 2 + AB 2 = 
 2BC.BD + AC 2 ; that is, AC 2 is less than 
 BC 2 +AB 2 by 2BC.BD. 
 
 B D C 
 
 Secondly, let AD fall without the triangle ABC :* Then because the 
 angle at D is a right angle, the angle ACB is greater (16. 1.) than a right 
 angle, and AB 2 = (12. 2.) AC 2 +BC 2 4-2BCCD. Add BC 2 to each; 
 then AB 2 +BC 2 =:A.C 2 -r-2BC 2 +2BC.CD. But because BD is divided 
 into two parts in C, BC 2 +BC.CD=(3. 2.) BC.BD, and 2BC 2 +2BC.CD 
 =2BC.BD: therefore AB 2 + BC 2 =AC 2 + 2BC.BD ; and AC 2 is les* 
 than AB 2 +BC 2 , by 2BD.BC. 
 
 Lastly, let the side AC be perpendicular 
 *.o BC ; then is BC the straight line between 
 the perpendicular and the acute angle at B ; 
 and it is manifest that (47. 1.) AB 2 +BC 2 = 
 AC 2 +2BC 2 =AC 2 +2BC.BC. 
 
 PROP. XIV. PROB. 
 
 To describe a square that shall be equal to a given rectilineal figure. 
 
 Let A be the given rectilineal figure ; it is required to describe a square 
 that shall be equal to A. 
 
 Describe (45. 1.) the rectangular parallelogram BCDE equal to the 
 rectilineal figure A. If then the sides of it, BE, ED are equal to one an- 
 other, it is a square, and what was required is done ; but if they are not 
 equal, produce one of them, BE to F, and make EF equal to ED, and bi 
 sect BF in G ; and from the centre G, at the distance GB, or GF, de- 
 scribe the semicircle BHF, and produce DE to H, and join GH. There 
 fore, because the straight line BF is divided into two equal parts in tl 
 point G, and into two unequal in the point E, the rectangle BE.EF, to 
 gether with the square of EG, is equal (5. 2.) to the square of GF : 
 W GF is equal to GH ; therefore the rectangle BE, EF, together 
 mth the square of EG, is equal to the square of GH : But the squares of 
 
 * See figure of the last Proposition
 
 OF GEOMETRY. BOOK II. 
 
 59 
 
 HE and EG are equal (47. 
 1.) to the square of GH : 
 Therefore also the rectangle 
 BE.EF, together with the 
 square of EG, is equal to 
 the squares of HE and EG. 
 Take away the square of 
 EG, which is common to 
 both, and the remaining 
 rectangle BE.EF is equal 
 to the square of EH : But 
 BD is the rectangle con- 
 tained by BE and EF, because EF is equal to ED ; therefore BD is equal 
 to the square of EH ; and BD is also equal to the rectilineal figure A ; 
 therefore the rectilineal figure A is equal to the square of EH : Where- 
 fore a square has been made equal to the given rectilineal figure A, viz. 
 the square described upon EH. 
 
 PROP. A. THEOR. 
 
 If one side of a triangle be bisected, the sum of the squares of the other two 
 sides is double of the square of half the side bisected, and of the square 
 of the line drawn from the point of bisection to the opposite angle of tht 
 triangle. 
 
 Let ABC be a triangle, of which the side BC is bisected in D, and DA 
 drawn to the opposite angle ; the squares of BA and AC are togethei 
 double of the squares of BD and DA. 
 
 From A draw AE perpendicular to BC, and because BE A is a right an 
 gle, AB 2 =(47. 1.) BE 2 +AE 2 and AC 2 = 
 
 CE 2 +AE*; wherefore AB 2 +AC 2 = BE 2 A 
 
 + CE 2 +2AE 2 . But because the line 
 BC is cut equally in D, and unequally 
 in E, BE 2 + CE 2 = (9. 2.) 2BD 2 + 
 2DE 2 ; therefore AB 2 + AC 2 =2BD 2 + 
 2DE 2 .2AE 2 . 
 
 Now DE 2 +AE 2 =(47. 1.) AD 2 , and 
 2DE 2 +2AE 2 =2AD 2 ; wherefore AB 2 + 
 AC 2 =2BD 2 +2AD a . 
 
 PROP. B. THEOR. 
 
 The sum of the squares of the diameters of any parallelogram is equat to 
 the sum of the squares of the sides of the parallelogram. 
 
 Let ABCD be a parallelogram, of which the diameters are AC and BD ; 
 the sum of the squares of AC and BD is equal to the sum of the squarei 
 of AB, BC, CD, DA. 
 
 Let AC and BD intersect one another in E • and because the vertical 
 angle* AED, CKB are equal (15. 1.), and also the alternate angles EAD,
 
 60 
 
 ELEMENTS, &c. 
 
 ECB (29. 1.), tho triangles ADE, CEB have two angles in the one equal 
 
 to two angles in the other, each to each ; but the sides AD and BC, which 
 
 are opposite to equal angles in 
 
 these triangles, are also equal 
 
 (34. 1.); therefore the other 
 
 6ides which are opposite to the 
 
 equal angles are also equal (26. 
 
 1.), viz. AE to EC, and ED to 
 
 EB. 
 
 Since, therefore, BD is bi- 
 sected in E, AB 2 +AD 2 =(A. 
 2.) 2BE 2 +2AE 2 ; and for the 
 same reason, CD 2 + BC 2 = 
 2BE 2 +2EC 2 =2BE 2 +2AE 2 , because EC = AE. Therefore AB 2 + AD 2 
 + DC 2 +BC 2 =4BE 2 +4AE 2 . But 4BE 2 =BD 2 , and 4AE 2 =AC 2 (2. 
 Cor. 8. 2.) because BD and AC are both bisected in E ; therefore AB 2 + 
 AD 2 +CD 2 +BC 2 =BD 2 +AC 2 . 
 
 Cor. From this demonstration, it is manifest that the diameters of every 
 parallelogram bisect one another. 
 
 SCHOLIUM. 
 
 In the case of the rhombus, the sides AB, BC, being equal, the triangles 
 BEC, DEC, have all the sides of the one equal to the corresponding sides 
 of the other, and are therefore equal : whence it follows that the angles 
 BEC, DEC, are equal ; and, therefore, that the two diagonals of a rham- 
 bua cut each other at right angles.
 
 ELEMENTS 
 
 OF 
 
 GEOMETRY. 
 
 BOOK III. 
 
 DEFINITIONS. 
 
 A . The radius of a circle is the straight line drawn from the centre to tj»« 
 circumference. 
 
 1. A straight line is said to touch 
 a circle, when it meets the cir- 
 cle, and being produced does 
 not cut it. 
 And that line which has but 
 
 one point in common with 
 the circumference, is called a 
 tangent, and the point in com- 
 mon, the point of contact. 
 
 2. Circles are said to touch one 
 another, which meet, but do not 
 cut one another. 
 
 3. Straight lines are said to be equally dis- 
 tant from the centre of a circle, when the 
 perpendiculars drawn to them from the centre 
 are equal. 
 
 4. And the straight line on which the greater 
 perpendicular falls, is said to be farther from 
 the centre. 
 
 B. Any portion of the circumference is called an arc. 
 The chord or subtense of an arc is the straight line which joins its two ex- 
 tremities. 
 
 C. A straight line is said to be inscribed in a circle, when the extremities of 
 it are in the circumference of the circle. And any straight line which 
 meets the circle in two points, is called a secant. 
 
 5. A segment of a circle is the figure con- 
 tained by a straight line, and the arc which 
 it cuts off.
 
 62 
 
 ELEMENTS 
 
 6. An angle in a segment is the angle contained 
 by two straight lines drawn from any point in 
 the circumference of the segment, to the extre- 
 mities of the straight line which is the base of 
 the segment. 
 
 An inscribed triangle, is one which has its three 
 angular points in the circumference. 
 
 And, generally, an inscribed figure is one, of 
 which all the angles are in the circumference. 
 The circle is said to circumscribe such a figure. 
 
 7. And an angle is said to insist or stand upon 
 the arc intercepted between the straight lines 
 which contain the angle. 
 
 This is usually called an angle at the centre. The 
 angles at the circumference and centre, are 
 both said to be subtended by the chords or 
 arcs which their sides include. 
 
 8. The sector of a circle is the figure contained 
 by two straight lines drawn from the centre, and 
 the arc of the circumference between them. 
 
 9. Similar segments of a circle, 
 are those in which the angles are 
 equal, or which contain equal an- 
 gles. 
 
 PROP. I. PROB. 
 To find the centre of a given circle. 
 
 Let ABC be the given circle ; it is required to find its centre. 
 
 Draw within it any straight line AB, and bisect (10. 1.) it in D ; 
 from the point D draw (11. 1.) DC at right angles to AB, and produce it 
 to E, and bisect CE in F : the point F is the centre of the cirele ABC 
 
 For, if it be not, let, if possible, G be the centre, and join GA, GD, GB : 
 Then, because DA is equal to DB, and DG common to the two triangles 
 ADG, BDG, the two sides AD, DG are equal to 
 the two BD, DG, each to each ; and the base 
 GA is equal to the base GB, because they are 
 radii of the same circle : therefore the angle 
 ADG is equal (8. 1.) to the angle GDB : But 
 when a straight line standing upon another 
 straight line makes the adjacent angles equal to 
 one another, each of the angles is a right angle 
 (7. def. 1.) Therefore the angle GDB is a right 
 angle : But FDB is likewise a right angle : 
 wherefore the angle FDB is equal to the angle 
 GDB, the greater to the less which is impos-
 
 OF GEOMETRY. BOOK III. 
 
 53 
 
 sible : Therefore G is not the centre of the circle ABC : In the Bame 
 manner it can be shown that no other point but F is the centre : that is 
 F is the centre of the circle ABC. 
 
 Cor. From this it i3 manifest that if in a circle a straight line bisect 
 another at right angles, the centre of the circle is in the line which bisects 
 the other. 
 
 PROP. II. THEOR. 
 
 If any two points be taken in the circumference of a circle, the straight line 
 which joins them shall fall within the circle. 
 
 Let ABC be a circle, and A, B any two points in the circumference ; 
 he straight line drawn from A to B shall fall 
 within the circle. 
 
 Take any point in AB as E ; find D (1. 3.) 
 »he centre of the circle ABC ; join AD, DB 
 and DE, and let DE meet the circumference 
 in F. Then, because DA is equal to DB, the 
 angle DAB is equal (5. 1.) to the angle DBA ; 
 and because AE, a side of the triangle DAE, 
 is produced to B, the angle DEB is greater 
 (16. 1.) than the angle DAE ; but DAE is 
 equal to the angle DBE ; therefore the angle DEB is greater than the 
 angle DBE: Now to the greater angle the greater side is opposite (19. 
 1.) ; DB is therefore greater than DE : but BD is equal to DF ; where- 
 fore DF is greater than DE, and the point E is therefore within the circle. 
 The same may be demonstrated of any other point between A and B, 
 therefore AB is within the circle. 
 
 Cor. Every point, moreover, in the production of AB, is farther from the 
 centre than the circumference. 
 
 PROP. III. THEOR. 
 
 If a straight line drawn through the centre of a circle bisect a straight line in 
 the circle, which does not pass through the centre, it will cut that line at right 
 angles ; and if it cut it at right angles, it will bisect it. 
 
 Let ABC be a circle, and let CD, a straight line drawn through the 
 centre, bisect any straight line AB, which does not pass through the 
 centre, in the point F ; it cuts it also at right angles. 
 
 Take (1. 3.) E the centre of the circle, and join EA, EB. Then be- 
 cause AF is equal to FB, and FE common to the 
 two triangles AFE, BFE, there are two sides in the 
 one equal to two sides in the other : but the base 
 EA is equal to the base EB ; therefore the angle 
 AFE is equal (8. 1.) to the angle BFE. And 
 when a straight line standing upon another makes 
 tho adjacent angles equal to one another, each of 
 them is a right (7. Def. 1.) angle : Therefore each 
 of the angles AFE, BFE is a right angle ; where- 
 fore the straight line CD, drawn through the centre
 
 64 ELEMENTS 
 
 bisecting AJB, which does not pass through the centre, cuts AB at ngfn 
 angles. 
 
 Again, let CD cut AB at right angles ; CD also bisects AB, that is, AF 
 is equal to FB. 
 
 The same construction being made, because the radii EA, EB are equal 
 to one another, the angle EAF is equal (5. 1.) to the angle EBF ; and 
 the right angle AFE is equal to the right angle BFE : Therefore, in the 
 two triangles EAF, EBF, there are two angles in one equal to two angles 
 in the other ; now the side EF, which is opposite to one of the equal an- 
 gles in each, is common to both ; therefore the other sides are equal to 
 (28. 1.) : AF therefore is equal to FB. 
 
 Cor. 1. Hence, the perpendicular through the middle of a chord, passes, 
 through the centre ; for this perpendicular is the same as the one let fall 
 from the centre on the same chord, since both of them passes through the 
 middle of the chord. 
 
 Cor. 2. It likewise follows, that the perpendicular drawn through the 
 middle of a chord, and terminated both ways by the circumference of the circle, 
 is a diameter, and the middle point of that diameter is therefore the centre of 
 the circle. 
 
 PROP. IV. THEOR. 
 
 If in a circle two straight lines cut one another, which do not both pass through 
 the centre, they do not bisect each other. 
 
 Let ABCD be a circle, and AC, BD two straight lines in it, which cut 
 one another in the point E, and do not both pass through the centre : AC, 
 BD do not bisect one another. 
 
 For if it is possible, let AE be equal to EC, and BE to ED ; if one of the 
 lines pass through the centre, it is plain that it 
 cannot be bisected by the other, which does not 
 pass through the centre. But if neither of them 
 pass through the centre, take (1. 3.) F the centre 
 of the circle, and join EF : and because FE, a 
 straight line through the centre, bisects another 
 AC, which does not pass through the centre, it 
 must cut it at right (3. 3.) angles ; wherefore 
 FEA is a right angle. Again, because the 
 straight line FE bisects the straight line BD, which does not pass through 
 the centre, it must cut it at right (3. 3.) angles ; wherefore FEB is a right 
 angle : and FEA was shown to be a right angle : therefore FEA is eaual 
 to the angle FEB, the less to the greater, which is impossible ; therefore 
 AC, BD, do not bisect one another. 
 
 PROP. V. THEOR. 
 If two circles cut one another, they cannot have the same centre. 
 
 Let the two circles ABC, CDG cut one another in the points B, C ; 
 they have not the same centre.
 
 OF GEOMETRY. BOOK III. 
 
 For, if it be possible, let E be their 
 centre : join EC, and draw any straight line 
 EFG meeting the circles in F and G : and 
 because E is the centre of the circle ABC, 
 CE is equal to EF : Again, because E is 
 the centre of the circle CDG, CE is equal to 
 EG : but CE was shown to be equal to EF, 
 therefore EF is equal to EG, the less to the 
 greater, which is impossible : therefore E 
 is not the centre of the circles, ABC, CDG. 
 
 PROP. VI. THEOR. 
 
 If two circles touch one another internally, they cannot have the same centre 
 
 Let the two circles ABC, CDE, touch one another internally in the 
 point C ; they have not the same centre. 
 
 For, if they have, let it be F ; join FC, and 
 draw any straight line FEB meeting the circles 
 in E and B ; and because F is the centre of 
 the circle ABC, CF is equal to FB ; also, be- 
 cause F is the centre of the circle CDE, CF 
 is equal to FE : but CF was shown to be equal 
 to FB ; therefore FE is equal to FB, the less 
 to the greater, vhich is impossible ; Where- 
 fore F is not the centre of the circles ABC, 
 CDE. 
 
 PROP. VII. THEOR. 
 
 If any point be taken in the diameter of a circle which is not the centre, oj aHl 
 the straight lines which can be drawn from it to the circumference, the great- 
 est is that in which the centre is, and the other part of that diameter is the 
 least ; and, of any others, that which is nearer to the line passing through 
 the centre is always greater than one more remote from it ; And from the 
 same point there can be drawn only two straight lines that are equal to or,t 
 another, one upon each side of the shortest line. 
 
 Let ABCD be a circle, and AD its diameter, in which let any point h 
 be taken which is not the centre : let the centre be E ; of all the straight 
 iines FB, FC, FG, &c. that can be drawn from F to the circumference. 
 FA is the greatest ; and FD, the other part of the diameter AD, is the 
 least ; and of the others, FB is greater than FC, and FC than FG. 
 
 Join BE, CE, GE ; and because two sides of a triangle are greatej 
 (20. 1.) than the third, BE, EF are greater than BF ; but AE is equal to 
 KB ; therefore AE and EF, that is, AF, is greater than BF : Again, be 
 •ause BE is equal to CE, and FE common to the triangles BEF, CEF, 
 
 9
 
 66 ELEMENTS 
 
 the two sides BE, EF are equal to the two 
 CE EP; but the angle BE F is greater than 
 the angle CEF ; therefore the base BF is 
 greater (24. 1 .) than the base FC ; for the same 
 reason, CF is greater than GF. Again, be- 
 cause GF, FE are greater (20. 1.) than EG, 
 and EG is equal to ED ; GF, FE are greater 
 than ED ; take away the common part FE, 
 and the remainder GF is greater than the re- 
 mainder FD : therefore FA is the greatest, and 
 FD the least of all the straight lines from F to 
 the circumference ; and BF is greater than CF, and CF than GF. 
 
 Also there can be drawn only two equal straight lines from the point F 
 to the circumference, one upon each side of the shortest line FD : at the 
 point E in the straight line EF, make (23. 1.) the angle FEH equal to the 
 angle GEF, and join FH : Then, because GE is equal to EH, and EF com- 
 mon to the two triangles GEF, HEF ; the two sides GE, EF are equal 
 to the two HE, EF; and the angle GEF is equal to the angle HEF ; 
 therefore the base FG is equal (4. 1.) to the base FH : but besides FH, 
 no straight line can be drawn from F to the circumference equal to 
 FG : for, if there can, let it be FK ; and because FK is equal to FG, and 
 FG to FH, FK is equal to FH ; that is, a line nearer to that which passes 
 through the centre, is equal to one more remote, which is impossible. 
 
 PROP. VIII. THEOR. 
 
 tf any point be taken without a circle, and straight lines be drawn from it to 
 the circumference, whereof one passes through the centre , of those which 
 fall upon the concave circumference, the greatest is that which passes through 
 the centre ; and of the rest that which is nearer to that through the centre 
 is always greater than the more remote ; But of those which fall upon the 
 convex circumference, the least is that between the point without the circle, 
 and the diameter ; and of the rest, that which is nearer to the least is al- 
 ways less than the more remote : And only two equal straight lines can be 
 drawn from the point unto the circumference, one upon each side of the least. 
 
 Let ABC be a circle, and D any point without it, from which let the 
 straight lines DA, DE, DF, DC be drawn to the circumference, whereof DA 
 passes through the centre. Of those which fall upon the concave part of the 
 circumference AEFC, the greatest is AD, which passes through the cen- 
 tre ; and the line nearer to AD is always greater than the more remote, 
 viz. DE than DF, and DF than DC ; but of those which fall upon the con- 
 vex circumference HLKG, the least is DG, between the point D and the 
 diameter AG ; and the nearer to it is always less than the more remote, 
 viz. DK than DL, and DL than DH. 
 
 Take (1. 3.) M the centre of the circle ABC, and join ME, MF, MC, 
 MK, ML, MH: And because AM is equal to ME, if MD be added to 
 each, AD is equal to EM and MD; but EM and MD are greater (20. 1.) 
 than ED : therefore also AD is greater than ED. Again, because ME is 
 equal to MF, and MD common to the triangles EMD, FMD; EM, MD
 
 OF GEOMETRY. BOOK III. 
 
 6? 
 
 are equal to FM, MD ; but the angle EMD is greater than the angle 
 
 FMD ; therefore the base ED is greater 
 
 (24. 1.) than the base FD. In like manner 
 
 it may be shewn that FD is greater than 
 
 CD. Therefore DA is the greatest ; and 
 
 DE greater than DF, and DF than DC. 
 
 And because MK, KD are greater (20. 
 1 ) than MD, and MK is equal to MG, the 
 remainder KD is greater (5. Ax.) than the 
 remainder GD, that is, GD is less than 
 KD : And because MK, DK are drawn to 
 the point K within the triangle MLD from 
 M, D, the extremities of its side MD ; MK, 
 K I ) are less (21.1.) than ML, LD, whereof 
 MK is equal to ML ; therefore the remain- 
 der DK is less than the remainder DL : 
 In like manner, it may be shewn that DL 
 is less than DH : Therefore DG is the 
 least, and DK less than DL, and DL 
 than DH. 
 
 Also there can be drawn only two equal straight lines from the point D 
 to the circumference, one upon each side of the least; at the point M, in 
 the straight line MD, make the angle DMB equal to the angle DMK, and 
 join DB ; and because in the triangles KMD, BMD, the side KM is equal 
 to the side BM, and MD common to both, and also the angle KMD equal 
 to the angle BMD, the base DK is equal (4. l.)to the base DB. But, 
 besides DB, no straight line can be drawn from D to the circumference, equal 
 to DK ; for, if there can, let it be DN ; then, because DN is equal to DK, 
 and DK equal to DB, DB is equal to DN ; that is, the line nearer to DG, 
 the least, equal to the more remote, which has been shewn to be impossible. 
 
 PROP. IX. THEOR. 
 
 If a point be taken within a circle, from which there fall more than two equal 
 straight lines upon the circumference, that point is the centre of the circle. 
 
 Let the point D be taken within the circle ABC, from which there fall 
 on the circumference more than two equal straight lines, viz. DA, DB, DC, 
 the point D is the centre of the circle. 
 
 For, if not, let E be the centre, join DE, and produce it to the circum- 
 ference in F, G ; then FG is a diameter of 
 the circle ABC : And because in FG, the di- 
 ameter of the circle ABC, there is taken the 
 point D which is not the centre, DG is the 
 greatest line from it to the circumference, and 
 DC greater (7. 3.) than DB, and DB than 
 DA ; but they are likewise equal, which is 
 /mpossible : Therefore E is not the centre of 
 iho circle ABC : In like manner it may be 
 demonstrated, that no other point but D is the 
 centre : D therefore is the centre.
 
 68 
 
 ELEMENTS 
 
 PROP. X. THEOR. 
 
 One circle cannot cut another in more than two points. 
 
 if it be possible, let the circumference FAB cut the circumference DEF 
 in more than two points, viz. in B, G, F ; take the centre K of the circ'e 
 ABC, and join KB, KG, KF ; and because within the circle DEF there 
 is taken the point K, from which more than two 
 equal straight lines, viz. KB, KG, KF fc fall on 
 the circumference DEF, the point K is (9. 3.) 
 the centre of the circle DEF ; but K is also the 
 centre of the circle ABC ; therefore the same 
 point is the centre of two circles that cut one 
 aiiother, which is impossible (5. 3.). There- 
 fore one circumference of a circle cannot cut 
 another in more than two points. 
 
 PROP. XI. THEOR. 
 
 If two circles touch each other internally, the straight line which joins then 
 centres being produced, will pass through the point of contact. 
 
 Let the two circles ABC, ADE, touch each other internally in the point 
 A, and let F be the centre of the circle ABC, and G the centre of the cir- 
 cle ADE ; the straight line which joins the cen- 
 tres F, G, being produced, passes through the 
 point A. 
 
 For, if not, let it fall otherwise, if possible, as 
 FGDH, and join AF, AG : And because AG, 
 GF are greater (20. 1.) than FA, that is, than 
 FH, for FA is equal to FH, being radii of the 
 same circle ; take away the common part FG, 
 and the remainder AG is greater than the re- 
 mainder GH. But AG is equal to GD, there- 
 fore GD is greater than GH ; and it is also less, 
 which is impossible. Therefore the straight line 
 
 which joins the points F and G cannot fall otherwise than on the point 
 A ; that is, it must pass through A. 
 
 Cor. 1. If two circles touch each other internally, the distance be- 
 tween their centre must be equal to the difference of their radii : for the 
 circumferences pass through the same point in the line joining the centres. 
 
 Cor. 2. And, conversely, if the distance between the centres be equal 
 »o the difference of the radii, the two circles will touch each other inter- 
 nally.
 
 OF GEOMETRY. BOOK III. 
 
 PROP. XII. THEOR. 
 
 If two circles touch each other externally, the straight line which joins then 
 centres will pass through the point of contact. 
 
 Let the two circles ABC, ADE, touch each other externally in the point 
 A ; and let F be the centre of the circle ABC, and G the centre of ADE ; 
 the straight line which joins the points F, G shall pass through the point 
 of contact. 
 
 For, if not, let it pass otherwise, if possible, FCDG, and join FA, AG : 
 and because F is the centre of the circle ABC, AF is equal to FC : Also 
 because G is the centre of the 
 circle, ADE, AG is equal to 
 GD. Therefore FA, AG are 
 equal to FC, DG ; wherefore 
 the whole FG is greater than 
 FA, AG ; but it is also less 
 (20. IX which is impossible : 
 Therefore the straight line 
 which joins the points F, G 
 cannot pass otherwise than 
 through the point of contact A ; that is, it passes through A. 
 
 Cor. Hence, if two circles touch each other externally, the distance 
 between their centres will be equal to the sum of their radii. 
 
 And, conversely, if the distance between the centres be equal to the sum 
 of the radii, the two circles will touch each other externally. 
 
 PROP. XIII. THEOR. 
 
 One circle cannot touch another in more points than one, whether it louche: 
 it on the inside or outside. 
 
 For, if it be possible, let the circle EBF touch the circle AbC in more 
 points than one, and first on the inside, in the points B, D ; join BD, and 
 draw (10. 11. 1.) GH, bisecting BD at right angles : Therefore because 
 'he points B, D are in the circumference of each of the circles, the straight 
 
 line BD faL'a within each (2. 3.) of them : and therefore their centres are 
 (Cor 1. 3) in the straight line GH which bisects BD at right angles:
 
 70 
 
 ELEMENTS 
 
 therefoiv Gil passes through the point of contact (11. 3 ), but it does 
 not pass through it, because the points B, D are without the straight line 
 GH, which is absurd: therefore one circle cannot touch another in thp 
 inside in more points than one. 
 
 Nor can two circles touch one another on the outside in more than one 
 point : For, if it be possible, let the circle ACK 
 touch the circle ABC in the points A, C, and join 
 AC : therefore, because the two points A, C are 
 in the circumference of the circle ACK, the straight 
 line AC which joins them shall fall within the 
 circle ACK : And the circle ACK is without the 
 circle_ ABC : and therefore the straight line AC is 
 also without ABC ; but, because the points A, C 
 are in the circumference of the circle ABC, the 
 straight line AC must be within (2. 3.) the same 
 circle, which is absurd : therefore a circle cannot 
 touch another on the outside in more than one 
 point : and it has been shewn, that a circle cannot 
 touch another on the inside in more than one point. 
 
 PROP. XIV. THEOR. 
 
 Equal straight lines in a circle are equally distant from the centre ; aiA thost 
 which are equally distant from the centre, are equal to one another. 
 
 Let the straight lines AB, CD, in the circle ABDC, be equal to one 
 another : they are equally distant from the centre. 
 
 Take E the centre of the circle ABDC, and from it draw EF, EG, per- 
 pendiculars to AB, CD ; join AE and EC. Then, because the straight 
 line EF passing through the centre, cuts the 
 straight line AB, which does not pass through 
 the centre at right angles, it also bisects (3. 
 3.) it : Wherefore AF is equal to FB, and 
 x\B double of AF. For the same reason, 
 CD is double of CG : But AB is equal to 
 CD ; therefore AF is equal to CG : And be- 
 cause AE is equal toEC, the square of AE is 
 equal to the square of EC : Now the squares 
 of AF, FE are equal (47. 1.) to the square 
 
 of AE, because the angle AFE is a right angle ; and, for the like reason, 
 the squares of EG, GC are equal to the square of EC : therefore the 
 squares of AF, FE are equal to the squares of CG, GE, of which the 
 square of AF is equal to the square of CG, because AF is equal to CG ; 
 therefore the remaining square of FE is equal to the remaining square of 
 EG, and the straight line EF is therefore equal to EG : But straight lines 
 in a circle are said to be equally distant from the centre when the perpen 
 diculars drawn to them from the centre are equal (3. Def. 3.) : therefore 
 AB, CD are equally distant from the centre. 
 
 Next, if the straight lines AB, CD be equally distant from the centre 
 'hat is, if FE be equal to EG, AB is equal to CD. For, the same con
 
 ler 
 er to 
 any sti 
 BC grec 
 
 From ti 
 to BC, FG, . 
 AE is equal U 
 toEB, EC: Bu 
 than BC ; wheretc 
 BC. 
 
 And, because BC is i. 
 3.) than EK ; But, as was 
 of BH, and FG double of FK, 
 the squares of EK, KF, of which tn'O :. 
 of EK, because EH is less than EK ; there*., 
 than the square of FK, and the straight line Bn fa . .^.. 
 therefore BC is greater than FG. *** 
 
 Next, let BC be greater than FG ; BC is nearer to the centre than FG : 
 that is, the same construction being made, EH is less than EK ; because 
 BC is greater than FG, BH likewise is greater than KF : but the squares 
 of BH, HE are equal to the squares of FK, KE, of which the square of 
 BH is greater than the square of FK, because BH is greater than FK ; 
 therefore the square of EH is less than the square of EK, and the strai«ht 
 line EH less than EK. 
 
 Cor. The shorter the chord is, the farther it is from the centre ; and, 
 conversely, the farther the chord is from the centre, the shorter it is. 
 
 PROP. XVI. THEOR. 
 
 The straight line drawn at right angles to the diameter of a circle, from the 
 extremity of it, falls without the circle ; and no straight line can be drawn 
 between that straight line and the circumference, from the extremity of th* 
 diameter, so as not to cut the circle. 
 
 Let ABC be a circle, the centre of which is D, and the diameter AB 
 »nd let AE be drawn from A perpendicular to AB, AE shall fall without 
 die circle
 
 ai 
 
 o.). 
 
 .at one straight line which touches the 
 
 , - perpendicular at the extremity of a diameter is a tan- 
 ^». iu me circle ; and, conversely, a tangent to a circle is perpendicular 
 to the diameter drawn from the point of contact. 
 
 Cor. 3. It follows, likewise, that tangents at each extremity of the 
 diameter are parallel (Cor. 28. B. 1.); and, conversely, parallel tangents 
 are both perpendicular to the same diameter, and have their points of con- 
 tact at its extremities. 
 
 PROP. XVII. PROB. 
 
 To draw a straight line from a given point either without or in the circum- 
 ference, which shall touch a given circle. 
 
 First, let A be a given point without the given circle BCD ; it is re- 
 quired to draw a straight line from A which shall touch the circle. 
 
 Find (1.3.) the centre E of the circle, and join AE ; and from the cen- 
 tre E, at the distance EA, describe the circle AFG ; from the point D 
 draw (11. 1.) DF at right angles to E A, join EBF, and draw AB. AB 
 touches the circle BCD. 
 
 Because E is the centre of the circles BCD, AFG, EA is equal to 
 RF, and ED to EB ; therefore the two sidss AE EB are equal to t\f
 
 OF GEOMETRY. BOOK III. 
 
 7j 
 
 two FE, ED, and they contain the angle at E common to the twj trian- 
 gles AEB, FED; therefore the base DF 
 is equal to the base AB, and the triangle 
 FED to the triangle AEB, and the other 
 angles to the other angles (4. 1.); there- Cy 
 fore the angle EBA is equal to the angle 
 EDF; but EDF is a right angle, where- 
 fore EBA is a right angle; and EB is a 
 line drawn from the centre : but a straight 
 line drawn from the extremity of a diame- 
 ter, at right angles to it, touches the circle 
 (1 Cor. 16.3.) : therefore AB touches the 
 circle ; and is drawn from the given point A. 
 
 But if the given point be in the circumference of the circle, as the poin' 
 D, draw DE to the centre E, and DF at right angles to DE ; DF touchei 
 the circle (1 Cor. 16. 3.) 
 
 SCHOLIUM. 
 
 When the point A lies without the circle, there will evidently be always 
 two equal tangents passing through the point A. For, by producing the 
 tangent FD till it meets the circumference AG, and joining E and the poir 
 of intersection, and also A and the point where this last line will intersex 
 the circumference DC ; there will be formed a right angled triangle equal 
 to ABE (46. 1.). 
 
 PROP. XVIII. THEOR. 
 
 If a straight line touch a circle, the straight line drawn from the centre to 
 the point of contact, is perpendicular to the line touching the circle. 
 
 Let the straight line DE touch the circle ABC in the point C ; take 
 the centre F, and draw the straight line FC : FC is perpendicular to DE. 
 
 For, if it be not, from the point F draw FBG perpendicular to DE ; and 
 because FGC is a right angle, GCF must 
 be (17. 1.) an acute angle ; and to the great- 
 er angle the greater side (19. 1.) is oppo- 
 site ; therefore FC is greater than FG ; 
 but FC is equal to FB ; therefore FB is 
 greater than FG, the less than the greater, 
 which is impossible ; wherefore FG is not 
 perpendicular to DE : in the same manner 
 it may be shewn, that no other line but FC 
 can be perpendicular to DE ; FC is there- 
 tore perpendicular to DE. 
 
 10
 
 74 
 
 ELEMENTS 
 
 PROP. XIX. THEOR. 
 
 If a straight lint touch a circle, and from the point of contact a straight line 
 be drawn at right angles to the touching line, the centre of the circle is in 
 that line. 
 
 Let the straight line DE touch the circle ABC, in C, and from C let 
 CA be drawn at right angles to DE ; the centre of the circle is in CA. 
 
 For, if not, let F be the centre, if possible, 
 and join CF. Because DE touches the cir- 
 cle ABC, and FC is drawn from the centre 
 to the point of contact, FC is perpendicular 
 (18. 3 ) to DE ; therefore FCE is a right 
 angle ; but ACE is also a right angle ; 
 therefore the angle FCE is equal to the an- 
 gle ACE, the less to the greater, which is 
 impossible ; Wherefore F is not the centre 
 of the circle ABC : in the same manner it 
 may be shewn, that no other point which is 
 not in CA, is the centre ; that is, the centre 
 is in CA. 
 
 PROP. XX. THEOR 
 
 The angle at the centre of a circle is double of the angle at the circumjer 
 ence, upon the same base, that is, upon the same part of the circumfe* 
 ence. 
 
 Let ABC be a circle, and BDC an angle at the centre, and BAC an 
 angle at the circumference which have the same circumference BC for 
 the base ; the angle BDC is double of the angle BAC. 
 
 First, let D, the centre of the circle, be within the angle BAC, arid join 
 AD, and produce it to E : because DA is equal 
 to DB, the angle DAB is equal (5. 1.) to the 
 angle DBA : therefore the angles DAB, DBA 
 together are double of the angle DAB ; but the 
 angle BDE is equal (32. 1.) to the angles DAB, 
 DBA ; therefore also the angle BDE is double 
 of the angle DAB ; for the same reason, the an- 
 gle EDC is double of the angle DAC : there- 
 fore the whole angle BDC is double of the whole 
 angle BAC. 
 
 Again, let D, the centre of the circle, be 
 without the angle BAC ; and join AD and pro- 
 duce it to E. It may be demonstrated, as in 
 the first case, that the angle EDC is double 
 of the angle DAC, and that EDB, a part of 
 the first, is double of DAB, a part of the 
 other ; therefore the remaining angle BDC is 
 double of the remaining angle BAC.
 
 OF GEOMETRY. BOOK III. 
 
 73 
 
 PROP. XXI. THEOR. 
 Fhe anglt s in the same segment of a circle are equal to one another 
 
 Let ABCD be a circle, and BAD, BED 
 angles in the same segment BAED : the an- 
 gles BAD, BED are equal to one another. 
 
 Take F the centre of the circle ABCD : 
 And, first, let the segment BAED be greatei 
 than a semicircle, and join BF, FD : and be- 
 cause the angle BFD is at the centre, and the 
 angle BAD at the circumference, both having 
 the same part of the circumference, viz. BCD, 
 for their base ; therefore the angle BFD is 
 double (20. 3.) of the angle BAD: for the 
 same reason, the angle BFD is double of the 
 angle BED : therefore the angle BAD is equal 
 to the angle BED. 
 
 But, if the segment BAED be not greater 
 than a semicircle, let BAD, BED be angles 
 in it , these also are equal to one another. 
 Draw AF to the centre, and produce to C, and 
 join CE : therefore the segment BADC is 
 greater than a semicircle ; and the angles in 
 it, BAC, BEC are equal, by the first case : 
 for the same reason, because CBED is great- 
 er than a semicircle, the angles CAD, CED 
 are equal ; therefore the whole angle BAD is 
 equal to the whole angle BED. 
 
 PROP. XXII. THEOR. 
 
 The opposite angles of any quadrilateral figure described in a circle, ate 
 together equal to two right angles. 
 
 Let ABCD be a quadrilateral figure in the circle ABCD ; any two of 
 its opposite angles are together equal to two right angles. 
 
 Join AC, BD. The angle CAB is equal (21. 3.) to the anglo 
 CDB, because they are in the same segment 
 BADC, and the angle ACB is equal to the an- 
 gle ADB, because they are in the same seg- 
 ment ADCB ; therefore the whole angle ADC 
 is equal to the angles CAB, ACB : to each of 
 these equals add the angle ABC ; and the an- 
 gles ABC, ADC, are equal to the angles ABC, 
 CAB, BCA. But ABC, CAB, BCA are equal 
 to two right angles (32. 1.) ; therefore also the 
 ■ngles ABC, ADC are equal to two right an- 
 gles ; in the same manner, the angles BAD, 
 DCB may be shewn to be equal to two right angle*.
 
 76 ELEMENTS 
 
 Cor. 1. If an} side of a quadrilateral be produced, the exterior angle 
 will be equal to the interior opposite angle. 
 
 Cor. 2. It follows, likewise, that a quadrilateral, of which the op- 
 posite angles are not equal to two right angles, cannot be inscribed in a 
 circle. 
 
 , PROP. XXIII. THEOR. 
 
 Upon the same straight line, and upon the same side of it, there cannot be 
 two similar segments of circles, not coinciding with one another. 
 
 If it be possible, let the two similar segments of circles, viz. ACB, ADB, 
 be upon the same side of the same straight line AB, not coinciding with 
 one another ; then, because the circles ACB, ADB, cut one another in 
 the two points A, B, they cannot cut one another in any other point (10. 
 3.) : one of the segments must therefore fall 
 within the other: let ACB fall within ADB, 
 draw the straight line BCD, and join CA, DA : 
 avid because the segment ACB is similar to the 
 segment ADB, and similar segments of circles 
 contain (9. def. 3.) equal angles, the angle 
 ACB is equal to the angle ADB, the exterior 
 to the interior, which is impossible (16. 1.). 
 
 PROP. XXIV. THEOR. 
 Similar segments of circles upon equal straight li?ies are equal to one another. 
 
 Let AEB, CFD be similar segments of circles upon the equal straight 
 lines AB, CD ; the segment AEB is equal to the segment CFD. 
 
 For, if the segment AEB be applied to the segment CFD, so as the 
 point A be on C, and the —, t-, 
 
 straight line AB upon CD, 
 the point B shall coincide 
 with the point D, because 
 AB is equal to CD : there- 
 fore the straight line AB A. 33 C X) 
 coinciding with CD, the segment AEB must (23. 3.) coincide with &e 
 segment CFD, and therefore is equal to it. 
 
 PROP. XXV. PROB. 
 
 A segment of a circle being given, to describe the circle of which it is the 
 
 segment. 
 
 Let ABC be the given segment of a circle ; it is required to describe 
 the circle of which it is the segment. 
 
 Bisect (10. 1.) AC in D, and from the point D draw (11. 1.) DB at 
 right angles to AC, and join AB : First, let the angles ABD, BAD be 
 equal to one another; then the straight line BD is equal (6. 1.) to DA, 
 and therefore to DC ; and because the three straight lines DA, DB DC,
 
 OF GEOMETRY. BOOK 111. 
 
 77 
 
 are all equal ; D is the centre of the circle (9. 3.) ; from the cemre D, at 
 the distance of any of the three DA, DB, DC, describe a circle ; this shall 
 pass through the other points ; and the circle of which ABC is a segment 
 
 is described : and because the centre D is in AC, the segment ABC is 
 semicircle. Next, let the angles ABD, BAD be unequal ; at the point A, it 
 the straight line AB, make (23. 1.) the angle BAE equal to the angle ABD 
 and produce BD, if necessary, to E, and join EC : and because the angle 
 ABE is equal to the angle BAE, the straight line BE is equal (G. 1.) to 
 EA : and because AD is equal to DC, and DE common to the triangles 
 ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each 
 to each ; and the angle ADE is equal to the angle CDE, for each of them 
 is a right angle ; therefore the base AE is equal (4. 1.) to the base EC : 
 but AE was shewn to be equal to EB, wherefore also BE is equal to EC : 
 and the three straight lines AE, EB, EC are therefore equal to one another; 
 wherefore (9. 3.) E is the centre of the circle. From the centre E, at 
 the distance of any of the three AE, EB, EC, describe a circle, this shall 
 pass through the other points ; and the circle of which ABC is a segment 
 is described : also, it is evident, that if the angle ABD be greater than the 
 angleBAD, the centre E falls without the segment ABC, which therefore 
 is less than a semicircle ; but if the angle ABD be less than BAD, the cen- 
 tre E falls within the segment ABC, which is therefore greater than a semi- 
 circle : Wherefore, a segment of a circle being given, the circle is de- 
 scribed of which it is a segment. 
 
 PROP. XXVI. THEOR. 
 
 In equal circles, equal angles stand upon equal arcs, whether they be at the 
 centres or circumferences. 
 
 r,et ABC, DEF be equal circles, and the equal angles BGC, EHF at 
 their centres, andBAC, EDF at their circumferences : the arc BKC i» 
 equal to the arc ELF.
 
 78 ELEMENTS 
 
 Join BC, EF ; and because the circles ABC, DEF are equal, the straigh 
 lines drawn from their centres are equal : therefore the two sides BG, 
 GC, are equal to the two EH, HF ; and the angle at G is equal to the an- 
 gle at H ; therefore the base BC is equal (4. 1.) to the base EF : and be- 
 cause the angle at A is equal to the angle at D, the segment BAC is similar 
 (9. def. 3.) to the segment EDF ; and they are upon equal straight lines 
 BC, EF ; but similar segments of circles upon equal straight lines are 
 equal (24. 3.) to one another, therefore the segment BAC is equal to ihe 
 segment EDF : but the whole circle ABC is equal to the whole DEF ; 
 therefore the remaining segment BKC is equal to the remaining segment 
 ELF, and the arc BKC to the arc ELF. 
 
 PROP. XXVII. THEOR. 
 
 In equal circles, the angles which stand upon equal arcs are equal to one 
 another, whether they be at the centres or circumferences. 
 
 Let the angles BGC, EHF at the centres, and BAC, EDF at the cir- 
 cumferences of the equal circles ABC, DEF stand upon the equal arcs 
 BC, EF : the angle BGC is equal to the angle EHF, and the angle BAG 
 to the angle EDF. 
 
 If the angle BGC be equal to the angle EHF, it is manifest (20. 3.) 
 that the angle BAC is also equal to EDF. But, if not, one of them is the 
 greater : let BGC be the greater, and at the point G, in the straight line 
 BG, make the angle (23. 1.) BGK equal to the angle EHF. And because 
 equal angles stand upon equal arcs (26. 3.), when they are at the centre, 
 
 the arc BK is equal to the arc EF : but EF is equal to BC ; therefore 
 also BK is equal to BC, the less to the greater, which is impossible. There- 
 fore the angle BGC is not unequal to the angle EHF ; that is, it is equal 
 to it : and the angle at A is half the angle BGC, and the angle at D half 
 of the angle EHF ; therefore the angle at A is equal to the angle at D. 
 
 PROP. XXVIII. THEOR. 
 
 In equal circles, equal straight lines cut off equal arcs, the greater equal to 
 the greater, and the less to the less. 
 
 Let ABC, DEF be equal circles, and BC, EF equal straight lines in 
 them, which cut off the two greater arcs BAC EDF, and the two lesa
 
 OF GEOxMETRY. BOOK HI. 
 
 7<« 
 
 BGC, EHF : the greater BAC is equal to the greater EDF, and the lesa 
 BGC to the less EHF. 
 
 Take (1. 3.) K, L, the centres of the circles, and join BK, KC, EL, 
 LF ; and because the circles are equal, the straight lines from their centres 
 
 *re equal ; therefore BK, KC are equal to EL, LF ; but the base BC is 
 also equal to the base EF ; therefore the angle BKC is equal (8. 1.) to the 
 angle ELF : and equal angles stand upon equal (26. 3.) arcs, when they 
 are at the centres; therefore the arc BGC is equal to the arc EHF. 
 But the whole circle ABC is equal to the whole EDF ; the remaining part, 
 therefore, of the circumference viz. BAC, is equal to the remaining part 
 EDF. 
 
 PROP. XXIX. THEOR. 
 
 In equal circles equal arcs are subtended by equal straight lines. 
 
 Let ABC, DEF be equal circles, and let the arcs BGC, EHF also be 
 equal ; and join BC, EF : the straight line BC is equal to the straight line 
 EF. 
 
 Take (1. 3.) K, L the centres of the circles, and join BK, KC, EL, LF : 
 and because the arc BGC is equal to the arc EHF, the angle BKC is 
 equal (27. 3.) to the angle ELF : also because the circles ABC, DEF are 
 eoual their radii are equal : therefore BK, KC are equal to EL, LF : and 
 
 C E 
 
 they contain equal angles ; therefore the base BC is equal (4. 1.) to th* 
 bas* EF
 
 80 
 
 ELEMENTS 
 
 PROP. XXX. THEOR. 
 To bisect a given arc, that is, to divide it into two equal parts. 
 
 Let ADB be the given arc ; it is required to bisect it. 
 
 Join AB, and bisect (10. 1.) it in C ; from the point C draw CD at right 
 angles to AB, and join AD, DB : the arc ADB is bisected in the point D. 
 
 Because AC is equal to CB, and CD common to the triangle A CD, 
 BCD, the two sides AC, CD are equal to the J) 
 
 two BC, CD ; and the angle ACD is equal to 
 the angle BCD, because each of them is a 
 right angle : therefore the base AD is equal 
 (4. 1.) to the base BD. But equal straight 
 lines cut off equal arcs, (28. 3.) the greater xV. C 23 
 
 equal to the greater, and the less to the less ; and AD, DB are each of 
 them less than a semicircle, because DC passes through the centre (Cor. 
 1. 3.) ; wherefore the arc AD is equal to the arc DB : and therefore the 
 given arc ADB is bisected in D. 
 
 SCHOLIUM. 
 
 By the same construction, each of the halves AD, DB may be divided 
 into two equal parts ; and thus, by successive subdivisions, a given arc 
 may be divided into four, eight, sixteen, &c. equal parts. 
 
 PROP. XXXI. THEOR. 
 
 In a circle, the angle in a semicircle is a right angle ; but the angle in a seg- 
 ment greater than a semicircle is less than a right angle ; and the angle in 
 a segment less than a semicircle is greater than a right angle. 
 
 Let A BCD be a circle, of which the diameter is BC, and centre E ; 
 draw CA dividing the circle into the segments ABC, ADC, and join BA, 
 AD, DC ; the angle in the semicircle BAC is a right angle ; and the an- 
 gle in the segment ABC, which is greater than a semicircle, is less than a 
 right angle ; and the angle in the segment ADC, which is less than a semi- 
 circle, is greater than a right angle. 
 
 Join AE, and produce BA to F ; and because BE is equal to EA, the 
 angle EAB is equal (5. 1.) to EBA : also 
 because AE is equal to EC, the angle EAC 
 is equal to EC A ; wherefore the whole an- 
 gle BAC is equal to the two angles ABC, 
 ACB. But FAC, the exterior angle of the 
 triangle ABC, is also equal (32. 1.) to the 
 two angles ABC, ACB ; therefore the an- 
 gle BAC is equal to the angle FAC, and 
 each of them is therefore a right angle (7. 
 def. 1 .) ; wherefore the angle BAC in a semi- 
 circle is a right angle.
 
 OF GEOMETRY. BOOK III. 81 
 
 And because the two angles ABC, BAG of the triangle ABC are to- 
 gether less (17. 1.) than two right angles, and BAC is a right angle, ABC 
 must be less than a right angle ; and therefore the angle in a segment 
 ABC, greater than a semicircle, is less than a right angle. 
 
 Also because A BCD is a quadrilateral figure in a circle, any two of iui 
 opposite angles are equal (22. 3.) to two right angles ; therefore the angles 
 ABC. ADC are equal to two right angles ; and ABC is less than a right 
 angle ; wherefore the other ADC is greater than a right angle. 
 
 Cor. From this it is manifest, that if one angle of a triangle be equal to 
 the other two, it is a right angle, because the angle adjacent to it is equal 
 to the same two ; and when the adjacent angles are equal, they are right 
 angles. 
 
 PROP. XXXII. THEOR. 
 
 // a straight line touch a circle, and from the poi?it of contact a straight 
 line be drawn cutting the circle, the angles made by this line with the line 
 which touches the circle, shall be equal to the angles in the alternate seg- 
 ments of the circle. 
 
 Let the straight line EF touch the circle ABCD in B, and from the 
 point B let the straight line BD be drawn cutting the circle : the angles 
 which BD makes with the touching line EF shall be equal to the angles 
 in the alternate segments of the circle : that is, the angle FBD is equal to 
 the angle which is in the segment DAB, and the angle DBE to the angle 
 in the segment BCD. 
 
 From the point B draw (11. 1.) BA at right angles to EF, and take any 
 point C in the arc BD, and join AD, DC, CB ; and because the st/aight 
 line EF touches the circle ABCD in the point B, and BA is drawn at right 
 angles to the touching line, from the point of contact B, the centre of the 
 circle is (19. 3.) in BA ; therefore the an- 
 gle ADB in a semicircle, is a right an- 
 gle (31. 3.), and consquently the other two 
 angles, BAD, ABD, are equal (32, 1.) to 
 a right angle ; but A BF is likewise a right 
 angle ; therefore the angle ABF is equal 
 to the angles BAD, ABD: take from 
 these equals the common angle ABD, 
 and there will remain the angle DBF 
 equal to the angle BAD, which is in the 
 alternate segment of the circle. And be- 
 cause ABCD is a quadrilateral figure in 
 a circle, the opposite angles BAD, BCD are equal (22. 3.) to two riglr 
 angles ; therefore the angles DBF, DBE, being likewise equal (13 1.) t<* 
 two right angles, are equal to the angles BAD, BCD ; and DBF has been 
 proved equal to BAD : therefore the remaining angle DBE is equal to the 
 angle BCD in the alternate segment of the circle. 
 
 ' 11
 
 ELEMENTS 
 
 PROP. XXXIII. PROB. 
 
 Upon a given straight line to describe a segment of a circle, containing an 
 angle equal to a given rectilineal angle- 
 
 Let AB be the given straight line, and the angle at C the given recti- 
 lineal angle ; it is required to describe upon the given straight line AB a 
 segment of a circle, containing an angle equal to the angle C. 
 
 First, let the angle at C be a right angle ; bisect (10. 1.) AB in F„and 
 from the centre F, at the distance FB, 
 describe the semicircle AHB ; the an- 
 gle AHB being in a semicircle is (31. 
 3.) equal to the right angle at C. 
 
 But if the angle C be not a right an- 
 gle at the point A, in the straight line 
 AB, make (23. 1.) the angle BAD equal 
 
 to the angle C, and from the point A draw (11. 1.) AE at right angles to 
 AD ; bisect (10. 1.) AB in F, and 
 from F draw (11. 1.) FG at right 
 angles to AB, and join GB : then 
 because AF is equal to FB, and 
 FG common to the triangles AFG, 
 BFG, the two sides AF, FG are 
 equal to the two BF, FG ; but the 
 angle AFG is also equal to the 
 angle BFG ; therefore the base AG 
 is equal (4.1.) to the base GB ; and 
 the circle described from the centre 
 G, at the distance GA, shall pass 
 through the point B ; let this be the circle AHB: and because from the 
 point A the extremity of the diameter AE, AD is drawn at right angles to 
 AE, therefore AD (Cor. 1.16. 3.) touches 
 the circle ; and because AB, drawn from 
 the point, of contact A, cuts the circle, 
 the angle DAB is equal to the angle in 
 the alternate segment AHB (32. 3.) ; 
 out the angle DAB is equal to the angle 
 C, therefore also the angle C is equal to 
 the angle in the segment AHB : Where- 
 fore, upon the given straight line AB 
 the segment AHB of a circle is describ- 
 ed which contains an angle equal to the given angle at C
 
 OF GEOMETRY. BOOK III. 
 
 83 
 
 PROP. XXXIV. PROB. 
 
 To cut off a segment from a given circle which shall contain an angle equal 
 to a given rectilineal angle. 
 
 Let ABC be the given circle, and D the given rectilineal angle ; it is 
 required to cut off a segment from the circle ABC that shall contain an 
 angle equal to the angle D. 
 
 Draw (17. 3.) the straight line EF touching the circle ABC in the point 
 B and at the point B, in the straight 
 line BF make (23. I.) the angle FBC 
 equal to the angle D ; therefore, be- 
 cause the straight line EF touches 
 the circle ABC, and BC is drawn 
 from the point of contact B, the an- 
 gle FBC is equal (32. 3.) to the an- 
 gle in the alternate segment BAC ; 
 hut the angle FBC is equal to the an- 
 gle D : therefore the angle in the 
 segment BAC is equal to the angle 
 D : wherefore the segment BAC is cut off from the given circle ABC 
 containing an angle equal to the given angle D. 
 
 PROP. XXXV. THEOR. 
 
 If two straight lines within a circle cut one another, the rectangle contained 
 by the segments of one of them is equal to the rectangle contained by the 
 segments of the other. 
 
 Let the two straight lines AC, BD, within the circle ABCD, cut one 
 another in the point E ; the rectangle contained by AE, EC is equal tc 
 the rectangle contained by BE, ED. 
 
 If AC, BD pass each of them through the cen- 
 tre, so that E is the centre, it is evident that AE, 
 EC, BE, ED, being all equal, the rectangle AE. 
 EC is likewise equal to the rectangle BE. ED. 
 
 But let one of them BD pass through the cen- J$\ 
 tre, and cut the other AC, which does not pass 
 through the centre, at right angles in the point E ; 
 then, if BD be bisected in F, F is the centre of 
 the circle ABCD ; join AF : and because BD, which passes through the 
 centre, cuts the straight line AC, which does not 
 pass through the centre at right angles, in E, AE, 
 EC are equal (3. 3.) to one another; and because 
 the straight line BD is cut into two equal parts 
 in the point F, and into two unequal in the point 
 E, BE.ED (5. 2.) + EF 2 = FB 2 = AF 2 . But 
 AF 2 = AE 2 + (47. I.) EF 2 , therefore BE.ED + 
 EF 2 , -= AE 2 + EF 2 , and taking EF 2 from each, 
 BE.EP=AE 2 = AE.EC. 
 
 Next, let BD, which passes through the centre, 
 cut the other AC, which does not pass through
 
 fr% ELEMENTS 
 
 rtie ceiurc, in E, but not at right angles ; then, as before, if BD be bisect 
 ed in F, F is the centre of the circle. Join AF, 
 ar.d from F draw (12. 1.) FG perpendicular to 
 AC ; therefore AG is equal (3. 3.) to GC ; where- 
 fore AE.EC + (5. 2.) EG 2 = AG 2 , and adding 
 GF 2 to both, AE.EC + EG 2 + GF 2 = AG 2 +GF 2 . 
 Now EG 2 +GF 2 = EF 2 , and AG 2 +GF 2 =AF 2 ; 
 therefore AE.EC + EF 2 =AF 2 =FB 2 . But FB 2 
 = BE.ED + (5 2.) EF 2 , therefore AE.EC + EF a 
 =BE.ED + EF 2 , and taking EF 2 from both, AE. 
 EC = BE.ED. 
 
 Lastly, let neither of the straight lines AC, 
 BD pass through the centre : take the centre F, 
 and through E, the intersection of the straight 
 lines AC, DB, draw the diameter GEFH : and 
 because, as has been shown, AE.EC = GE.EH, 
 and BE.ED = GE.EH; therefore AE.EC = BE. 
 ED. 
 
 B G 
 
 PROP. XXXVI. THEOR. 
 
 If from any point without a circle two straight lines be drawn, one of which 
 cuts the circle, and the other touches it ; the rectangle contained by the whole 
 line which cuts the circle, and the part of it without the circle, is equal to the 
 square of the line which touches it. 
 
 Let D be any point without the circle ABC, and DCA, DB two straight 
 lines drawn from it, of which DCA cuts the circle, and DB touches it • 
 the rectangle AD. DC is equal to the square of DB. 
 
 Either DCA passes through the centre, or it 
 does not ; first, let it pass through the centre E, 
 and join EB ; therefore the angle EBD is a 
 right angle (18. 3.) : and because the straight 
 line AC is bisected in E, and produced to the 
 point D, AD.DC + EC 2 =ED 2 (6. 2.). But 
 EC = EB, therefore AD.DC + EB 2 = ED 2 . 
 Now ED 2 = (47. 1.) EB 2 + BD 2 , because EBD 
 is a right angle ; therefore AD.DC + EB 2 = 
 EB 2 -I- BD 2 , and taking EB 2 from each, AD.DC 
 =BD 2 . 
 
 But, if DCA does not pass through the cen- 
 tre of the circle ABC, take (1.3.) the centre E, 
 and draw EF perpendicular (12. 1.) to AC, and 
 ioin EB, EC, ED ; and because the straight 
 ine EF, which passes through the centre, cuts
 
 OF GEOMETRY. BOOK III. 
 
 8d 
 
 the straight line AC, which does not pass 
 through the centre, at right angles, it likewise 
 bisects it (J. 3.) ; therefore AF is equal to FC ; 
 and because the straight line AC is bisected in 
 F, and produced to D (6. 2.), AD.DC + FC 2 = 
 FD 2 ; add FE" to both, then AD.DC + FC 2 -*- 
 FE 2 r=FD 3 +FE 2 . But (47. 1.) EC 2 = FC 2 + 
 FE 2 , and ED*-FD 2 +FE 2 , because DFE is 
 aright angle; therefore AD.DC + EC 2 =ED 2 . 
 Now, because EI3D is a right angle, ED 2 = 
 EB 2 +BD 2 =EC 2 +BD 2 , and therefore, AD. 
 DC + EC 2 =EC 2 +BD 2 , and AD.DC = BD 2 . 
 
 Cor. 1. If from any point without a circle, 
 there be drawn two straight lines cutting it, as 
 AB, AC, the rectangles contained by the whole 
 lines and the parts of them without the circle, 
 are equal to one another, viz. BA.AE = CA. 
 AF ; for each of these rectangles is equal to 
 the square of the straight line AD, which touch- 
 es the circle. 
 
 Cor. 2. It follows, moreover, that two tan- 
 gents drawn from the same point are equal. 
 
 Cor. 3. And since a radius drawn to the 
 point of contact is perpendicular to the tangent, 
 it follows that the angle included by two tangents, 
 drawn from the same point, is bisected by a line 
 drawn from the centre of the circle to that point ; 
 for this line forms the hypotenuse common to 
 two equal right angled triangles. 
 
 PROP. XXXVII. THEOR. 
 
 If from a point without a circle there be drawn two straight lines, one oj 
 which cuts the circle, and the other meets it ; if the rectangle contained by 
 the whole line, which cuts the circle, and the part of it without the circle, 
 be equal to the square of the line which meets it, the line which meets shall 
 touch the circle. 
 
 Let any point D be taken without the circle ABC. and from it let two 
 straight lines DCA and DB be drawn, of which DCA :uts the circle, and 
 DB meets it ; if the rectangle AD.DC, be equal to the square of DB, DB 
 touches the circle. 
 
 Draw (17. 3.) the straight line DE touching the circle ABC ; find the 
 centre F, and join FE, FB, FD ; then FED is a right angle (18. 3.) : and 
 because DE touches the circle ABC, and DCA cuts it, the rectangle AD 
 DC is equal (36. 3.) to the square of DE ; but the rectangle AD.DC is 
 by hypothesis, equal to he square of DB : therefore the square of DE is
 
 86 
 
 ELEMENTS 
 
 equal *.o the square of DB ; and the straight line 
 DE equal to the straight line DB : but FE is 
 equal to FB, wherefore DE.EF are equal to DB, 
 BF ; an i the base FD is common to the two trian- 
 gles DEF, DBF; therefore the angle DEF is 
 equal (8. 1.) to the angle DBF; and DEF is a 
 right angle, therefore also DBF is a right angle : 
 but FB, if produced, is a diameter, and the straight 
 line which is drawn at right angles to a diame- 
 ter, from the extremity of it, touches (16- 3.) the 
 circle : therefore DB touches the circle ABC. 
 
 ADDITIONAL PROPOSITIONS. 
 
 PROP. A. THEOR. 
 
 A diameter divides a circle and its circumference into two equal parts ; and, con 
 versely, the line which divides the circle into two equal parts is a diameter 
 
 Let AB be a diameter of the circle 
 AEBD, then AEB, ADB are equal in 
 surface and boundary. 
 
 Now, if the figure AEB be applied to 
 the figure ADB, their common base AB 
 retaining its position, the curve line AEB 
 must fall on the curve line ADB ; other- 
 wise there would, in the one or the other, 
 be points unequally distant from the cen- 
 tre, which is contrary to the definition of 
 a circle. 
 
 Conversely. The line dividing the circle into two equal parts is a diameter 
 
 For, let AB divide the circle into two equal parts ; then, if the centre is 
 not in AB, let AF be drawn through it, which is therefore a diameter, and 
 consequently divides the circle into two equal parts ; hence the portion 
 AEF is equal to the portion AEFB, which is absurd. 
 
 Cor. The arc of a circle whose chord is a diameter, is a semicircum- 
 ference, and the included segment is a semicircle. 
 
 PROP. B. THEOR. 
 
 Through three given points which are not in the same straight line, one cir- 
 cumference of a circle may be made to pass, and but one. 
 
 Let A, B, C, be three points not in the same straight line : they shall 
 all lie in the same circumference of a circle.
 
 OF GEOMETRY. BOOK III. 
 
 87 
 
 For, let the distances AB, BC be bisected by the perpendiculars DF 
 EF, which must meet in some point F ; for if they were parallel, the lint/s 
 DB, CB, perpendicular to them would also be parallel (Cor. 2. 29. 1.), o 
 else form but one straight line : but thev meet in B, and ABC is not a 
 straight line by hypothesis. 
 
 Let then, FA, FB, and FC be drawn ; then, 
 because FA, FB meet AB at equal distances 
 from the perpendicular, they are equal. For 
 similar reasons FB, FC, are equal ; hence 
 the points A, B, C, are all equally distant 
 from the point F, and consequently lie in the 
 circumference of the circle, whose centre is 
 F, and radius FA. 
 
 It is obvious, that besides this, no other 
 circumference can pass through the same 
 points ; for the centre, lying in the perpen- 
 dicular DF bisecting the chord AB, and at the same time in the perpen- 
 dicular EF bisecting the chord BC (Cor. 1. 3. 3.), must be at the intersec- 
 tion of these perpendiculars ; so that, as there is but one centre, there can 
 be but one circumference. 
 
 PROP. C. THEOR. 
 
 If two circles cut each other, the line which passes through their centres will be 
 perpendicular to the chord which joins the points of intersection, and will 
 divide it into two equal parts. 
 
 Let CD be the line which passes through the centres of two circles cut- 
 ting each other, it will be perpendicular to the chord AB, and will divide it 
 into two equal parts. 
 
 For the line AB, which joins the points of intersection, is a chord com- 
 
 mon to the two circles. And if a perpendicular be erected from the middle 
 of this chord, it will pass (Cor. 1. 3. 3.) through each of the two centres C 
 and D. But no more than one straight line can be drawn through two 
 points ; hence, the straight line which passes through the centres will bi- 
 sect the chord at right angles. 
 
 Cor. Hence, the line joining the intersections of the circumferences of 
 two circles, will be perpendicular to the line which joins their centres. 
 
 SCHOLIUM. 
 
 1. If two circles cut each other, the distance between their centres will 
 be less than the sum of their radii, and the greater radius will be also Irs*
 
 88 
 
 ELEMENTS 
 
 thau the sum of the smaller and the distance between the centres. For, 
 CD is less (20. 1.) than CA+AD, and for the same reason, AD^AC-f- 
 CD 
 
 2. And. conversely, if the distance between the centres of two circles 
 be less than the sum of their radii, the greater radius being at the same time 
 less than the sum of the smaller and the distance between the centres, 
 the two circles will cut each other. 
 
 For, to make an intersection possible, the triangle CAD must be possi- 
 ble. Hence, not only must we have CD < AC + AD, but also the greater 
 radius AD<AC-|-CD ; And whenever the triangle CAD can be con- 
 structed, it is plain that the circles described from the centres C and D, 
 will cut each other in A and B. 
 
 Cor. 1. Hence, if the distance between the centres of two circles be 
 greater than the sum of their radii, the two circles will not intersect eacl 
 other. 
 
 Cor. 2. Hence, also, if the distance between the centres be less thar 
 the difference of the radii, the two circles will not cut each other. 
 
 For, AC + CD>AD; therefore, CD>AD — AC ; that is, any side oi 
 a triangle exceeds the difference between the other two. Hence, the tn 
 angle is impossible when the distance between the centres is less than the 
 difference of the radii ; and consequently the two circles cannot cut eaca 
 other. 
 
 PROP. D. THEOR. 
 
 In the same circle, equal angles at the centre are subtended by equal arcs , 
 and, conversely, equal arcs subtend equal angles at the centre. 
 
 Let C be the centre of a circle, and let the angle ACD be equal to the 
 angle BCD ; then the arcs AFD, DGB, subtending these angles, are 
 equal. 
 
 Join AD, DB ; then the triangles ACD, 
 BCD, having two sides and the included an- 
 gle in the one, equal to two sides and the 
 included angle in the other, are equal : so 
 that, if ACD be applied to BCD, there shall 
 be an entire coincidence, the point A coin- 
 ciding with B, and D common to both arcs ; 
 the two extremities, therefore, of the arc 
 AFD, thus coinciding with those of the arc 
 BGD, all the intermediate parts must coin- 
 cide, inasmuch as they are all equally dis- 
 tant from the centre. 
 
 Conversely. Let the arc AFD be equal to the arc BGD ; then the a - 
 gle ACD is equal to the angle BCD. 
 
 For, if the arc AFD be applied to the arc BGD, they would coincide ; 
 so that the extremities AD of the chord AD, would coincide with those of 
 the chord BD ; these chords are therefore equal : hence, the angle ACD 
 's equal to the angle BCD (8. 1.). 
 
 Cor. 1. It follows, moreover, that equal angles at the centre are sul
 
 OF GEOMETRY. BOOK III 
 
 89 
 
 .ended by equal chords : and, conversely, equal chords subtend equal an- 
 gles at the centre. 
 
 Cor. 2. It is also evident, that equal chords subtend equal arcs . and. 
 conversely, equal arcs are subtended by equal chords. 
 
 Cor. 3. If the angle at the centre of a circle be bisected, both the ar 
 and the chord which it subtends shall also be bisected. 
 
 Cor. 4. It follows, likewise, that a perpendicular through the middle 
 cf the chord, bisects the angle at the centre, and passes through the middle 
 of the arc subtended by that chord. 
 
 SCHOLIUM. 
 
 The centre C, the middle point E of the chord AB, and the middle point 
 D of the arc subtended by this chord, are three points situated in the same 
 line perpendicular to the chord. But two points are sufficient to determine 
 the position of a straight line ; hence every straight line which passes 
 through two of the points just mentioned, will necessarily pass through the 
 third, and be perpendicular to the chord. 
 
 PROP. E. THEOR. 
 
 The arcs of a circle intercepted by two parallels are equal ; and, conversely, if 
 two straight lines intercept equal arcs of a circle, and do not cut each other 
 within the circle, the lines will be parallel. 
 
 There may be three cases : 
 
 First. If the parallels are tangents 
 to the circle, as AB, CD ; then, each 
 of the arcs intercepted is a semi-cir- 
 cumference, as their points of contact 
 (Cor. 3. 16. 3.) coincide with the ex- 
 tremities of the diameter. 
 
 Second. When, of the two parallels 
 AB, GH, one is a tangent, the other 
 a chord, which being perpendicular to 
 FE, the arc GEH is bisected by FE 
 (Cor. 4. Prop. D. Book 3.) ; so that in 
 this case also, the intercepted arcs 
 GE, EH are equal. 
 
 Third. If the two parallels are chords, as GH, JK ; let the diameter 
 FE be perpendicular to the chord GH, it will also be perpendicular to JK, 
 since they are parallel ; therefore, this diameter must bisect each of the 
 arcs which they subtend : that is, GE = EH, and JE = EK ; therefore, 
 JE — GE = EK— EH ; or, which amounts to the same thing, JG is equal 
 toHK. 
 
 Conversely. If the two lines be AB, CD, which touch the circumfer- 
 ence, and if, at the same time, the intercepted arcs EJF, EKF are equal, 
 EF must be a diameter (Prop. A. Book 3.) ; and therefore AB, CD (Cor. ' 
 3. 16. 3.), are parallel. 
 
 But if only one of the lines, as AB, touch, while the other, GH, cuts the 
 circumference, making the arcs EG, EH equal; then the diameter FE 
 
 12
 
 90 
 
 ELEMENTS, &c. 
 
 whi<;h bisects the arc GEH, is perpendicular (Schol. D. 3.) to its chord 
 GH : it is also perpendicular to the tangent AB ; therefore AB, GH are 
 parallel. 
 
 If both lii.es cut the circle, as GH, JK, and intercept equal arcs GJ, 
 HK ; let the diameter FE bisect one of the chords, as GH : it will also 
 bisect the arc GEH, so that EG is equal to EH ; and since GJ is (by hyp.) 
 equal to HK, the whole arc EJ is equal to the whole arc EK ; therefore 
 the chord JK is bisected by the diameter FE : hence, as both chords are 
 bisected by the diameter FE, they are perpendicular to it ; that is, they are 
 parallel (Cor. 28 1.). 
 
 SCHOLIUM. 
 
 The restriction in the enunciation of the converse proposition, namely, 
 that the lines do not cut each other within the circle, is necessary ; for 
 lines drawn through the points G, K, and J, H, will intercept equal arcs 
 GJ, HK, and yet not be parallel, since they will intersect each other within 
 the circle. 
 
 PROP. F. PROB. 
 To draw a tangent to any point in a circular arc, without finding the centre 
 
 From B the given point, take two equal 
 distances BC, CD on the arc ; join BD, 
 and draw the chords BC, CD : make (23. 
 1.) the angle CBG=CBD, and the straight 
 line BG will be the tangent required. 
 
 For the angle CBD = CDB ; and there- 
 fore the angle GBC (32. 3.) is also equa* 
 to CDB, an angle in .he alternate segment ; 
 hence, BG is a tangent at B.
 
 ELEMENTS 
 
 OF 
 
 GEOMETRY. 
 
 BOOK IV. 
 
 DEFINITIONS. 
 
 1 A rectilineal figure is said to be inscribed in another rectilinet. 
 figure, when all the angles of the inscribed 
 figure are upon the sides of the figure in which 
 it is inscribed, each upon each. 
 
 2 In like manner, a figure is said to be described 
 about another figure, when all the sides of the 
 circumscribed figure pass through tho angular 
 points of the figure about which it is described, 
 each through each. 
 
 3 A rectilineal figure is said to be inscribed in 
 a circle, when all the angles of the inscribed 
 figure are upon the circumference of the cir- 
 cle. 
 
 4. A rectilineal figure is said to be described 
 about a circle, when each side of the circum- 
 scribed figure touches the circumference of the 
 circle. 
 
 5. In like manner, a circle is said to be inscrib- 
 ed in a rectilineal figure, when the circum- 
 ference of the circle touches each side of the 
 figure. 
 
 6. A circle is said to be described about a recti- 
 lineal figure, when the circumference of tho 
 circle passes through all the angular points of 
 the figure about which it is described. 
 
 7. A straight line is said to be placed in a circle, 
 when tho extremities of it are in the circum- 
 ference of the circle.
 
 92 ELEMENTS 
 
 8. Polygons of five sides are called pentagons; those of six sides, hexa* 
 gons ; those of seven sides, heptagons ; those of eight sides, octagons ; 
 and so on. 
 
 9 A polygon, which is at once equilateral and equiangular, is called a 
 regular polygon. 
 
 Regular polygons may have any number of sides ; the equilateral tri 
 angle is one of hree sides ; and the square is one of four sides. 
 
 LEMMA. 
 
 Any regular polygon may be inscribed in a circle, and circumscribed about one. 
 
 Let ABCDE, &c. be a regular polygon : describe a circle through the 
 thiee points A, B, C, the centre being 0, and OP the perpendicular let fall 
 from it, to the middle point of BC : join AO and OD. 
 
 If the quadrilateral OPCD be placed upon 
 the quadrilateral OPBA, they will coincide ; 
 for the side OP is common : the angle OPC= 
 OPB, being right ; hence the side PC wi.{ ap- 
 ply to its equal PB, and the point C will fall 
 on B ; besides, from the nature of the polygon, 
 the angle PCD=PBA; hence CD will take 
 the direction BA, and since CD=BA,the point 
 D will fall on A, and the two quadrilaterals 
 will entirely coincide. 
 
 The distance OD is therefore equal to AO ; ™ 
 
 and consequently the circje which passes through the three points A, B, C, 
 will also pass through the point D. By the same mode of reasoning, it 
 might be shown that the circle which passes through the points B, C, D, 
 will also pass through the point E ; and so of all the rest : hence the cir- 
 cle which passes through the points A, B, C, passes through the vertices 
 of all the angles in the polygon, which is therefore inscribed in this circle. 
 
 Again, in reference to this circle, all the sides AB, BC, CD, &c. are 
 equal chords ; they are L before equally distant from the centre (Th. 14. 
 3.) : hence, if from the point O with the distance OP, a circle be describ- 
 ed, it will touch the side BC, and all the other sides of the polygon, each 
 in its middle point, and the circle will be inscribed in the polygon, or the 
 polygon circumscribed about the circle. 
 
 Cor. 1. Hence it is evident that a circle may be inscribed in, or cir- 
 cumscribed about, any regular polygon, and the circles so described have a 
 common centre. 
 
 Cor. 2. Hence it likewise follows, that if from a common centre, circles 
 can be inscribed in, and circumscribed about a polygon, that polygon is regu- 
 lar. For, supposing those circles to be described, the inner one will touch 
 all the sides of the polygon ; these sides are therefore equally distant from 
 its centre ; and, consequently, being chords of the circumscribed circle, 
 they are equal, and therefore include equal angles. Hence the polygon is 
 at once equilateral and equiangular; that is (Def. 9. B. IV.), it is regular
 
 OF GEOMETRY. BOOK IV. 93 
 
 SCHOLIUMS. 
 
 1. The point O, the common centre of the inscribed and circumscribed 
 circles, may also be regarded as the centre of the polygon ; and upon this 
 principle the angle AOB is called the angle at the centre, being formed by 
 two radii drawn to the extremities of the &ame side AB. 
 
 Since all the chords are equal, all the angles at the centre must evident* 
 ly be equal likewise ; and therefore the value of each will be found by di- 
 viding four right angles by the number of the polygon's sides. 
 
 2. To inscribe a regular polygon of a certain number of sides in a given 
 circle, we have only to divide the circumference into as many equal parts 
 as the polygon has sides : for the arcs being equal (see fig. Prop. XV. B. 4.), 
 the chords AB, BC, CD, &c. will also be equal ; hence, likewise, the tri- 
 angles ABG, BGC, CGD, &c. must be equal, because they are equian- 
 gular ; hence all the angles ABC, BCD, CDE, &c. will be equal, and con- 
 sequently the figure ABCD, &c. will be a regular polygon. 
 
 PROP. I. PROB. 
 
 In a given circle to place a straight line equal to a given straight line, not 
 greater than the diameter of tJie circle. 
 
 Let ABC be the given circle, and D the given straight line, not greater 
 than the diameter of the circle. 
 
 Draw BC the diameter of the circle 
 ABC ; then, if BC is equal to D, the 
 thing required is done ; for in the circle 
 ABC a straight line BC is placed equal 
 to D ; But, if it is not, BC is greater 
 than D ; make CE equal (Prop. 3. 1.) 
 to D, and from the centre C, at the dis- 
 tance CE, describe the circle AEF, and 
 join CA : Therefore, because C is the 
 centre of the circle AEF, CA is equal 
 to OF ; but D is equal to CE ; there- 
 fore D is equal to CA : Wherefore, in the circle ABC, a straight line is 
 placed, equal to the given straight line D, which is not greater than the 
 diameter of the circle. 
 
 PROP. II. PROB. 
 In a given circle to inscribe a triangle equiangular to a given triangle. 
 
 Let ABC be the given circle, and DEF the given triangle ; i/ is re- 
 quired to inscribe in the circle ABC a triangle equiangular to the triangle 
 
 Draw (Prop. 17. 3.) the straight line GAH touching the circle in the ooint 
 A, and at the point A, in the straight line AH, make (Prop. 23. l.)the an- 
 gle HAC equal to the angle DEF ; and at the point A, in the straight line
 
 94 
 
 ELEMENTS 
 
 AG, make the angle GAB equal 
 to the angle DFE, and join 
 BC. Therefore, because HAG 
 touches the circle ABC, and AC 
 is drawn from the point of con- 
 tact, the angle HAC is equal 
 (32. 3.) to the angle ABC in the 
 alternate segment of the circle : 
 But HAC is equal to the angle 
 DEF ; therefore also the angle 
 ABC is equal to DEF ; for the 
 same reason, the angle ACB is 
 equal to the angle DFE ; therefore the remaining angle BAC is equal 
 (4. Cor. 32. 1.) to the remaining angle EDF : Wherefore the triangle ABC 
 is equiangular to the triangle DEF, and it is inscribed in the circle ABC 
 
 PROP. III. PROB. 
 About a given circle to describe a triangle equiangular to a given triangle. 
 
 Let ABC be the given circle and DEF the given triangle ; it is requir- 
 ed to describe a triangle about the circle ABC equiangular to the triangle 
 DEF. 
 
 Produce EF both ways to the points G, H, and find the centre K of the 
 circle ABC, and from it draw any straight line KB ; at the point K in the 
 straight line KB, make (Prop. 23 1.) the angle BKA equal to the angle 
 DEG, and the angle BKC equal to the angle DFH ; and through the 
 points A, B, C, draw the straight lines LAM, MBN, NCL touching (Prop. 
 17. 3.) the circle ABC : Therefore, because LM, MN, NL touch the circle 
 ABC in the points A, B, C, to which from the centre are drawn KA, KB, 
 KC, the angles at the points A, B, C, are right (18. 3.) angles. And be- 
 cause the four angles of" the quadrilateral figure AMBK are equal to four 
 right angles, for it can be divided into two triangles ; and because two of 
 
 FH 
 
 them, KAM, KBM, are right angles, the other two AKB, AMB are equal 
 to two right angles : But the angles DEG, DEF are likewise equal (13.1.) 
 to two right angles ; therefore the angles AKB, AMB are equal to the an 
 gles DEG, DEF, of which AKB is equal to DEG ; wherefore the remain-
 
 OF GEOMETRY. BOOK IV. 
 
 95 
 
 mg angle AMB is equal to the remaining angle DEF. In like manner, 
 the angle LNM may be demonstrated to be equal to DFE ; and therefore 
 the remaining angle MLN is equal (32. I.J to the remaining angle EDF : 
 Wherefore the triangle LMN is equiangular to the triangle DEF : and it 
 is described about the circle ABC. 
 
 PROP. IV. PROB. 
 
 To inscribe a circle in a given triangle. 
 
 Let the given triangle be ABC ; it is required to inscribe a circle in 
 ABC. 
 
 Bisect (9. 1.) the angles ABC, BCA by the straight lines BD, CD meet- 
 ing one another in the point D, from which draw (12. 1.) DE, DF, DG 
 perpendiculars to AB, BC, CA. Then be- 
 cause the angle EBD is equal to the angle 
 FBD, the angle ABC being bisected by 
 BD ; and because the right angle BED, is 
 equal to the right angle BFD, the two tri- 
 angles EBD, FBD have two angles of the 
 one equal to two angles of the other ; and 
 the side BD, which is opposite to one of 
 the equal angles in each, is common to 
 both ; therefore their other sides are equal 
 (26. 1.); wherefore DE is equal to DF. 
 For the same reason, DG is equal to 
 DF , therefore the three straight lines DE, DF, DG, are equal to one 
 another, and the circle described from the centre D, at the distance of anj 
 of them, will pass through the extremities of the other two, and will toucl 
 the straight lines AB, BC, CA, because the angles at the points E, F, G 
 are right angles, and the straight line which is drawn from the extremity 
 of a diameter at right angles to it, touches (1 Cor. 16. 3.) the circle. There- 
 fore the straight lines AB, BC, CA, do each of them touch the circle, and 
 the circle EFG is inscribed in the triangle ABC. 
 
 PROP. V. PROB. 
 
 To describe a circle about a given triangle. 
 
 Let the given triangle be ABC ; it is required to describe a circle abou*. 
 ABC. 
 
 Bisect (10. 1.) AB, AC in the points D, E, and from these points draw
 
 96 
 
 ELEMENTS 
 
 DF, EF at right angles (11. 1.) to AB, AC ; DF, EF produced will meet 
 one another ; for, if they do not meet, they are parallel, wherefore, AB, 
 AC, which are at right angles to them, are parallel, which is absurd : let 
 them meet in F, and join FA ; also, if the point F be not in BC, join BF, 
 CF : then, because AD is equal to BD, and DF common, and at right an 
 gles to AB, the base AF is equal (4. 1 .) to the base FB. In like manner, 
 it may be shewn that CF is equal to FA ; and therefore BF is equal to 
 FC ; and FA, FB, FC are equal to one another ; wherefore the circle de- 
 scribed from the centre F, at the distance of one of them, will pass 
 through the extremities of the other two, and be described about the trian- 
 gle ABC. 
 
 Cor. When the centre of the circle falls within the triangle, each of 
 its angles is less than a right angle, each of them being in a segment great- 
 er than a semicircle ; but when the centre is in one of the sides of the 
 triangle, the angle opposite to this side, being in a semicircle, is a right an- 
 gle : and if the centre falls without the triangle, the angle opposite to the 
 side beyond which it is, being in a segment less than a semicircle, is greater 
 than a right angle. Wherefore, if the given triangle be acute angled, the 
 centre of the circle falls within it ; if it be a right angle triangle, the cen- 
 tre is in the side opposite to the right angle ; and if it be an obtuse angled 
 triangle, the centre falls without the triangle, beyond the side opposite to the 
 obtuse angle. 
 
 SCHOLIUM. 
 
 1. From the demonstration it is evident that the three perpendiculars 
 bisecting the sides of a triangle, meet in the same point ; that is, the centre 
 of the circumscribed circle. 
 
 2. A circular segment arch of a given span and rise, may be drawn by 
 a modification of the preceding problem. 
 
 Let AB be the span and SR the rise. 
 
 loin AR, BR, and at their respective points of bisection, M, N, erect 
 the perpendicular MO, NO to AR, BR ; they 
 will intersect at 0, the centre of the circle. 
 That OA = OR = OB, is proved as before. 
 
 The joints between the arch-stones, or 
 voussoirs, are only continuations of radii 
 drawn from the centre O of the circle. 
 
 PROP. VI. PROB. 
 To inscribe a square in a given circle. 
 
 Let ABCD be the given circle ; it is required to inscribe a square in 
 ABCD. 
 
 Draw the diameters, AC, BD at right angles to one another, and join 
 AB, BC, CD, DA ; because BE is equal to ED, E being the centre, and
 
 OF GEOMETRY. BOOK IV. 
 
 V* 
 
 because EA is at right angles to BD, and 
 common to the triangles ABE, ADE ; the 
 base B A is equal (4. 1 .) to the base A D ; and, 
 for the same reason, BC, CD are each of 
 them equal to B A or AD ; therefore che quad- 
 rilateral figure A BCD is equilateral. It is 
 also rectangular ; for the straight line BD be- 
 ing a diameter of the circle A BCD, BAD is 
 a semicircle ; wherefore the angle BAD is a 
 right angle (31.3.); for the same reason each 
 of the angles ABC, BCD, CDA is a right an- 
 gle ; therefore the quadrilateral figure ABCD 
 is rectangular, and it has been shewn to be 
 equilateral ; therefore it is a square ; and 
 ABCD. 
 
 SCHOLIUM. 
 
 it is inscribed in the circle 
 
 Since the triangle AED is right angled and isosceles, we have (Cor. 2. 
 47. 1) AD : AE : : -y/2 : 1 ; hence the side of the inscribed square is to 
 the radius, as the square root of 2, is to unity. 
 
 PROP. VII. PROB. 
 
 To describe a square about a given circle. 
 
 Let ABCD be the given circle; it is required to describe a square about it. 
 
 Draw two diameters AC, BD of the circle ABCD, at right angles to 
 one another, and through the points A, B, C, D draw (17. 3.) FG, GH, HK, 
 KF touching the circle ; and because FG touches the circle ABCD, and 
 EA is drawn from the centre E to the point of contact A, the angles at A 
 are right angles (18. 3.) ; for the same reason, the angles at the points B, 
 C, D, are right angles; and because the angle AEB is a right angle, as 
 likewise is EBG, GH is parallel (28. 1.) to AC ; for the same reason, AC 
 is parallel to FK, and in like manner, GF, 
 HK may each of them be demonstrated to be 
 
 parallel to BED; therefore the figures GK, Gr_ A- 
 
 GC, AK, FB, BK are parallelograms ; and 
 GF is therefore equal (34. 1 .) to HK, and GH 
 to FK ; and because AC is equal to BD, 
 and also to each of the two GH, FK ; and 
 BD to each of the two GF, HK : GH, FK 
 are each of them equal to GF or HK ; there- 
 fore the quadrilateral figure FGHK is equi- 
 lateral. It is also rectangular; for GBEA 
 
 being a parallelogram, and AEB a right an- (^ ~'K. 
 
 gle, AGB (34. 1.) is likewise a right angle : 
 
 in the same manner, it may be shewn that the angles at H, K, F are right 
 angles; therefore the quadrilateral figure FGHK is rectangular; and it 
 ♦vas demonstrated to be equilateral ; therefore it is a square ; and it is de 
 scribed about the circle ABCD. 
 
 13 
 
 B 
 
 D
 
 $8 ELEMENTS 
 
 PROP. VIII. PROB. 
 To inscribe a circle in a given square. 
 
 Let ABCD be the given square ; it is required to inscribe a circle m 
 ABCD. 
 
 Bisect (10. 1.) each of the sides AB, AD, in the points F, E, and 
 through E draw (31. 1.) EH parallel to AB or DC, and through F draw 
 FK parallel to AD or BC ; therefore each of the figures, AK, KB, AH, 
 HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are 
 equal (34. 1.) ; and because that AD is equal to AB, and that AE is the 
 half of AD, and AF the half of AB, AE is equal to AF ; wherefore the 
 sides opposite to these are equal, viz. FG to GE ; in the same manner it 
 may be demonstrated, that GH, GK, are each 
 of them equal to FG or GE ; therefore the 
 four straight lines, GE, GF, GH, GK, are 
 equal to one another ; and the circle described 
 from the centre G, at the distance of one of 
 them, will pass through the extremities of the 
 other three ; and will also touch the straight 
 lines AB, BC, CD, DA, because the angles 
 at the points E, F, H, K, are right angles 
 (29. 1.), and because the straight line which 
 is drawn from the extremity of a diameter at 
 right angles to it, touches the circle (16. 3.) ; 
 therefore each of the straight lines AB, BC, 
 
 CD, DA touches the circle, which is therefore inscribed in the squares 
 ABCD. 
 
 PROP. IX. PROB. 
 To describe a circle about a given square. 
 
 Let ABCD be the given square ; it is required to describe a circle 
 about it. 
 
 Join AC, BD, cutting one another in E ; and because DA is equal to 
 AB, and AC common to the triangles DAC, BAC, the two sides DA, AC 
 are equal to the two BA, AC, and the base DC is equal to the base BC ; 
 wherefore the angle DAC is equal (8. 1.) to the 
 angle BAC, and the angle DAB is bisected by 
 the straight line AC. In the same manner it may 
 be demonstrated, that the angles ABC, BCD, 
 CDA are severally bisected by the straight lines 
 BD, AC ; therefore, because the angle DAB is 
 equal to the angle ABC, and the angle EAB is 
 the half of DAB, and EB A the half of ABC ; the 
 angle EAB is equal to the angle EBA : and the 
 side EA (6. 1.) to the side EB. In the same 
 manner, it may be demonstrated, that the straight 
 
 lines EC, ED are each of them equal to EA, or EB ; therefore the four 
 straight lines EA, EB, EC, ED, are equal to one another; and the circle 
 described from the centre E, at the distance of one of them, must pas*
 
 OF GEOMETRY. BOOK IV 99 
 
 through the extremities of the other three, and be described about the 
 square ABCD. 
 
 PROP. X. PROB. 
 
 To describe an isosceles triangle, having each of the angles at the base double 
 of the third angle. 
 
 Take any straight line AB, and divide (11. 2.) it in the point C, so 
 that the rectangle AB.BC may be equal to the square of AC ; and from 
 the centre A, at the distance AB, describe the circle BDE, in which 
 place (1.4.) the straight line BD equal to AC, which is not greater 
 than the diameter of the circle BDE ; join DA, DC, and about the tri- 
 angle ADC describe (5. 4.) the circle ACD ; the triangle ABD is such 
 as is required, that is, each of the angles ABD, ADB is double of the an- 
 gle BAD. 
 
 Because the rectangle AB.BC is equal to the square of AC, and AC 
 equal to BD, the rectangle AB.BC is 
 equal to the square of BD ; and because 
 from the point B without the circle ACD 
 two straight lines BCA, BD are drawn 
 to the circumference, one of which cuts, 
 and the other meets the circle, and the 
 rectangle AB.BC contained by the whole 
 of the cutting line, and the part of it 
 without the circle, is equal to the square 
 of BD, which meets it ; the straight line 
 BD touches (37. 3.) the circle ACD. 
 And because BD touches the circle, and 
 DC is drawn from the point of contact 
 D, the angle BDC is equal (32. 3.) to 
 the angle DAC in the alternate segment 
 of the circle, to each of these add the angle CDA ; therefore the whole 
 angle BDA is equal to the two angles CDA, DAC ; but the exterior angle 
 BCD is equal (32. 1.) to the angles CDA, DAC ; therefore also BDA is 
 equal to BCD ; but BDA is equal (5. 1.) to CBD, because the side AD 
 is equal to the side AB ; therefore CBD, or DBA is equal to BCD ; and 
 consequently the three angles BDA, DBA, BCD, are equal to one another. 
 And because the angle DBC is equal to the angle BCD, the side BD is 
 equal (6. 1.) to the side DC ; but BD was made equal to CA ; therefore 
 also CA is equal to CD, and the angle CDA equal (5. 1.) to the angle 
 DAC ; therefore the angles CDA, DAC together, are double of the angle 
 DAC; but BCD is equal to the angles CDA, DAC (32. 1.) ; therefore 
 nlso BCD is double of DAC. But BCD is equal to each of the angles 
 BDA, DBA, and therefore each of the angles BDA, DBA, is double of 
 the angle DAB ; wherefore an isosceles triangle ABD is described, hav- 
 ing each of the angles at the base double of the third angle. 
 
 " Cor. 1. The angle BAD is the fifth part of two right angles. 
 " For since each of the angles ABD and ADM is equal to twice the an- 
 cle BAD, they are together equal to four times BAD, and therefore all 
 "'he three angles ABD ADB, BAD, taken together, are equal to five
 
 100 ELEMENTS 
 
 " times the angle BAD. But the three angles ABD, ADB, BAD are 
 '' equal to two right angles therefore five times the angle BAD is equal to 
 " two right angles ; or BAD is the fifth part of two right angles." 
 
 " Cor. 2. Because BAD is the fifth part of two, or the tenth part of 
 " four right angles, all the angles about the centre A are together equal to 
 " ten times the angle BAD, and may therefore be divided into ten parts 
 " each equal to BAD. And as these ten equal angles at the centre, must 
 "stand on ten equal arcs, therefore the arc BD is one-tenth of the cir- 
 " cumference ; and the straight line BD, that is, AC, is therefore equal to 
 "the side of an equilateral decagon inscribed in the circle BDE." 
 
 PROP. XI. PROB. 
 
 To inscribe an equilateral and equiangular pentagon in a given circle. 
 
 Let ABCDE be the given circle, it is required to inscribe an equilateral 
 and equiangular pentagon in the circle ABCDE. 
 
 Describe (10. 4.) an isosceles triangle FGH, having each of the angles 
 at G, H, double of the angle at F ; and in the circle ABCDE inscribe (2. 
 4.) the triangle ACD equiangular to the triangle FGH, so that the angle 
 CAD be equal to the angle at F, and each of the angles ACD, CDA equal 
 to the angle at G or H : where- 
 fore each of the angles ACD, 
 CDA is double of the angle 
 CAD. Bisect (9. 1.) the angles 
 ACD, CDA by the straight lines 
 CE,DB; andjoinAB,BC,ED, 
 EA. ABCDE is the pentagon 
 required. 
 
 Because the angles ACD, 
 CDA are each of them double 
 of CAD, and are bisected by the 
 straight lines CE, DB,the five angles DAC, ACE, ECD, CDB, BDA are 
 equal to one another ; but equal angles stand upon equal arcs (26. 3.) ; 
 therefore the five arcs AB, BC, CD, DE, E A are equal to one another ; and 
 equal arcs are subtended by equal (29. 3.) straight lines ; therefore the 
 five straight lines AB, BC, CD, DE, E A are equal to one another. Where- 
 fore the pentagon ABCDE is equilateral. It is also equiangular ; be- 
 cause the arc AB is equal to the arc DE ; if to each be added BCD, the 
 whole ABCD is equal to the whole EDCB ; and the angle A ED stands 
 on the arc ABCD, and the angle BAE on the arc EDCB : therefore the 
 angle BAE is equal (27. 3.) to the angle AED : for the same reason, each 
 of the angles ABC, BCD, CDE is equal to the angle BAE or AED : there- 
 fore the pentagon ABCDE is equiangular; and it has been shewn that it 
 is equilateral. Wherefore, in the given circle, an equilateral and equian 
 gular pentagon has been inscribed. 
 
 Otherwise. 
 
 " Divide the radius of the given circle, so that the rectangle contained 
 ** by the whole and one of the parts may be equal to the square of the other
 
 OF GEOMETRY. BOOK IV. 
 
 10i 
 
 "(11. 2.). Apply in the circle, on each side of a given po'nt, a line 
 •' equal to the greater of these parts ; then (2. Cor. 10. 4.), each of thu 
 " arcs cut off will be one-tenth of the circumference, and therefore the 
 " arc made up of both will be one-fifth of the circumference ; and if the 
 " straight line subtending this arc be drawn, it will be the side of an 
 " equilateral pentagon inscribed in the circle." 
 
 PROP. XII. PROB. 
 To describe an equilateral and equiangular pentagon about a given circle. 
 
 Let ABODE be the given circle, it is required to describe an equilateral 
 and equiangular pentagon about the circle ABODE. 
 
 Let the angles of a pentagon, inscribed in the circle, by the last pro- 
 position, be in the points A, B, C, D, E, so that the arcs AB, BC, CD, 
 DE, EA are equal (11. 4.) ; and through the points A, B, C, D, E, draw 
 GH, HK,.KL, LM, MG, touching (17. 3.) the circle ; take the centre F. 
 and join FB, FK, FC, FL, FD. And because the straight line KL touch- 
 es the circle ABCDE in the point C, to which FC is drawn from the cen- 
 tre F, FC is perpendicular (18. 3.) to KL ; therefore each of the angles 
 at C is a right angle ; for the same reason, the angles at the points B, D are 
 right angles ; and because FCK is a right angle, the square of FK is equal 
 (47. 1.) to the squares of FC, CK. For the same reason, the square of 
 FK is equal to the squares of FB, BK : therefore the squares of FC, I iv 
 are equal to the squares of FB, BK, of which the square of FC is equal to 
 the square of FB ; the remaining square of CK is therefore equal to the 
 remaining square of BK, and the straight line CK equal to BK : and be- 
 cause FB is equal to FC, and FK common to the triangles BFK, CFK, 
 the two BF, FK are equal to the two CF, FK ; and the base BK is equal 
 to the base KC ; therefore the angle BFK is equal (8. 1.) to the angle 
 KFC, and the angle BKF to FKC ; wherefore the angle BFC is double 
 of the angle KFC, and BKC double of FKC : for the same reason, the an- 
 gle CFD is double of the angle CFL, and CLD double of CLF : and be- 
 cause the arc BC is equal to the arc CD, the angle BFC is equal (27. 3.) 
 to the angle CFD: and BFC is double of the angle KFC, and OKI) 
 double of CFL ; therefore the angle 
 KFC is equal to the angle CFL : 
 now the right angle FCK is equal to 
 the right angle FCL ; and therefore, 
 in the two triangles FKC, FLC, there 
 are two angles of one equal to two an- 
 gles of the other, each to each, and the 
 side FC, which is adjacent to the 
 equal angles in each, is common to 
 both ; therefore the other sides are 
 equal (26. 1 .) to the other sides,and the 
 third angle to the third angle ; there- 
 fore the straight line KC is equal to 
 CL, and the angle FKC to the angle 
 FLC : and because KC is equal to CL, KL is double of KC : in the same 
 manner, it may b« ihewn that HK is double of BK ; and because BK n
 
 103 
 
 ELEMENTS 
 
 equal to KC, as was demonstrated, and KL is do.ible of KC, and HK double 
 of BR, HK is equal toKL ; in like manner, it may be shewn that GH, GM, 
 ML aie each of them equal to HK or KL: therefore the pentagon GHKLM 
 is equilateral. It is also equiangular ; for, since the angle FKC is equal to 
 the angle FLC, and the angle HKL double of the angle FKC, and KLM 
 double of FLC, as was before demonstrated, the angle HKL is equal to 
 KLM ; and in like manner it may be shewn, that each of the angles KHG, 
 HGM, GML is equal to the angle HKL or KLM ; therefore the five an- 
 gles GHK, HKL, KLM, LMG, MGH being equal to one another, the pen- 
 tagon GHKLM is equiangular ; and it is equilateral as was demonstra 
 ted : and it is described about the circle ABCDE. 
 
 PROP. XIII. PROB. 
 
 To inscribe a circle in a given equilateral and equiangular pentagon. 
 
 Let ABCDE be the given equilateral and equiangular pentagon ; it is 
 required to inscribe a circle in the pentagon ABCDE. 
 
 Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, DF, and 
 from the point F, in which they meet, draw the straight lines FB, FA, 
 FE ; therefore, since BC is equal to CD, and CF common to the trian- 
 gles BCF, DCF, the two sides BC, CF are equal to the two DC, CF ; 
 and the angle BCF is equal to "the angle DCF : therefore the base BF is 
 equal (4. 1.) to the base FD, and the other angles to the other angles, to 
 which the equal sides are opposite ; therefore the angle CBF is equal, to 
 the angle CDF : and because the angle CDE is double of CDF, and CDE 
 equal to CBA, and CDF to CBF ; CBA is also double of the angle CBF 
 therefore the angle ABF is equal to the 
 angle CBF ; wherefore the angle ABC 
 is bisected by the straight line BF : in 
 the same manner, it may be demonstra- 
 ted that the angles BAE, AED, are bi- 
 sected by the straight lines AF, EF : 
 from the point F draw (12. 1.) FG, 
 FH, FK, FL, FM perpendiculars to 
 the straight lines AB, BC, CD, DE, 
 EA ; and because the angle HCF is 
 equal to KCF, and the right angle 
 FHC equal to the right angle FKC ; in 
 the triangles FHC, FKC there are two 
 angles of one equal to two angles of the other, and the side FC, which is 
 opposite to one of the equal angles in each, is common to both ; therefore, 
 the other sides shall be equal (26. 1.), each to each ; wherefore the per- 
 pendicular FH is equal to the perpendicular FK : in the same manner it 
 may be demonstrated, that FL, FM, FG are each of them equal to FH, or 
 FK ; therefore the five straight lines FG, FH, FK, FL, FM are equal to 
 one another ; wherefore the circle described from the centre F, at the dis- 
 tance of one of these five, will pass through the extremities of the other 
 four, and touch the straight lines AB, BC, CD. DE, E A, because that the 
 angles at the points G, H, K, L, M are right angles, and that a straight line 
 diawn from tne extremity of the diameter of a circle at right angles \o if
 
 OF GEOMETRY. BOOK IV. 10) 
 
 euuches (1. Cor. 16. 3.) the circle ; therefore each of the straight lines AB- 
 BC, CD, DE, EA touches the circle ; wherefore the circle is inscribed in 
 the pentagon ABCDE. 
 
 PROP. XIV. PROB. 
 To describe a circle about a given equilateral and equiangular pentagon. 
 
 Let ABCDE be the given equilateral and equiangular pentagon ; it is 
 required to describe a circle about it. 
 
 Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, FD, and 
 from the point F, in which they meet, draw 
 the straight lines FB, FA, FE to the points 
 B, A, E. It may be demonstrated, in the 
 same manner as in the preceding proposition, 
 that the angles CB A, BAE, AED are bisect- 
 ed by the straight lines FB, FA, FE : and 
 because that the angle BCD is equal to the 
 angle CDE, and that FCD is the half of the 
 angle BCD, and CDF the half of CDE ; the 
 angle FCD is equal to FDC ; wherefore the 
 side CF is equal (6. 1 .) to the side FD : in 
 like manner it may be demonstrated, that FB, 
 FA, FE are each of them equal to FC, or FD : therefore the five straight 
 lines FA, FB, FC, FD, FE are equal to one another; and the circle de- 
 scribed from the centre F, at the distance of one of them, will pass through 
 the extremities of the other four, and be described about the equilateral 
 and equiangular pentagon ABCDE. 
 
 PROP. XV. PROB. 
 To inscribe an equilateral and equiangular hexagon in a given circle. 
 
 Let ABCDEF be the given circle ; it is required to inscribe an equi- 
 lateral and equiangular hexagon in it. 
 
 Find the centre G of the circle ABCDEF, and draw the diameter AGD : 
 and from D, as a centre, at the distance DG, describe the circle EGCH, 
 join EG, CG, and produce them to the points B, F ; and join AB, BC, 
 CD, DE, EF, FA : the hexagon ABCDEF is equilateral and equiangular. 
 
 Because G is the centre of the circle ABCDEF, GE is equal to GD : 
 and because D is the centre of the circle EGCH, DE is eqoal to DG ; 
 wherefore GE is equal to ED, and the triangle EGD is equilateral ; and 
 therefore its three angles EGD, GDE, DEG are equal to one another 
 (Cor. 5. 1.); and the three angles of a triangle are equal (32. 1.) to two 
 right angles; therefore the angle EGD is the third part of two right .in- 
 gles : in the same manner it may be demonstrated that the angle DGC is 
 also the third part of two right angles; and because the straight line GC 
 makes with EB the adjacent angles EGC, CGB equal (13. 1.) to two 
 right angles; the remaining angle CGB is the third part of two right 
 angles ; therefore the angles EGD, DGC, CGB, are equal to one an 
 other; aad also the angles vertical to them, BGA, AGF, FGE (15
 
 104 
 
 ELEMENTS 
 
 I.*; therefore the six angles EGD, DGC, 
 CGB, BG A, AGF, FGE are equal to one an- 
 other. But equal angles at the centre stand 
 upon equal arcs (26. 3.) : therefore the six 
 arcs AB, BC, CD, DE, EF, FA are equal 
 to one another : and equal arcs are subtend- 
 ed by equal (29. 3.) straight lines ; there- 
 fore the six straight lines are equal to one 
 another, and the hexagon ABCDEF is 
 equilateral. It is also equiangular ; for, 
 since the arc AF is equal to ED, to each of 
 these add the arc ABCD ; therefore the 
 whole arc FA BCD shall be equal to the 
 whole EDCBA : and the angle FED stands 
 upon the arc FABCD, and the angle AFE 
 upon EDCBA; therefore the angle AFE 
 is equal to FED : in the same manner it may be demonstrated, that the 
 other angles of the hexagon ABCDEF are each of them equal to the 
 angle AFE or FED ; therefore the hexagon is equiangular ; it is also 
 equilateral, as was shown ; and it is inscribed in the given circle ABCDEF. 
 
 Cor. From this it is manifest, that the side of the hexagon is equal to 
 the straight line from the centre, that is, to the radius of the circle. 
 
 And if through the points A, B, C, D, E, F, there be drawn straight 
 lines touching the circle, an equilateral and equiangular hexagon shall be 
 described about it, which may be demonstrated from what has been said 
 of the pentagon ; and likewise a circle may be inscribed in a given equi- 
 lateral and equiangular hexagon, and circumscribed about it, by a method 
 like to that used for the pentagon. 
 
 PROP. XVI. PROB. 
 
 To inscribe an equilateral and equiangular quindecagon in a given 
 
 circle. 
 
 Let ABCD be the given circle ; it is required to inscribe an equilateral 
 and equiangular quindecagon in the circle ABCD. 
 
 Let AC be the side of an equilateral triangle inscribed (2. 4.) in the 
 circle, and AB the side of an equilateral 
 and equiangular pentagon inscribed (11. 4.) 
 in the same ; therefore, of such equal parts 
 as the whole circumference ABCDF con- 
 tains fifteen, the arc ABC, being the third 
 part of the whole, contains five ; and the 
 arc A B, which is the fifth part of the whole, 
 contains three ; therefore BC their differ- 
 ence contains two of the same parts : bi- 
 sect (30. 3.) BC in E ; therefore BE, EC 
 are, each of them, the fifteenth part of the 
 whole circumference ABCD : therefore, if 
 the straight lines BE, EC be drawn, and
 
 OF GEOMETRY. BOOK IV. 103 
 
 straight lines equal to them be placed (1. 4.) around in the whole circle, 
 an equilateral and equiangular quindecagon will be inscribed in it. 
 
 And in the same manner as was done in the pentagon, if through the 
 points of division made by inscribing the quindecagon, straight lines be 
 drawn touching the circle, an equilateral and equiangular quindecagon may 
 be described, about it : and likewise, as in the pentagon, a circle may be 
 inscribed in a given equilateral and equiangular quindecagon, and cir- 
 cumscribed about it. 
 
 SCHOLIUM. 
 
 Any regular polygon being inscribed, if the arcs subtended by its sides 
 be severally bisected, the chords of those semi-arcs will form a new regu- 
 lar polygon of double the number of sides : thus, from having an inscribed 
 square, we may inscribe in succession polygons of 8, 16, 32, 64, &c. sides ; 
 from the hexagon may be formed polygons of 12, 24, 48, 96, &c. sides ; 
 from the decagon polygons of 20, 40, 80, &c. sides ; and from the pente- 
 decagon we may inscribe polygons of 30, 60, &c. sides ; and it i& plain 
 that each polygon wdl exceed the preceding in surface or area. 
 
 It is obvious that any regular polygon whatever might be inscribed in u 
 circle, provided that its circumference could be divided into any proposed 
 number of equal parts ; but such division of the circumference like the tri- 
 section of an angle, which indeed depends on it, is a problem which has 
 not yet been effected. There are no means of inscribing in a circle a regu- 
 lar heptagon, or which is the same thing, the circumference of a circle can- 
 not be divided into seven equal parts, by any method hitherto discovered 
 
 It was long supposed, that besides the polygons above mentioned, no 
 other could be inscribed by the operations of elementary Geometry, or, 
 what amounts to the same thing, by the resolution of equations of the first 
 and second degree. But M. Gauss, of Gdttingen, at length proved, in a 
 work entitled Disquisitiones Arithmetics, Lipsie, 1801, that the circumfer- 
 ence of a circle could be divided into any number of equal parts, capable 
 of being expressed by the formula 2"-f-l, provided it be a prime number, 
 tnat is, a number that cannot be resolved into factors. 
 
 The number 3 is the simplest of this kind, it being the value of the 
 above formula when n=l ; the next prime number is 5, and this is also 
 contained in the formula; that is, when n=2. But polygons of 3 and 5 
 sides have already been inscribed. The next prime number expressed by 
 the formula is 17 ; so that it is possible to inscribe a regular polygon of 
 17 sides in a circle. . 
 
 For the investigation of Gauss's theorem, which depend tqxm the the- 
 ory of algebraical equations, the student may consul Ba ~us Theory of 
 Numbers. 
 
 14
 
 ELEMENTS 
 
 GEOMETRY 
 
 BOOK V. 
 
 In the demonstrations of this book there are certain " signs or characters* 
 which it has been found convenient to employ. 
 
 • 1. The letters A, B, C, &c. are used to denote magnitudes of any kind. 
 "The letters m, n, p, q, are used to denote numbers only. 
 
 It is to be observed, that in speaking of the magnitudes A, B, C, &c, 
 we mean, in reality, those which these letters are employed to repre- 
 sent ; they may be either lines, surfaces, or solids. 
 
 ' 2. When a number, or a letter denoting a number, is written close to 
 " another letter denoting a magnitude of any kind, it signifies that the 
 " magnitude is multiplied by the number. Thus, 3A signifies three 
 " times A ; mB, m times B, or a multiple of B by m. When the num- 
 " ber is intended to multiply two or more magnitudes that follow, it is 
 " written thus, m(A+B), which signifies the sum of A and B taken m 
 "times ; m(A — B) is m times the excess of A above B. 
 
 • Also, when two letters that denote numbers are written close to one an- 
 
 " other, they denote the product of those numbers, when multiplied into 
 " one another. Thus, mn is the product of m into n ; and inn A. is A mul- 
 " tiplied by the product of m into n. 
 
 DEFINITIONS. 
 
 1 A less magnitude is said to be a part of a greater magnitude, when the 
 less measures the greater, that is, when the less is contained a certain 
 number of times, exactly, in the greater. 
 
 2. A greater magnitude is said to be a multiple of a less, when the greater 
 is measured by the less, that is, when the greater contains the less a cer- 
 tain number of times exactly. 
 
 3. Ratio is a mutual relation of two magnitudes, of the same kind, to one 
 another, in respect of quantity.
 
 OF GEOMETRY. BOOK V. 107 
 
 4. Magnitudes are said to be of the same kind, when the less can be mul« 
 tiplied so as to exceed the greater ; and it is only such magnitudes tLlt 
 are said to have a ratio to one another. 
 
 5. If there be four magnitudes, and if any equimultiples whatsoever be 
 taken of the first and third, and any equimultiples whatsoever of the se- 
 cond and fourth, and if, according as the multiple of the first is greater 
 than the multiple of the second, equal to it, or less, the multiple of the 
 third is also greater than the multiple of the fourth, equal to it, or less ; 
 then the first of the magnitudes is said to have to the second the same 
 ratio that the third has to the fourth. 
 
 b. Magnitudes are said to be proportionals, when the first has the same 
 ratio to the second that the third has to the fourth ; and the third to the 
 fourth the same ratio which the fifth has to the sixth, and so on whatever 
 be their number. 
 
 When four magnitudes, A, B, C, D are proportionals, it is usual to say 
 "that A is to B as C to D, and to write them thus, A : B::C : D, or 
 "thus, A : B=C : D." 
 
 7. When of the equimultiples of four magnitudes, taken as in the fifth 
 definition, the multiple of the first is greater than that of the second, 
 but the multiple of the third is not greater than the multiple of the fourth : 
 then the first is said to have to the second a greater ratio than the third 
 magnitude has to the fourth : and, on the contrary, the third is said to 
 have to the fourth a less ratio than the first lias to the second. 
 
 8 When there is any number of magnitudes greater than two, of which 
 the first has to the second the same ratio that the second has to the 
 third, and the second to the third the same ratio which the third has to 
 the fourth, and so on, the magnitudes are said to be continual propor- 
 tionals. 
 
 9. When three magnitudes are continual proportionals, the second is said 
 to be a mean proportional between the other two. 
 
 10. When there is any number of magnitudes of the same kind, the first 
 is said to have to the last the ratio compounded of the ratio which the 
 first has to the second, and of the ratio which the second has to the 
 third, and of the ratio which the third has to the fourth, and so on unto 
 the last magnitude. 
 
 For example, if A, B, C, D, be four magnitudes of the same kind, the 
 first A is said to have to the last D, the ratio compounded of the ratio 
 of A to B, and of the ratio of B to C, and of the ratio of C to D ; or, 
 the ratio of A to D is said to be compounded of the ratios of A to B, 
 B to C, and C to D. 
 
 And if A : B::E : F; and B : C::G : H.andC : D::K : L, then, since 
 by this definition A has to D the ratio compounded of the ratios of A to 
 B, B to C, C to D ; A may also be said to have to D the ratio compounded 
 ol the ratios which are the same with the ratios of E to F, G to H 
 and K to L.
 
 108 ELEMENTS 
 
 In like manner, the same things being supposed, if M has to N the same 
 ratio which A has to D, then, for shortness' sake, M is said to have to 
 N a ratio compounded of the same ratios which compound the ratio of 
 A to D ; that is, a ratio compounded of the ratios of E to F, G to H, 
 and K to L. 
 
 11. If three magnitudes are continual proportionals, the ratio of the first 
 to the third is said to be duplicate of the ratio of the first to the second 
 
 * Thus, if A be to B as B to C, the ratio of A to C is said to be duplicate 
 " of the ratio of A to B. Hence, since by the last definition, the ratio 
 *' of A to C is compounded of the ratios of A a B, and B to C, a ratio, 
 " which is compounded of two equal ratios, is duplicate of either of 
 " these ratios." 
 
 12. If four magnitudes are continual proportionals, the ratio of the first 
 to the fourth is said to be triplicate of the ratio of the first to the second, 
 or of the ratio of the second to the third, &c. 
 
 ' So also, if there are five continual proportionals ; the ratio of the first 
 " to the fifth is called quadruplicate of the ratio of the first to the se- 
 cond ; and so on, according to the number of ratios. Hence, a ratio 
 " compounded of three equal ratios, is triplicate of any one of those ra- 
 " tios ; a ratio compounded of four equal ratios quadruplicate," &c. 
 
 1 3. In proportionals, the antecedent terms are called homologous to one 
 another, as also the consequents to one another. 
 
 Geometers make use of the following technical words to signify certain 
 ways of changing either the order or magnitude of proportionals, so as 
 that they continue still to be proportionals. 
 
 14. Permutando, or alternando, by permutation, or alternately ; this word 
 is used when there are four proportionals, and it is inferred, that the first 
 has the same ratio to the third which the second has to the fourth ; or 
 that the first is to the third as the second to the fourth : See Prop. 16. 
 of this Book. 
 
 15. Invertendo, by inversion : When there are four proportionals, and it is 
 inferred, that the second is to the first, as the fourth to the third. Prop 
 A. Book 5. 
 
 16. Componendo, by composition : When there are four proportionals, and 
 it is inferred, that the first, together with the second, is to the second as 
 the third, together with the fourth, is to the fourth. 18th Prop. Book 5. 
 
 17. Dividendo, by division; when there are four proportionals, and it is 
 inferred that the excess of the first above the second, is to the second, 
 •as the excess of the third above the fourth, is to the fourth. 17th Prop. 
 Book 5. 
 
 1 8. Convertendo by conversion ; when there are four proportionals, and 
 it is inferred, that the first is to its excess above the second, as the third 
 
 o its excess above the fourth. Prop. D. Book 5.
 
 Of GEOMETRY. BOOK V. 109 
 
 iy. Ex aequali (sc. distantia), or ex aequo, from equality of distance 
 when there is any number of magnitudes more than two, and as many 
 others, so that they are proportionals when taken two and two of each 
 rank, and it is inferred, that the first is to the last of the first rank of 
 magnitudes, as the first is to the last of the others ; Of this there are the 
 two following kinds, which arise from the different order in which the 
 magnitudes are taken two and two. 
 
 20. Ex aequali, from equality ; this term is used simply by itself, when 
 the first magnitude is to the second of the first rank, as the first to the 
 second of the other rank ; and as the second is to the third of the first 
 rank, so is the second to the third of the other ; and so on in order, and 
 the inference is as mentioned in the preceding definition ; whence this 
 is called ordinate proportion. 
 
 It is demonstrated in the 22d Prop. Book 5. 
 
 21. Ex aequali, in proportione perturbata, seu inordinata : from equality, in 
 perturbate, or disorderly proportion ; this term is used when the first 
 magnitude is to the second of the first rank, as the last but one is to the 
 last of the second rank ; and as the second is to the third of the first 
 rank, so is the last but two to the last but one of the second rank ; and 
 as the third is to the fourth of the first rank, so is the third from the last, 
 to the last but two, of the second rank ; and so on in a cross, or inverse, 
 order ; and the inference is as in the 19th definition. It is demonstrated 
 in the 23d Prop, of Book 5. 
 
 AXIOMS. 
 
 1. Equimultiples of the same, or of equal magnitudes, are equal to one 
 another. 
 
 2. Those magnitudes of which the same, or equal magnitudes, are equi- 
 multiples, are equal to one another. 
 
 3. A multiple of a greater magnitude is greater than the same multiple of 
 a less. 
 
 4. That magnitude of which a multiple is greater than the same multi- 
 ple of another, is greater than that other magnitude. 
 
 PROP. I. THEOR. 
 
 If any number of magnitudes be equimultiples of as many others, each of 
 each what multiple soever any one of the first is of its part, the same mul- 
 tiple is the sum of all the first of the sum of all the rest. 
 
 Let any number of magnitudes A, B, and C be equimultiples of as many 
 others, D, E, and F, each to each, A-f-B+C is the same multiple of D+ 
 E-f-F, that A is of D. 
 
 Let A contain D, B contain E, and C contain F, each the same number 
 of times, as, for instance, three times
 
 J10 ELEMENTS 
 
 Then, because A contains D three times, A=D-f D + D. 
 
 For the same reason, B=E-|-E-i-E ; 
 
 And also, C=F+F+F. 
 
 Therefore, adding equals to equals (Ax. 2. 1.), A+B + C is equal to 
 
 D4*E-f F, taken three times. In the same manner, if A, B, and C were 
 
 each any other equimultiple of D, E, and F, it would be shown that A+ 
 
 B+C was the same multiple of D+E-f-F. 
 
 Cor. Hence, if m be any number, mD -r-"*E4-wF=7n(D+E-f-F). 
 For mD, mE, and mF are multiples of D, E, and F by m, therefore their 
 sum is also a multiple of D+E-f-F by m. 
 
 PROP. II. THEOR. 
 
 If to a multiple of a magnitude by any number, a multiple of the same mag- 
 nitude by any number be added, the sum will be the same multiple of that 
 magnitude that the sum of the two numbers is of unity. 
 
 Let A=mC, and B=nC ; A + B=(m+n)C. 
 
 For, since A=mC, A=C+C + C+&c. C being repeated m times. For 
 the same reason, B=C+C4-&c. C being repeated n times. Therefore, 
 adding equals to equals, A-f-B is equal to C taken m-^-n times ; that is, 
 A+B=(m-|-7^C. Therefore A+B contains C as oft as there are units 
 in m-\-n. 
 
 Cor. 1. In the same way, if there be any number of multiples what- 
 soever, as A=wiE, B=nE, C=j»E, it is shown, that A+B-fC=(m+n 
 +/>)E. 
 
 Cor. 2. Hencealso, since A-f-B + C=(wi+n+p)E, andsinceA=mE, 
 B=;iE, and C=pE, mE+nE+pE=(m+n+p)E. 
 
 PROP. III. THEOR. 
 
 If the first of three magnitudes contain the second as often as there are units 
 in a certain number, and if the second contain the third also, as often as 
 there are units in a certain number, the first will contain the third as often 
 as there are units in the product of these two numbers. 
 
 Let A=»iB, and B=nC ; then A=ffmC. 
 
 Since B=nC, mB=nC-\-nC-\-&c. repeated m times. But nC-j-nC, 
 &c. repeated m times is equal to C (2. Cor. 2. 5.), multiplied by n+n-\-&c. 
 ri being added to itself m times ; but n added to itself m times, is n multi- 
 plied by 771, or 77i7i. Therefore nC-f-7tC-f-&c. repeated m times=77mC; 
 whence also mBrrrmnC, and by hypothesis A=77iB, therefore A=mit»C
 
 OF GEOMETRY. BOOK V. 11] 
 
 PROP. IV. THEOR. 
 
 If ths first of four magnitudes has the same ratio to the second which the third 
 has to the Jour lh, and if any equimultiples whatever be taken of the first and 
 third, and any whatever of the second and fourth ; the multiple of the first 
 shall have the same ratio to the multiple of the second, that the multiple oj 
 the third has to the multiple of the fourth. 
 
 Let A : B : : C : D, and let m and n be any two numbers ; mA : nB : : 
 mC : nD. 
 
 Take of mA and mC equimultiples by any number^), and of nB and nD 
 equimultiples by any number q. Then the equimultiples of mA, and mC 
 by p, are equimultiples also of A and C, for they contain A and C as oft as 
 there are units in pm (3. 5.), and are equal to pm A and pmC For the same 
 reason the multiples of nB and nD by q, are qnB, qnt). Since, therefore, 
 A : B : : C : D, and of A and C there are taken any equimultiples, viz. pm A 
 and pmC, and of B and D, any equimultiples ^nB, qnD, if pmA be greater 
 than qnB,pmC must be greater than qnD (def. 5. 5.) ; if equal, equal ; and 
 if less, less. But pmA, pmC are also equimultiples of mA. and mC, and 
 qnB, qnD are equimultiples of nB and nD, therefore (def. 5. 5.), mA : nB 
 : : mC : nD. 
 
 Cor. In the same manner it may be demonstrated, that if A : B : : C : 
 D, and of A and C equimultiples be taken by any number m, viz. mA and 
 «nC, mA : B : : mC : D. This may also be considered as included in the 
 proposition, and as being the case when n=l. 
 
 PROP. V. THEOR. 
 
 If one magnitude be the same multiple of another, which a magnitude taken 
 from the first is of a magnitude taken from the other ; the remainder is the 
 same multiple of the remainder, that the whole is of the whole 
 
 Let mA and mB be any equimultiples of the two magnitudes A and B, 
 of which A is greater than B ; mA— mB is the same multiple of A— B 
 that mA is of A, that is, mA — mB=m(A— B). 
 
 Let D be the excess of A above B, then A — B=D, and adding B to 
 both, A=D+B. Therefore (1. 5.) mA=mD-f mB ; take mB from both, 
 and mA — mB=mD ; but D=A — B, therefore mA— mB=m(A— B). 
 
 PROP. VI. THEOR. 
 
 If from a multiple of a magnitude by any number a multiple of the same mag- 
 nitude by a less number be taken away, the remainder will be the same mul 
 tiple of that magnitude that the difference of the numbers is of unity. 
 
 Let mA and nA be multiples of the magnitude A, by the numbers m on 1 
 ii, and let m be greater than n ; mA — nA contains A as oft as m—n con- 
 tains uniw, or mA— nA = (m— n)A.
 
 112 ELEMENTS 
 
 Let tr.—n=q; then m=n-\-q. Therefore (2. 5.) mA=nk-{-qA ; take 
 rA from both, and mA — nA=qA. Therefore mA — nA contains A as oft 
 as there are units in q, that is, in m — n, or mA— nA = (m — n)A. 
 
 Cor. When the difference of the two numbers is equal to unity or m • 
 n=l, then mA — nA=A. 
 
 PROP. A. THEOR. 
 
 If four magnitudes be proportionals, they are proportionals also when taken 
 
 inversely. 
 
 If A : B : : C : D, then also B : A : : D : C. 
 
 Let mA and mC be any equimultiples of A and C ; nB and nD any equi- 
 multiples of B and D. Then, because A : B : : C : D, if mA be less than 
 nB, mC will be less than nD (def. 5. 5.), that is, if nB be greater than mA, 
 nD will be greater than mC. For the same reason, if nB=mA, nD=mC, 
 and if nB^/mA, nD /mC. But nB, nD are any equimultiples of B and D, 
 and mA, mC any equimultiples of A and C, therefore (def. 5. 5.), B : A • 
 D: C. 
 
 PROP. B. THEOR. 
 
 If the first be the same multiple of the second, or the same part of it, that the 
 third is of the fourth ; the first is to the second as the third to the fourth. 
 
 First, if mA, mB be equimultiples of the magnitudes A and B, mA : A : 
 mB : B. 
 
 Take of mA and mB equimultiples by any number n ; and of A and B 
 equimultiples by any number p ; these will be nmA (3. 5.),pA, nmB (3. 5.) 
 pB. Now, if nmA be greater than pA, nm is also greater than p ; and u 
 nm is greater than p, nmB is greater than pB, therefore, when nmA is great 
 er than pA, nmB is greater than pB. In the same manner, if nmA—pA 
 nmB=joB, and if nmA/_pA, nmB/_pB. Now, nmA, nmB are any equi 
 multiples of mA and mB ; and pA, pB are any equimultiples of A and B 
 therefore mA : A : : mB : B (def. 5. 5.). 
 
 Next, Let C be the same part of A that D is of B ; then A is the same 
 multiple of C that B is of D, and therefore, as has been demonstrated, A : 
 : : B : D and inversely (A. 5.) C : A : : D : B. 
 
 PROP. C. THEOR. 
 
 If the first be to the second as the third to the fourth; and if the first be a 
 multiple or a part of the second, the third is the same multiple or the same 
 part of the fourth. 
 
 Let A : B : : C : D, and first, let A be a multiple of B, C .'s the same 
 multiple of D, that is, if A=mB, C=mD. 
 
 Take of A and C equimultiples by any number as 2, viz. 2 A and 2C ; 
 and of B and D, take equimultiples by the number 2m, viz. 2mB, 2mD (3
 
 OF GEOMETRY. BOOK V. 113 
 
 5.) ; then,because A=mB,2A=2mB ; and since A : B : : C : D, and since 
 2A=2mB, therefore 2C=2mD (del - . 5. 5.), and C=mD, that is, C contains 
 D, m times, or as often as A contains B. 
 
 Next, Let A be a part ot B, C is the same part of D. For, since A : B 
 • • C : D, inversely (A. 5.), B : A : : D : C. But A being a part of B, B is 
 a multiple of A ; and therefore, as is shewn above, D is the same multiple 
 of C, and therefore C is the same part of D that A is of B. 
 
 PROP. VII. THEOR. 
 
 Equal magnitudes have the same ratio to the same magnitude ; and the same 
 has the same ratio to equal magnitudes. 
 
 Let A and B be equal magnitudes, and C any other ; A : C : : B : C. 
 
 Let mA, mB, be any equimultiples of A and B ; and nC any multiple 
 of C. 
 
 Because A=B, mA=mB (Ax. 1. 5.) ; wherefore, if mA be greater than 
 ♦»C, mB is greater than nC ; and ifmA=nC, mB=»C ; or, ifmA^nC, mB 
 /71C. But mk and mB are any equimultiples of A and B, and nC is any 
 multiple of C, therefore (def. 5. 5.) A : C : : B : C. 
 
 Again, if A=B, C : A : : C : B ; for, as has been proved, A : C : : B • 
 C, and inversely (A. 5.), C : A : : C : B. 
 
 PROP. VIII. THEOR. 
 
 Of unequal magnitudes, the greater has a greater ratio to the same than the less 
 has ; and the same magnitude has a greater ratio to the less than it has to 
 the greater. 
 
 Let A -J- B be a magnitude greater than A, and C a third magnitude, 
 A+B has to C a greater ratio than A has to C ; and C has a greater ratio 
 to A than it has to A-j-B. 
 
 Let m be such a number that mA and mB are each of them greater than 
 C ; and let nC be the least multiple of C that exceeds m A +mB ; then nC 
 — C, that is (n— 1)C (1. 5.) will be less than mA-j-mB, or mA-f-mB, that 
 is, m(A+B) is greater than (n — 1)C. But because nC is greater than 
 mk-\~mB, and C less than mB, nC — C is greater than mA, or mA is less 
 than nC— C, that is, than (n— 1)C. Therefore the multiple of A-f-B by 
 m exceeds the multiple of C by n — 1, but the multiple of A by m does not 
 exceed the multiple of C by n — 1 ; therefore A-f-B has a greater ratio to 
 •} than A has to C (def. 7. 5.). 
 
 Again, because the multiple of C by n — 1, exceeds the multiple of A by 
 m, but does not exceed the multiple of A-fB by m, C has a greater ratio to 
 k than it has to A+B (def. 7. 5.). 
 
 15
 
 114 ELEMENTS 
 
 PROP. IX. THEOR. 
 
 Magnitudes which have the same ratio to the same magnitude are equal to one 
 another ; and those to which the same magnitude has the same ratio are equa, 
 to one another. 
 
 If A : C :: B : C, A=B. 
 
 For if not, let A be greater than B ; then because A js greater than B, 
 two numbers, m and n, may be found, as in the last proposition, such that 
 mA shall exceed nC, while mB does not exceed nC. But because A : C 
 : : B : C ; and if mA exceed raC, mB must also exceed nC (def. 5. 5.) : and 
 it is also shewn that mB does not exceed nC, which is impossible. There- 
 fore A is not greater than B ; and in the same way it is demonstrated that 
 B is not greater than A ; therefore A is equal to B. 
 
 Next, let C : A : : C : B, A=B. For by inversion (A. 5.) A. : C : : B : 
 C ; and therefore, by the first case, A=B. 
 
 PROP. X. THEOR. 
 
 That magnitude, which has a greater ratio ttian another has to the same magni- 
 tude, is the greatest of the two : And that magnitude, to which the same has 
 a greater ratio than it has to another magnitude, is the least of the two. 
 
 If the ratio of A to C be greater than that of B to C, A is greater than B. 
 
 Because A : C/^B : C, two numbers m and n may be found, such that 
 mA^nC, and mB/»C (def. 7. 5.). Therefore also mAymB, and A/B 
 (Ax. 4. 5.). 
 
 Again, let C : B/C : A; B/A. For two numbers, m and n maybe 
 found, such that mC/nB, and mCj/nA (def. 7. 5.). Therefore, since »B 
 is less, and nA greater than the same magnitude mC,nB/_nA, and there- 
 fore BZ A. 
 
 PROP. XI. THEOR 
 
 Ratios tnat are equal to the same ratio are equal to one another. 
 
 If A : B : : C : D ; and also C : D : : E : F ; then A : B : : E : F. 
 
 Take mA, mC, mE, any equimultiples of A, C, and E ; and nB, »D, nF, 
 any equimultiples of B, D, and F. Because A : B : : C : D, if mAy nB, 
 mC7nD (def. 5.5.); but if mC7nD, mE7«F (def. 5. 5.), because C : D 
 : : E : F ; therefore if mAynB, mE ~/nY. In the same manner, if mA= 
 i*B,mE=nF; and if mA/_nB, mE^nF. Now, mA, mE are any equi- 
 multiples whatever of A and E ; and nB, nF any whatever of B and F \ 
 therefore A : B : : E : F (def. 5. 5.).
 
 OF GEOMETRY. BOOK V. 115 
 
 PROP. XII. THEOR. 
 
 If any number of magnitudes be proportionals, as one of the antecedents is to 
 its consequent, so are all the antecedents, taken together, to all the conse- 
 quents. 
 
 I" A : B : C : D, and C . D : . E : F ; then also, A : B : : A+C+E : 
 B+D+F. 
 
 Take mA, mC, mE any equimultiples of A, C, and E ; and nB, nD, nF, 
 any equimultiples of B, D, and F. Then, because A : B : : C : D, if mA 
 7 «B,mC7"D (def. 5. 5.) ; and when wiC/nD, wiE/nF, because C : D 
 :: E : F. Therefore, if mA7nB,mA+mC + mE7nB + nD + nF : In the 
 same manner, if mA=«B, mA+mC+mE=nB + nD + nF ; and if mk£ 
 nB, ?/»A + 7/iC + mE/nB + nD+nF. Now, mA+mC+mE=m(A + C + 
 E) (Cor. 1. 5.), so that mA and mA+mC + niE are any equimultiples of 
 A, and of A + C + E. And for the same reason nB, and nB+nD+nF are 
 any equimultiples of B, and of B+D+F ; therefore (def. 5. 5.) A : B : : 
 A+C+E : B + D+F. 
 
 PROP. XIII. THEOR. 
 
 If the first have to the second the same ratio which the third has to thejourth, 
 but the third to the fourth a greater ratio than the fifth has to the sixth; 
 the first has also to the second a greater ratio than the fifth has to the sixth. 
 
 If A : B :: C : D; but C : D7E : F; then also, A : B7E : F. 
 
 Because C : D7E : F, there are two numbers m and n, such that mC / 
 nD, but ffiE/nF(def. 7. 5.). Now,i( mC/nD, rn Ay nB, because A : B 
 : : C : D. Therefore mkynB, and mE/nF, wherefore, A : B7E : F 
 (def. 7. 5.). 
 
 PROP. XIV. THEOR. 
 
 If the first have to the second the same ratio which the third has to the fourth, 
 and if the first be greater than the third, the second shall be greater than 
 thejourth; if equal, equal ; and if less, less. 
 
 If A : B : : C : D; then if A7C, B7D; if A=C,B=D; and if A^ 
 C B^D. 
 
 First, let A7C ; then A : B7C : B (8. 5.), but A : B : : C • D, there- 
 fore C : D7C : B (13. 5.), and therefore B7D (10. 5.). 
 
 In the Ramo manner, it is proved, that if A=C, B=D ; and if A./.C, 
 B/D. 
 
 PROP. XV. THEOR. 
 Magnitudes hive the same ratio to one another which their equimultiples hav*. 
 
 If A and B be two magnitudes, and m any number, A : B . : mA : m\i. 
 Because A • B : : A : B (7. 5.) ; A : B : : A+ A : B + B (12. 5.), or A ■
 
 i!6 ELEMENTS 
 
 B : . 2A : 2B. And in the same manner, since A : B : : 2A : 2B, A : B 
 : : A+2A : B+2B (12. 5.), or A : B : : 3A : 3B ; and so on, for all the 
 equimultiples of A and B. 
 
 PROP. XVI. THEOR. 
 
 If four magnitudes of the same kind be proportionals, they will also be pro- 
 portionals when taken alternately. 
 
 If A : B : : C : D, then alternately, A : C . : B : D. 
 
 Take mk, mB any equimultiples of A and B, and nC, nD any equimul 
 tiples of C and D. Then (15. 5.) A : B : : mk : mB ; now A : B : : C . 
 D, therefore (11. 5.) C : D : : mk : mB. But C : D : : nC : nD (15. 5.) ; 
 therefore mk : mB : : nC : nD (11. 5.) : wherefore if mkynC, mBynD 
 (14. 5.); if mA=nC, mB=nD, or if mk/_nC, 7nB,/nD; therefore (def 
 5. 5.) A : C : : B : D. 
 
 PROP. XVII. THEOR. 
 
 If magnitudes, taken jointly, be proportionals, they will also be proportionals 
 when taken separately ; that is, if the first, together with the second, have 
 to the second the same ratio which the third, together with the fourth, has to 
 the fourth, the first will have to the second the same ratio which the thira 
 has to the fourth. 
 
 If A+B : B : : C+D : D, then by division A : B : : C : D. 
 
 Take mk and nB any multiples of A and B, by the numbers m and n ; 
 and first, let mk"/nB : to each of them add mB, then mA + mB/mB + nB. 
 But mA+mB=m(A+B) (Cor. 1. 5.), and mB+nB=(m+n)B (2. Cor 2. 
 5.), therefore m(A+B)7(m+n)B. 
 
 And because A + B : B :: C + D : D, if m(A+B)7(m+n)B, m(C+D) 
 7(ra+n)D, or mC+mD7mD+nD, that is, taking mD from both, mCy 
 nD. Therefore, when mk is greater than nB, mC is greater than nD. In 
 like manner it is demonstrated, that if mA=nB, mC=nD, and if mk/_nB, 
 that mD/nD ; therefore A : B : : C : D (def. 5. 5.). 
 
 PROP. XVIII. THEOR. 
 
 If magnitudes, taken separately, be proportionals, they will also be proportion- 
 als when taken jointly, that is, if the first be to the second as the third to the 
 fourth, the first and second together will be to the second as the third and 
 fourth together to the fourth. 
 
 H A • B : : C : D, then, by composition, A+B : B : : C+D : D. 
 
 Take m(A + B), and nB any multiples whatever of A+B and B; and 
 first, let m be greater than n. Then, because A+B is also greater than 
 B, m(A+B)/nB. For the same reason, m(C+D)7nD. In this case, 
 therefore, that is, when my n, m(A + B) is greater than nB, andm(C + D) 
 is greater than nD. And in the same manner it may be proved, that wher 
 <n=n, m(A+B) is greater than nB, and m(CA-D) greater than nD-
 
 OF GEOMETRY. BOOK V. 117 
 
 Next, let m/_n, or n"/m, then m(A-f-B) maybe greater than 7jB, or niaj 
 be equal to it, or may be hess ; first, let m(A + B) be greater than nB ; then 
 also, mA-fw»B7«B ; take mB, which is less than nB, from both, and «A 
 ynB— mB,ox mky{n— m)B (6. 5.). But if mAy(n— m)B, mC7(n — m) 
 D, because A : B : : C : D. Now, (n—m)D=nl)—mD (6. 5.), therefore 
 mC/nD — mD, and adding /nD to both, mC + mDynD, that is (1. 5.), 
 «j(C+D)7«D. If, therefore, »i(A + B)7nB,m(C+D)7nD. 
 
 In the same manner it will be proved, that if »j(A-r-B) = nB, m(C-|-D) 
 =nD; and if m(A+B)/nB, m(C+D)/nD ; therefore (def. 5. 5.), A-f 
 B : B :: C+D : D 
 
 PROP. XIX. THEOR. 
 
 If a whole magnitude be to a whole, as a magnitude taken from the first is to a 
 magnitude taken from the other ; the remainder will be to the remainder as 
 the whole to the whole. 
 
 If A : B : : C : D, and if C be less than A, A-C : B-D : : A : B. 
 
 Because A : B : : C : D, alternately (16. 5.), A : C : : B: D ; and there- 
 fore by division (17. 5.) A— C : C : : B— D : D. Wherefore, again alter- 
 nately, A— C : B— D : : C : D ; but A : B : : C : D, therefore (11. 5.) A 
 -C: B-D:: A: D. 
 
 Cor. A-C : B-D :: C : D. 
 
 PROP. D. THEOR. 
 
 If four magnitudes be proportionals, they are also proportionals by conversion, 
 that is, the first is to its excess above the second, as the third to its excess 
 above the fourth. 
 
 If A : B : : C : D, by conversion, A : A — B : : C : C— D. 
 
 For, since A : B : : C : D, by division (17. 5.), A— B : B : : C— D : D. 
 and inversely (A. 5.) B : A— B : : D : 0— D ; therefore, by composition 
 (18. 5.), A : A-B :: C : C— D. 
 
 Cor. In the same way, it may be proved that A.A+B::C:C-f-D. 
 PROP. XX. THEOR. 
 
 If there be three magnitudes, and other three, which taken two and two, havn 
 the same ratio ; if the first be greater than the third, the fourth is greater 
 than the sixth ; if equal, equal ; and if less, less. 
 
 If there be three magnitudes, A, B, and C, and other three D, E, and F ; 
 *\A if A : B : : D : E ; and also B : C : : E : F, then 
 if A7C.D7F; if A=C, D = F; and if A/C.D 
 
 A. B, C, 
 D E, F. 
 
 First,letA7C; then A : B7C : B (8. 5.). But A : B : : D : E, thero. 
 fore alao D : E7C : E (13. 5.). Now B : C : : E : F, and inversely (A
 
 A, B, C, 
 D, E, F. 
 
 118 ELEMENTS 
 
 5.), C : ti : f : E ; and it has been shewn that D : E7C : B, therefore 
 D : E;T . E (13. 5.), and consequently D7F (10. 5.). 
 
 Next, let A =C ; then A : B : : C : B (7. 5.), but A : B : : D : E ; there- 
 fore, C : B : : D : E, but C : B : : F : E, therefore, D : E : : F : E (! :. 
 5.), and D=J (9. 5.). Lastly, let A/C. Then C 7 A, and because, as 
 was already shewn, C : B : : F : E, and B : A : : E : D ; therefore, by the 
 first case, if C 7 A, F7D, that is, if A /C, D/F. 
 
 PROP. XXI. THEOR. 
 
 If there be three magnitudes, and other three, which have the same ratio taken two 
 and two, but in a cross order; if the first magnitude be greater than the third, 
 the fourth is greater than the sixth ; if equal, equal ; and if less, less. 
 
 If there be three magnitudes, A, B, C, and other three, D, E, and F, 
 such that A : B : : E : F,andB : C : : D : E; if A7C,D7F; if A=C, 
 D=F; and if A/C, D/F. 
 
 First, let A 7 C. Then A : B 7 C : B (8. 5.), but 
 A:B :: E : F, therefore E : F 7 C : B(13.5.). Now, 
 B : C : : D : E, and inversely, C : B : : E : D ; there- 
 fore,E : F7E : D (13. 5.), wherefore, D7F (10. 5.). 
 
 Next, let A=C. Then (7. 5.) A : B : : C : B ; but A : B : : E : F, 
 therefore, C : B : : E : F (11. 5.) ; but B : C : : D : E, and inversely, C : 
 B : : E : D, therefore (11. 5.), E : F : : E : D, and, consequently, D=F 
 (9. 5.). 
 
 Lastly, let A/C. Then C7A, and, as was already proved, C : B : . 
 E : D ; and B : A : : F : E, therefore, by this first case, since C 7 A, F 7 
 
 D, that is, D^F. 
 
 PROP. XXII. THEOR. 
 
 If there be any number of magnitudes, and as many others, which, taken two ana 
 two in order, have the same ratio ; the first will have to the last of the first 
 magnitudes, the same ratio which the first of the other has to the last. 
 
 First, let there be three magnitudes, A, B, C, and other three, D, E, F, 
 which, taken two and two, in order, have the same ratio, viz. A : B : : D : 
 
 E, and B : C : : E : F ; then A : C : : D : F. 
 
 Take of A and D any equimultiples whatever, mk, mD ; and of B and 
 D any whatever, nB, n¥ : and of C and F any whatever, ^C, qF. Because 
 A : B : : D : E, mA : nB : : mD : nE (4. 5.') ; and 
 for the same reason, nB : qC : : nE : qF. Therefore 
 (20. 5.) according as mA is greater than qC, equal to 
 it, or less, mD is greater than qF, equal to it, or 
 less ; but mA, mD are any equimultiples of A and D : 
 
 A, B, C, 
 
 D, E, F, I 
 
 mA, »B, qC, - 
 
 mD, nE, qF. 
 
 and qC. qF are any equimultiples of C and F ; therefore (def. 5. 5.), A : C 
 1) : F. 
 Again, let there be four magnitudes, and other four which, taken two 
 
 * N. B. This proposition is usually cited by the words " ex aequaH,'"or " « x aequo."'
 
 A, B, C, D, ( 
 E, F, G, H. ) 
 
 OF GEOMETRY. BOOK V. 119 
 
 and two in order, have the same ratio, viz. A:B::E:F;B:C • F 
 G ; C : D : : G : H, then A : D : : E : H. 
 For, since A, B, C are three magnitudes, and 
 
 E, F, G other three, which, taken two and two, 
 have the same ratio, by the foregoing case, A : 
 C : : E : G. And because also C : D : : G : H, by that same case, A : D 
 : : E : H. m the same manner is the demonstration extended to any num 
 ber of magnitudes. 
 
 PROP. XXIII THEOR. 
 
 If there be any number of magnitudes, and as many others, which, taken two 
 and two, in a cross order, have the same ratio ; the first will have to the last 
 of the first magnitudes the same ratio which the first of the others has to 
 the last* 
 
 First, Let there be three magnitudes, A, B, C, and other three, D, E, and 
 
 F, which, taken two and two in a cross order, have the same ratio, viz. A 
 r B : : E : F, and B : C : : D : E, then A : C : : D : F. Take of A, B, 
 and D, any equimultiples mA, mB, mD ; and of C, E, F any equimultiples 
 nC, nE, nF. 
 
 Because A : B : : E : F, and because also A : B : : mA : mB (15. 5.), 
 and E : F : : nE : nF ; therefore, mA : mB : : nE : nF (11. 5.). Again, 
 because B : C : : D : E, mB : nC : : mD : nE (4. 
 5.) ; and it has been just shewn that mA : mB : : 
 nE : nF; therefore, if mA 7 nC,mD 7 nF (21.5.) ; 
 if mA=nC, mD=nF ; and if mA/nC, mD/nF. 
 Now, mA and wD are any equimultiples of A and 
 D, and nC, nF any equimultiples of C and F ; therefore, A : C : : D : F 
 (def. 5. 5.). 
 
 Next, Let there be four magnitudes, A, B, C, and D, and other four, E, 
 F, G, and H, which, taken two and two in a cross order, have the same 
 ratio, viz. A : B : : G : H ; B : C : : F : G, and 
 
 A, 
 
 B, 
 
 c, 
 
 D, 
 
 E, 
 
 F, 
 
 mA, 
 
 mB, 
 
 nC, 
 
 mD, 
 
 nB, 
 
 nF. 
 
 C : D : : E : F, then, A : D : : E : H. For,since A, B, C, D, 
 
 A, B, C, are three magnitudes, and F, G, H, other E, F, G, H. 
 
 three, which, taken two and two, in a cross order, 
 
 have the same ratio, by the first case, A : C : : F : H. But C : D : : E : 
 F, therefore, again, by the first case, A : D : : E : H. In the same manner 
 may the demonstration be extended to any number of magnitudes 
 
 PROP. XXIV. THEOR. 
 
 If the first has to the second the same ratio which the third has to tut fourth . 
 and the fifth to the second, the. same ratio which the sixth has to the fourth ; 
 the first and fifth together, shall have to the second, the same ratio which 
 the third and sixth together, have to the fourth. 
 
 Let A : B : : C : D, and also E : B : : F : D, then A + E • B : : C-f F : D. 
 
 • N. B. This proposition f I usually cited by tb> words " ex HquaU in pr<>|<ortione pertui 
 MU; n or, " ex equo Inversely."
 
 120 ELEMENTS, &c. 
 
 Because E : B : : F : D, by inversion, B : E : : D : F. But by hypo- 
 thesis. A : B : : C : D. therefore, ex aequali (22. 5.), A : E : : C : F ; and 
 by composition (18. 5.), A+E:E::C + F:F. And again by hypothe- 
 sis, E : B : : F : D, therefore, ex aequali (22. 5.), A+E : B : : C+F : D. 
 
 PROP. E. THEOR. 
 
 If four magnitudes be proportionals, the sum of the first two is to their diffe* 
 rence as the sum of the other two to their difference. 
 
 Let A : B : : C : D ; then if A7B, 
 
 A+B : A-B :: C + D : C-D; or if A/B 
 
 A+B : B-A :: C + D : D-C. 
 Fcr, if A/^B, then because A : B : : C : D, by division (17. 5.), 
 
 A — B : B . : C — D : D, and by inversion (A. 5.), 
 
 B : A— B : : D : C— D. But, by composition (18. 5.), 
 
 A+B : B : : C+D : D, therefore, ex aequali (22. 5.), 
 
 A+B : A-B :: C + D : C-D. 
 In the same manner, if B 7 A, it is proved, that 
 
 A+B : B-A:: C + D : D-C. 
 
 PROP. F. THEOR. 
 Ratios which are compounded of equal ratios, are equal to one anothet . 
 
 Let the ratios of A to B, and of B to C, which compound the ratio of A 
 to C, be equal, each to each, to the ratios of D to E, and E to F, which com- 
 pound the ratio of D to F, A : C : : D : F. 
 
 For, first, if the ratio of A to B be equal to that of 
 D to E, and the ratio of B to C equal to that of E to 
 F, ex aequali (22. 5.), A : C : : D : F. 
 
 And next, if the ratio of A to B be equal to that of E to F, and the ratio 
 of B to C equal to that of D to E, ex aequali inversely (23. 5.), A : C : : D 
 : F. In the same manner may the proposition be demonstrated, whatever 
 be the number of ratios. 
 
 PROP. G. THEOR. 
 
 If a magnitude measure each of two others, it will also measure their sum and 
 
 difference. 
 
 Let C measure A, or be contained in it a certain number of times ; 9 times 
 for instance : let C be also contained in B, suppose 5 times. Then A=9C, 
 and B=5C ; consequently A and B together must be equal to 14 times C, 
 so that C measures the sum of A and B ; likewise, since the difference of 
 A and B is equal to 4 times C, C also measures this difference. And had 
 any other numbers been chosen, it is plain that the results would have been 
 sinrilar. For, let A=»iC, and B=»C ; A+B = (7»+w)C, and A — B= 
 (to— n)C. 
 
 Cor. If C measure B, and also A — B, or A+B, it must measure A for 
 the sum of B and A — B is A, and the difference of B and £ \-B is also A 
 
 A, B, C, 
 D, E, F.
 
 ELEMENTS 
 
 OP 
 
 GEOMETRY. 
 
 BOOK VI. 
 
 DEFINITIONS. 
 
 1. Similar rectilineal figures are 
 
 those which have their several 
 
 angles equal, each to each, and 
 
 the sides about the equal angles 
 
 proportionals. 
 
 In two similar figures, the sides which lie adjacent to equal angles, are 
 called homologous sides. Those angles themselves are called homo- 
 logous angles. In different circles, similar arcs, sectors, and seg- 
 ments, are those of which the arcs subtend equal angles at the 
 centre. Two equal figures are always similar; but two similar 
 figures may be very unequal. 
 
 2 Two sides of one figure are said to be reciprocally proportional to 
 two sides of another, when one of the sides of the first is to one of the 
 sides of the second, as the remaining side of the second is to the re- 
 maining side of the first. 
 
 3. A straight line is said to be cut in extreme and mean ratio, when the 
 whole is to the greater segment, as the greater segment is to the less. 
 
 4. The altitude of a triangle is the straight line 
 
 drawn from its vertex perpendicular to the base. 
 
 The altitude of a paraHelogram is the perpendicu- 
 lar which measures the distance of two oppo- 
 site sides, taken as bases. And the altitude of 
 a trapezoid is the perpendiculardrawn between 
 its two parallel sides. 
 
 PROP. I. THEOR. 
 
 Triangles and parallelograms, of the same altitude, are one to another 
 
 as their bases. 
 
 Let the triangles ABC, ACD, and the parallelograms EC, CF have the 
 same altitude, viz. the perpendicular drawn from the point A to BD: Then, 
 
 16
 
 122 ELEMENTS 
 
 as the base BC, is to the base CD, so is the triangle ABC to tho triangle 
 ACD, and the parallelogram EC to the parallelogram CF. 
 
 Produce BD both ways to the points H, L, and take any number of 
 straight lines BG, GH, each equal to the base BC; and DK, KL, any 
 number of them, each equal to the base CD ; and join AG, AH, AK, AL. 
 Then, because CB, BG, GH are all equal, the triangles AHG, AGB, ABC 
 are all equal (38. 1.) ; Therefore, whatever multiple the base HC is of the 
 base BC, the same multiple is the triangle AHC of the triangle ABC. For 
 the same reason, whatever the base LC is of the base CD, the same mul- 
 tiple is the triangle ALC of 
 
 the triangle ADC. But if B j^ F 
 
 the base HC be equal to the 
 base CL, the triangle AHC 
 :s also equal to the triangle 
 ALC (38. I.): and if the 
 base HC be greater than the 
 oase CL, likewise the tria* 
 gle AHC is greater tnan tL* 
 triangle ALC ; and if less, jr 7tT xT 
 less. Therefore, since there 
 are four magnitudes, viz. the two bases BC, CD, and the two triangles 
 ABC, ACD ; and of the base BC and the triangle ABC, the first and third, 
 any equimultiples whatever have been taken, viz. the base HC, and the 
 triangle AHC ; and of the base CD and triangle ACD, the second and 
 fourth, have been taken any equimultiples whatever, viz. the base CL and 
 triangle ALC ; and since it has been shewn, that if the base HC be greater 
 Uian the base CL, the triangle AHC is greater than the triangle ALC ; 
 and if equal, equal ; and if less, less ; Therefore (def. 5. 5.), as the base 
 BC is to the base CD, so is the triangle ABC to the triangle ACD. 
 
 And because the parallelogram CE is double of the triangle ABC (41. 
 I.), and the parallelogram CF double of the triangle ACD, and because 
 magnitudes have the same ratio which their equimultiples have (15. 5.) ; 
 as the triangle ABC is to the triangle ACD, so is the parallelogram EC to 
 he parallelogram CF. And because it has been shewn, that, as the base 
 BC is to the base CD, so is the triangle ABC to the triangle ACD ; and 
 as the triangle ABC to the triangle ACD, so is the parallelogram EC to 
 the parallelogram CF ; therefore, as the base BC is to the base CD, so is 
 (11. 5.) the parallelogram EC to the parallelogram CF. 
 
 Cor. From this it is plain, that triangles and parallelograms that have 
 equal altitudes, are to one another as their bases. 
 
 Let the figures be placed so as to have their bases in the same straight 
 line ; and having drawn perpendiculars from the vertices of the triangles to 
 the bases, the straight line which joins the vertices is parallel to that in 
 which their bases are (33. 1.), because the perpendiculars are both equal 
 and parallel to one another. Then, if the same construction be made as in 
 the proposition, the demonstration will be the same.
 
 OF GEOMETRY. BOOK VI. 
 
 123 
 
 PROP. II. THEOR. 
 
 lj a straight line be drawn parallel to one of the sides of a triangle, it wilt cut 
 the other sides, or the other sides produced, proportionally : And if the 
 sides, or the sides produced, be cut proportionally, the straight line which 
 joins the points of section will be parallel to the remaining side of the tri- 
 angle. 
 
 Let DE be drawn parallel to BC, one of the sides of the triangle ABC ■ 
 BD is to DA as CE to EA. 
 
 Join BE, CD ; then the triangle BDE is equal to the triangle CDE (37. 
 1.), because they are on the same base DE and between the same paral- 
 lels DE, BC : but ADE is another triangle, and equal magnitudes have, 
 to the same, the same ratio (7. 5.) ; therefore, as the triangle BDE to the 
 triangle ADE, so is the triangle CDE to the triangle ADE ; but as the 
 triangle BDE to the triangle ADE, so is (1. 6.) BD to DA, because, hav- 
 ing the same altitude, viz. the perpendicular drawn from the point E to AB, 
 they are to one another as their bases ; and for the same reason, as the 
 triangle CDE to the triangle ADE, so is CE to EA. Therefore, as BD 
 to DA, so is CE to EA (11. 5.). 
 
 Next, let the sides AB, AC of the triangle ABC, or these sides produced, 
 
 be cut proportionally in the points D, E, that is, so that BD be to DA, as 
 CE to EA, and join DE ; DE is parallel to BC. 
 
 The same construction being made, because as BD to DA, so is CE to 
 EA; and as BD to DA, so is the triangle BDE to the triangle \DE (1. 6.): 
 and as CE to EA, so is the triangle CDE to the triangle ADE ; therefore 
 the triangle BDE, is to the triangle ADE, as the triangle CDE to the tri- 
 angle ADE ; that is, the triangles BDE, CDE have the same ratio to the 
 triangle ADE ; and therefore (9. 5.) the triangle BDE is equal to the tri- 
 angle CDE : And they are on the same base DE ; but equal triangles on 
 the same base are between the same parallels (39. 1 .) ; therefore DE is 
 parallel to BC.
 
 124 
 
 ELEMENTS 
 
 PROP. III. THEOR. 
 
 If the angle of a triangle be bisected by a straight line which also cuts the base ; 
 the segments of the base shall have the same ratio which the other sides of 
 the triangle have to one another ; And if the segments of the base have the 
 same ratio which the other sides of the triangle have to one another, the straight 
 line drawn from the vertex to the point of section, bisects the vertical angle. 
 
 Let the angle BAC, of any triangle ABC, be divided into two equal an- 
 gles, by the straight line AD ; BD is to DC as BA to AC. 
 
 Through the point C draw CE parallel (Prop. 31. 1.) to DA, and let BA 
 produced meet CE in E. Because the straight line AC meets the paral- 
 lels AD, EC, the angle ACE is equal to the alternate angle CAD (29. 1.) : 
 But CAD, by the hypothesis, is equal to the angle BAD ; wherefore BAD 
 is equal to the angle ACE. Again, 
 because the straight line BAE meets 
 the parallels AD, EC, the exterior an- 
 gle BAD is equal to the interior and 
 opposite angle AEC ; But the angle 
 ACE has been proved equal to the an- 
 gle BAD ; therefore also ACE is 
 equal to the angle AEC, and conse- 
 quently the side AE is equal to the 
 side (6. 1.) AC. And because AD is 
 drawn parallel to one of the sides of 
 the triangle BCE, viz. to EC, BD is 
 to DC, as BA to AE (2. 6.) ; but AE is equal to AC ; therefore, as BD to 
 DC, so is BA to AC (7. 5.). 
 
 Next, let BD be to DC, as BA to AC, and join AD ; the angle BAC is 
 divided into two equal angles, by the straight line AD. 
 
 The same construction being made ■ because, as BD to DC, so is BA 
 to AC ; and as BD to DC, so is BA 
 to AE (2. 6.), because AD is paral- 
 lel to EC : therefore AB is to AC, as 
 AB to AE (11. 5.) : Consequently 
 AC is equal to AE (9. 5.), and the 
 angle AEC is therefore equal to the 
 angle ACE (5. 1.). But the angle 
 AEC is equal to the exterior and op- 
 posite angle BAD ; and the angle 
 ACE is equal to the alternate angle 
 CAD (29. 1.): Wherefore also the 
 angle BAD is equal to the angle 
 CAD : Therefore the angle BAC is cut into two equal angles by the straight 
 line AD.
 
 OF GEOMETRY. BOOK VI. 125 
 
 PROP. A. THEOR. 
 
 I c the exterior angle of a triangle be bisected by a straight line which also cuts 
 the base produced ; the segments between the bisecting line and the cxlretnities 
 of the base have the same ratio which the other sides of the triangles have tn 
 one another ; And if the segments of the base produced have the same ratio 
 which the other sides of the triangles have, the straight line, drawn from the 
 vertex to the point of section, bisects the exterior angle of the triangle. 
 
 Let the exterior angle CAE, of any triangle ABC, be bisected by the 
 straight line AD which meets the base produced in D ; BD is to DC, as 
 BA to AC. 
 
 Through C draw CF parallel to AD (Prop. 31. 1.) : and because the 
 straight line AC meets the parallels AD, FC, the angle ACF is equal to 
 the alternate angle CAD (29. 1): But CAD is equal to the angle DAE 
 (Hyp.) : therefore also DAE is equal to the angle ACF. Again, because 
 the straight line FAE meets the parallels AD, FC, the exterior angle DAE 
 is equal to the interior and opposite angle CFA; But the angle ACF has 
 been proved to be equal to the an- 
 gle DAE ; therefore also the angle 
 ACF is equal to the angle CFA, 
 and consequently the side AF is 
 equal to the side AC (6. 1.) ; and, 
 because AD is parallel to FC, a 
 side of the triangle BCF, BD is to 
 DC, as BA to AF (2. 6.) ; but AF 
 is equal to AC ; therefore as BD 
 is to DC, so is BA to AC. 
 
 Now let BD be to DC, as BA to AC, and join AD ; the angle CAD is 
 equal to the angle DAE. 
 
 The same construction being made, because BD is to DC as BA to AC ; 
 and also BD to DC, BA to AF (2. 6. ) ; therefore BA is to AC, as BA to 
 AF (11. 5.), wherefore AC is equal to AF (9. 5.), and the angle AFC 
 equal (5. 1.) to the angle ACF : but the angle AFC is equal to the exte- 
 rior angle EAD, and the angle ACF to the alternate angle CAD ; there- 
 fore also EAD is equal to the angle CAD 
 
 PROP. IV. THEOR. 
 
 The sidesabout the equalangles of equiangular triangles are proportion als ; ana 
 those which are opposite to the equal angles are homologous sides, that is, ate 
 the antecedents or consequents of the ratios 
 
 Let ABC, DCE, be equiangular triangles, having the angle ABC equal 
 to the angle DCE, and the angle ACB to the angle DEC, and conse- 
 quently (4. Cor. 32. 1.) the angle BAC equal to the angle CDE. The 
 sides about the equal angles of the triangles ABC, DCE are proportionals , 
 and those are the homologous sides which are opposite to the equal an- 
 gles.
 
 126 
 
 ELEMENTS 
 
 Let the triangle DCE be placed, so that its side CE may be contiguous 
 to BC, and in the same straight line with it : And because the angles ABO, 
 ACB are together less than two right angles (17. 1.), ABC and DEC, 
 which is equal to ACB, are also less than 
 two right angles : wherefore BA, ED pro- 
 duced shall meet (1 Cr. 29. 1.); let them be pro- 
 duced and meet in the point F ; and because 
 the angle ABC is equal to the angle DCE, 
 BF is parallel (28. 1.) to CD. Again, be- 
 cause the angle ACB is equal to the angle 
 DEC, AC is parallel to FE (28. 1 .) : There- 
 fore FACD is a parallelogram ; and conse- 
 quently AF is equal to CD, and AC to FD 
 (34. 1.) : And because AC is parallel to FE, 
 one of the sides of the triangle FBE, BA 
 AF is equal to CD ; therefore (7. 5.) BA 
 nately, BA : BC : : DC : CE (16. 5.) 
 BF, BC : CE : : FD 
 
 O E 
 
 CE (2. 6.) : but 
 
 B 
 
 AF : : BC 
 
 CD : : BC : CE ; and alter- 
 
 Again, because CD is parallel to 
 
 DE (2. 6.) ; but FD is equal to AC ; therefore BC 
 
 CE : : AC : DE ; and alternately, BC : CA : : CE : ED. Therefore 
 because it has been proved that AB : BC : : DC : CE ; and BC ■ CA 
 CE : ED, ex aequali, BA : AC : : CD : DE. 
 
 PROP. V. THEOR. 
 
 If the sides of tuio triangles, about each of their angles, be proportionals, the 
 triangles shall be equiangular, and have their equal angles opposite to the 
 homologous sides. 
 
 Let the triangles ABC, DEF have their sides proportionals, so that AB 
 is to BC, as DE to EF ; and BC to CA, as EF to FD ; and consequently 
 ex aequali, BA to AC, as ED to DF ; the triangle ABC is equiangular to 
 the triangle DEF, and their equal angles are opposite to the homologous 
 sides, viz. the angle ABC being equal to the angle DEF, and BCA to 
 EFD, and also BAC to EDF. 
 
 At the points E, F, in the straight 
 line EF, make (Prop. 23. l.)the an- 
 gle FEG equal to the angle ABC, 
 and the angle EFG equal to BCA, 
 wherefore the remaining angle BAC 
 is equal to the remaining angle 
 EGF (4. Cor. 32. 1.), and the trian- 
 gle ABC is therefore equiangular to 
 the triangle GfEF ; and consequently 
 they have tneir sides opposite to the 
 equal angles proportionals (4. 6.). 
 Wherefore, 
 
 AB : BC : : GE : EF ; but by supposition, 
 
 AB : BC : : DE : EF, therefore, 
 
 DE : EF : : GE : EF Therefore (11.5) DE and GE riav*
 
 OF GEOMETRY. BOOK VI. 12? 
 
 <ha same ratio to EF, and consequently are equal (9. 5.). For the same 
 reason, DF is equal to FG : And because, in the triangles DEF, GEF 
 DE is equal to EG, and EF common, and also the base DF equal to the 
 base GF ; therefore the angle DEF is equal (8. 1.) to the angle GEF, and 
 the other angles to the other angles, which are subtended by the equal 
 sides (4. 1.). "Wherefore the angle DFE is equal to the angle GFE, and 
 EDF to EGF: and because the angle DEF is equal to the angle GEF, 
 and GEF to the angle ABC ; therefore the angle ABC is equal to the an- 
 gle DEF : For the same reason, the angle ACB is equal to the angle 
 DFE, and the angle at A to the angle at D. Therefore the triangle ABC 
 is equiangular to the triangle DEF. 
 
 PROP. VI. THEOR. 
 
 If two triangles have one angle of the one equal to one angle of the other, and 
 the sides about the equal angles proportionals, the triangles shall be equian- 
 gular, and shall have those angles equal which are opposite to the homolo- 
 gous sides. 
 
 Let the triangles ABO, DEF have the angle BAC in the one equal to 
 the angle EDF in the other, and the sides about those angles proportion- 
 als ; that is, BA to AC, as ED to DF ; the triangles ABC, DEF are equi- 
 angular, and have the angle ABC equal to the angle DEF, and ACB to 
 DFE. 
 
 At the points D, F, in the 
 straight line DF, makb (Prop. 
 23. 1.) the angle FDG equal to 
 either of the angles BAC, EDF ; 
 and the angle DFG equal to the 
 angle ACB ; wherefore the re- 
 maining angle at B is equal to 
 the remaining one at G (4. Cor. 
 32. 1.), and consequently the 
 triangle ABC is equiangular to 
 the triangle DGF ; and therefore 
 
 BA : AC : : GD (4. 6.) : DF. But by hypothesis, 
 
 BA : AC : : ED : DF ; and therefore 
 
 ED : DF : : GD : (11. 5.) DF ; wherefore ED is equal (9. 5.) to 
 DG ; and DF is common to the two triangles EDF, GDF ; therefore the 
 two sides ED, DF are equal to the two sides GD, DF ; but the angle 
 EDF is also equal to the angle GDF; wherefore the base EF is equal to 
 the base FG (4. IX and the triangle EDF to the triangle GDF, and the 
 remaining angles to the remaining angles, each to each, which are sub- 
 tended by the equal sides : Therefore the angle DFG is equal to the angle 
 DFE, and the angle at G to the angle at E : But the angle DFG is equal 
 to the angle ACB ; therefore the angle ACB is equal the angle DFE, and 
 the angle BAC is equal to the ang e EDF (Hyp.) ; wherefore also the re- 
 maining angle at B is equal to the remaining anj/le at E. Therefore th« 
 triangle ABC is equiangu.ar to the triangle DEF
 
 128 ELEMENTS 
 
 PROP. VII. THEOR. 
 
 If two tiiangles have one angle of the one equal to one angle of the other , and 
 the sides about two other angles proportionals, then, if each of the remaining 
 angles be either less, or not less, than a right angle, the triangles shah be 
 equiangular, and Have tkise angles equal about which the sides are propor- 
 tionals. 
 
 Let the two triangles ABC, DEF have one angle in the one equal to one 
 angle in the other, viz. the angle BAC to the angle EDF, and the sides 
 about two other angles ABC, DEF proportionals, so that AB is to BC, as 
 DE to EF ; and, in the first case, let each of the remaining angles at C, F, 
 be less than a right angle. The triangle ABC is equiangular to the tri- 
 angle DEF, that is, the angle ABC is equal to the angle DEF, and the 
 remaining angle at C to the remaining angle at F. 
 
 For, if the angles ABC, DEF be not equal, one of them is greater than 
 the other : Let ABC be the greater, and at the point B, in the straight 
 line AB, make the angle ABG equal 
 to the angle (Prop. 23. l.)DEF : and 
 because the angle at A is equal to the 
 angle at D, and the angle ABG to 
 the angle DEF ; the remaining an- 
 gle AGB is equal (4. Cor. 32. 1.) to 
 the remaining angle DFE ; There- 
 fore the triangle ABG is equiangular 
 to the triangle DEF ; 
 
 wherefore (4. 6.), AB : BG : : DE : EF ; but, 
 by hypothesis, DE : EF : : AB : BC, 
 therefore, ' AB : BC : : AB : BG (11. 5.), 
 
 and because AB has the same ratio to each of the lines BC, BG ; BC is 
 equal (9. 5.) to BG, and therefore the angle BGC is equal to the angle 
 BCG (5. 1.) ; But the angle BCG is, by hypothesis, less than a right an- 
 gle ; therefore also the angle BGC is less than a right angle, and the adja- 
 cent angle AGB must be greater than a right angle (13. 1.). But it was 
 proved that the angle AGB is equal to the angle at F ; therefore the angle 
 at F is greater than a right angle : But by the hypothesis, it is less than a 
 right angle ; which is absurd. Therefore the angles ABC, DEF are not 
 unequal, that is, they are equal : And the angle at A is equal to the angle 
 at D ; wherefore the remaining angle at C is equal to the remaining angle 
 at F ; Therefore the triangle ABC is equiangular to the triangle DEF. 
 
 Next, let each of the angles at C, F be not less than a right angle ; the 
 triangle ABC is also, in this case, equiangular to the triangle DEF. 
 
 The same construction being 
 made, it may be proved, in like 
 manner, that BC is equal to BG, 
 and the angle at C equal to the 
 angle BGC : But the angle at C 
 is not less than a right angle ; 
 therefore the angle BGC is not 
 less than a right anelc : vVhere-
 
 UF GEOxMETRY. BOOK VI. 
 
 129 
 
 fore, two angles of the triangle BGC are together not less than two right 
 angles, which is impossible (17. 1.) ; and therefore the triangle ABC may 
 be proved to be equiangular to the triangle DEF, as in the first case. 
 
 PROP. VIII. THEOR. 
 
 In a *ight angled triangle if a perpendicular be drawn from the right angle k M 
 the base ; the triangles on each side of it are similar to the whole triangU, 
 and to one another. 
 
 Let ABC be a right angled triangle, having the right angle BAC ; and 
 from the point A let AD be drawn perpendicular to the base BC : the trian- 
 gles ABD, ADC are similar to the whole triangle ABC, and to one another. 
 
 Because the angle BAC is equal to the angle ADB, each of them being 
 a right angle, and the angle at B com- 
 mon to the two triangles ABC, ABD ; 
 the remaining angle ACB is equal to 
 the remaining angle BAD (4. Cor. 32. 
 1.): therefore the triangle ABC is 
 equiangular to the triangle ABD, and 
 the sides about their equal angles are 
 proportionals (4. 6.) ; wherefore the 
 triangles are similar (def. 1. 6.). In 
 like manner, it may be demonstrated, that the triangle ADC is equiangulai and 
 similar to the triangle ABC : and the triangles ABD, ADC, being each equi- 
 angular and similar to ABC, and equiangular and similar to one another. 
 
 Cor. From this it is manifest, that the perpendicular, drawn from the 
 right angle of a right angled triangle, to the base, is a mean proportional 
 between the segments of the base ; and also that each of the sides is a mean 
 proportional between the base, and its segment adjacent to that side. For 
 \n the triangles BDA, ADC, 
 
 
 BD i 
 
 DA : 
 
 : DA 
 
 DC (4. 6.) 
 
 and in the 
 
 triangles ABC, 
 
 BDA, BC 
 
 BA : 
 
 : BA 
 
 BD (4. 6.) ; 
 
 and in the 
 
 triangles ABC, 
 
 ACD, BC 
 
 CA : 
 
 : CA 
 
 CD (4. 6.). 
 
 
 PROP. IX. PROB. 
 
 From a given straight line to cut off any part required, that is, a part whtJi 
 shall be contained in it a given number of times. 
 
 Let AB be the given straight line ; it is required 
 to cut ofF from AB, a part which shall be contained 
 in it a given number of times. 
 
 From the point A draw a straight line AC mak- 
 ing any angle with AB ; and in AC take any point 
 D, and tako AC such that it shall contain AD, as 
 oft as A B is to contain the part, which is to be cut 
 off from it ; join BC, and draw DE parallel to it: 
 then A E is the part required to be cut off. 
 
 Became BD is parallel to one of the sides of the 
 
 17
 
 130 
 
 ELEMENTS 
 
 triangle ABC, viz. toBC, CD : DA : : BE : EA (2.6.) ; and by composi- 
 tion (18. 5), CA : AD : : BA : AE : But CA is a multiple of AD ; there- 
 fore (C. 5.) BA is the same multiple of AE, or contains AE the same num- 
 ber of times that AC contains AD ; and therefore, whatever part AD is of 
 AC, AE is the same of AB ; wherefore, from the straight line AB the par* 
 required is cut off. 
 
 PROP. X. PROB. 
 
 To divide a given straight line similarly to a given divided straight line, that is, 
 into parts that shall have the same ratios to one another which the parts of 
 the divided given straight line have. 
 
 Let AB be the straight line given to be divided, and AC the divided line, 
 it is required to divide AB similarly to AC. 
 
 Let AC be divided in the points D, E ; and let AB, AC be placed so as 
 to contain any angle, and join BC, and through the points D, E, draw 
 (Prop. 31. 1.) DF, EG, parallel to BC ; and 
 through D draw DHK, parallel to AB ; there- 
 fore each of the figures FH, HB, is a parallelo- 
 gram : wherefore DH is equal (34. 1.) to FG, 
 and HK to GB : and because HE is parallel 
 to KC, one of the sides of the triangle DKC, 
 CE : ED : : (2. 6.) KH : HD ; But KH=BG, 
 and HD = GF ; therefore CE : ED : : BG : 
 GF ; Again, because FD is parallel to EG, 
 one of the sides of the triangle AGE, ED : DA 
 : : GF : FA ; But it has been proved that CE 
 :ED: 
 to AC 
 
 B K O 
 
 BG : GF; therefore the given straight line AB is divided similarly 
 
 PROP. XI. PROB. 
 
 To find a third proportional to two given straight lines. 
 
 Let AB, AC be the two given straight lines, and let them be placed so 
 as to contain any angle ; it is required to 
 find a third proportional to AB, AC. 
 
 Produce AB, AC to the points D, E ; and 
 make BD equal to AC ; and having joined 
 BC, through D draw DE parallel to it (Prop. 
 31.1.) 
 
 Because BC is parallel to DE, a side of 
 the triangle ADE, AB : (2. 6.) BD . : AC : 
 CE ; but BD=AC: therefore AB : AC : : 
 AC : CE. Wherefore to the two given 
 Htraight lines A73, AC a third proportional, 
 OS is found.
 
 OF GEOMETRY. BOOK VI. 
 
 231 
 
 PROP. XII. PROB. 
 To find a founh proportional to three given straight lines. 
 
 Let A, B, C be the three given straight lines ; it is required to find a 
 fourth proportional to A, B, C. 
 
 Take two straight lines DE, DF, containing any angle EDF ; and upon 
 these make DG equal to A, GE equal to B, and DH equal to C ; and hay- 
 ing joined GH, draw EF parallel (Prop. 31. 1.) to it through the point E 
 
 And because GH is parallel to EF, one of the sides of the triangle DEF, 
 DG : GE : : DH : HF (2. 6.) ; but DG = A, GE=B, and DH=C ; and 
 therefore A : B : : C : HF. Wherefore to the three given straight lines, 
 A. B, C, a fourth proportional HF is found. 
 
 PROP. XIII. PROB. 
 To find a mean proportional between two given straight lines. 
 
 Let AB, BC be the two given straight lines ; it is required to find a mean 
 proportional between them. 
 
 Place AB, BC in a straight line, and upon AC describe the semicircle 
 ADC, and from the point B (Prop. 1 1. 
 I.) draw BD at right angles to AC, and 
 join AD, DC. 
 
 Because the angle ADC in a semi- 
 circle is a right angle (31. 3.) and be- 
 cause in the right angled triangle ADC, 
 DB is drawn from the right angle, per- 
 pendicular to the base, DB is a mean 
 oroportional between AB, BC, the seg- 
 ments of the base (Cor. 8. 6.) ; therefore between the two given straight 
 lines AB, BC, a mean proportional DK is foun'l
 
 (* c 
 
 132 ELEMENTS 
 
 PROP. XIV. PROB. 
 
 Equal parallelograms which have one angle of the one equal to one an^le of 
 the other, have their sides about the equal angles reciprocally proportional : 
 And parallelograms which have one angle of the one equal to one angle of 
 the other, and their sides about the equal angles reciprocally proportional, 
 are equal to one another. 
 
 Let AB, BC be equal parallel- 
 ograms, which have the angles at B 
 equal, and let the sides DB, BE be 
 placed in the same straight line ; 
 wherefore also FB, BG are in one 
 straight line (14. 1.) ; the sides of the 
 parallelograms AB, BC, about the 
 equal angles, are reciprocally propor- 
 tional ; that is, DB is to BE, as GB 
 toBF. 
 
 Complete the parallelogram FE ; and because the parallelograms AB, 
 BC are equal, and FE is another parallelogram, 
 
 AB : FE : : BC : FE (7. 5.) : 
 but because the parallelograms AB, FE have the same altitude, 
 AB : FE : : DB : BE (1. 6.), also, 
 BC : FE : : GB : BF (1. 6.) ; therefore 
 DB : BE : : GB : BF (11. 5.). Wherefore, the sides 
 of the parallelograms AB, BC about their equal angles are reciprocally pro- 
 portional. 
 
 But, let the sides about the equal angles be reciprocally proportional, viz 
 as DB to BE, so GB to BF ; the parallelogram AB is equal to the parallel 
 ogram BC. 
 
 Because DB : BE : : GB : BF, and DB : BE : : AB : FE, and GB : 
 BF : : BC : EF, therefore, AB : FE : : BC : FE (11. 5.) : wherefore the 
 parallelogram AB is equal (9. 5.) to the parallelogram BC. 
 
 PROP. XV. THEOR. 
 
 Equal triangles which have one angle of the one equal to one angle of the 
 other have their sides about the equal angles reciprocally proportional ; And 
 triangles which have one angle in the one equal to one angle in the other, 
 and their sides about the equal angles reciprocally proportional, are equal 
 to one another. 
 
 Let ABC, ADE be equal triangles, which have the angle BAC equal to 
 the angle DAE : the sides about the equal angles of the triangles are re- 
 ciprocally proportional ; that is, CA is to AD, as EA to AB. 
 
 Let the triangles be placed so that their sides CA, AD be in one straight 
 tine ; wherefore also EA and AB are in one straight line (14. 1.) ; join BD. 
 Because the triangle ABC is equal to the triangle ADE, and ABD is an- 
 other triangle ; therefore, triangle CAB : triangle BAD : : triangle E A.D
 
 OF GEOMETRY. BOOK VI. 
 
 133 
 
 mangle BAD ; but CAB : 
 BAD ::CA : AD, and EAD: 
 BAD : : EA : AB ; therefore 
 CA: AD:: EA: AB(ll.o), 
 wherefore the sides of the trian- 
 gles ABC, ADE about the equal 
 angles are reciprocally propor- 
 tional. 
 
 But let the sides of the trian- 
 gles ABC, ADE, about the 
 equal angles be reciprocally 
 proportional, viz. CA to AD, as 
 EA to AB ; the triangle ABC is 
 equal to the triangle ADE. 
 
 Having joined BD as before ; because CA : AD : : EA : AB ; and since 
 CA : AD : : triangle ABC : triangle BAD (1. 6.) ; and also EA : AB : : 
 triangle EAD : triangle BAD (11. 5.) ; therefore, triangle ABC : triangle 
 BAD : : triangle EAD : triangle BAD ; that is, the triangles ABC, EAD 
 have the same ratio to the triangle BAD ; wherefore the triangle ABC i8 
 equal (9. 5.) to the triangle EAD. 
 
 PROP. XVI. THEOR. 
 
 If four straight lines be proportionals, the rectangle contained by the extremes is 
 equal to the rectangle contained by the means; And if the rectangle contained 
 by the extremes be equal to the rectangle contained by the means, the four 
 straight lines are proportionals. 
 
 Let the four straight lines, AB, CD, E, F, be proportionals, viz. as AB 
 to CD, so E to F ; the rectangle contained by AB, F is equal to the rect 
 angle contained by CD, E. 
 
 From the points A, C draw (11. l.)AG, CH at right angles to AB, CD ; 
 and make AG equal to F, and CH equal to E, and complete the parallel- 
 ograms BG, DH. Because AB : CD : : E : F ; and since E=CH, and 
 F=AG, AB : CD (7. 5.) : : CH : AG ; therefore the sides of the parallel- 
 ograms BG, DH about the equal angles are reciprocally proportional ; but 
 parallelograms which have their sides about equal angles reciprocally pro- 
 portional, are equal to one another (14. 6.); therefore the parallelogram 
 
 BG is equal to the parallelogram DH : TT 
 
 and the parallelogram BG is contain- 
 ed by the straight lines AB, F ; be- JT- 
 cause AG is equal to F ; and the pa- 
 rallelogram DH is contained by CD 
 and E, because CH is equal to E : 
 therefore the rectangle contained by 
 the straight lines AB, F is equal to that 
 which is contained by CD and E. 
 
 And if the rectangle contained by 
 ihe straight lines AB, F be equal to that which is contained by CD,E ; 
 •heae four lines are proportionals, <iz. AB is to CD as E to F-
 
 134 ELEMENTS 
 
 The same construction being made, because the rectangle contained by 
 the straight lines AB, F is equal to that which is contained by CD, E, and 
 the rectangle BG is contained by AB, F, because AG is equal to F ; and 
 the rectangle DH, by CD, E, because CH is equal to E ; therefore the pa- 
 rallelogram BG is equal to the parallelogram DH, and they are equiangu- 
 lar : but the sides about the equal angles of equal parallelograms are reci- 
 procally proportional (14. 6.) : wherefore AB : CD : : CH : AG ; but CH 
 =E, and AG=F; therefore AB : CD : : E : F. 
 
 PROP. XVII. THEOR. 
 
 If three straight lines be proportionals, the rectangle contained by the extremes is 
 equal to the square of the mean : And if the rectangle contained by the ex- 
 tremes be equal to the square of the mean, the three straight lines arepropor- 
 tionals. 
 
 Let the three straight lines, A, B, C be proportionals, viz. as A to B, so 
 B to C ; the rectangle contained by A, C is equal to the square of B. 
 
 Take D eqaal to B : and because as A to B, so B to C, and that B is 
 equal to D ; A is (7. 5.) to B, as D to C : but if four straight lines be pro- 
 portionals, the rectangle contained by the extremes is equal to that which 
 is contained by the means (16. 6.) ; therefore the 
 rectangle A.C = the rectangle B.D ; but the rect- fr ' ~" 
 
 angle B.D is equal to the square of B, because B= ~* 
 D ; therefore the rectangle A.C is equal to the p; 
 square of B. " 
 
 And if the rectangle contained by A, C be equal to the square of h ; A : 
 B : : B : C. 
 
 The same construction being made, because the rectangle contained by 
 A, C is equal to the square of B, and the square of B is equal to the rect- 
 angle contained by B, D, because B is equal to D ; therefore the rectangle 
 contained by A, C is equal to that contained by B, D ; but if the rectangle 
 contained by the extremes be equal to that contained by the means, the 
 four straight lines are proportionals (16. 6.) : therefore A : B : : D C, but 
 B=D t ; wherefore A : B : : B : C. 
 
 PROP. XVIII. PROS. 
 
 Upon a given straight line to describe a rectilineal figure similar, and similarly 
 situated to a given rectilineal figure. 
 
 Let AB be the given straight line, and CDEF the given rectilineal figure 
 of four sides ; it is required upon the given straight line AB to deccribe a 
 rectilineal figure similar, and similarly situated to CDEF. 
 
 Join DF, and at the points A, B in the straight line AB, make (Prop. 23. 
 1.) the angle BAG equal to the angle at C, and the angle ABG equal to 
 the angle CDF ; therefore the remaining angle CFD is equa/ to the re- 
 maining angle AGB (4. Cor. 32. 1.) : wherefore the triangle FCD is equi- 
 angular to the triangle GAB : Again, at the points G, B in the straight 
 line GB make (Prop. 23. 1.) the angle BGH equal to the angle DFE, and 
 ♦he angle GBH equal to FDE ; therefore the remaining angle FED is
 
 OF GEOMETRY. BOOK VI. 
 
 135 
 
 equal to the remaining angle GHB, and the triangle FDE equiangular to 
 the triangle GBH : then, because the angle AGB is equal to the angle 
 CFD BGH to DFE the whole angle AGH is equal to the whole CFE 
 
 L F 
 
 for the same reason, the angle ABH is equal to the angle CDE ; also the 
 angle at A is equal to the angle at C, and the angle GHB to FED ; There- 
 fore the rectilineal figure ABHG is equiangular to CDEF : but likewise 
 these figures have their sides about the equal angles proportionals : for the 
 triangles GAB, FCD being equiangular, 
 
 BA : AG : : DC : CF (4. 6.) ; for the same reason, 
 AG : GB : : CF : FD ; and because of the equian- 
 gular triangles BGH, DFE, GB : GH : : FD : FE ; therefore, 
 
 ex aequali (22. 5.) AG : GH : : CF : FE. 
 In the same manner, it may be proved, that 
 
 AB : BH : : CD : DE. Also (4. 6.), 
 GH : HB : : FE : ED. Wherefore, because the rectili- 
 neal figures ABHG, CDEF are equiangular, and have their sides about 
 the equal angles proportionals, they are similar to one another (def. 1. 6.). 
 
 Next, Let it be required to describe upon a given straight line AB, a 
 rectilineal figure similar, and similarly situated to the rectilineal figure 
 CDKEF. 
 
 Join DE, and upon the given straight line AB describe the rectilineal 
 figure ABHG similar, and similarly situated to the quadrilateral figure 
 CDEF, by the former case ; and at the points B, H in the straight line 
 BH, make the angle HBL equal to the angle EDK, and the angle BHL 
 equal to the angle DEK ; therefore the remaining angle at K is equal to 
 the remaining angle at L; and because the figures ABHG, CDEF are 
 similar, the angle GHB is equal to the angle FED, and BHL is equal to 
 DEK ; wherefore the whole angle GHL is equal to the whole angle FEK ; 
 for the same reason the angle ABL is equal to the angle CDK : therefore 
 the five-sided figures AGHLB, CFEKD are equiangular ; and because 
 the figures AGHB, CFED are similar, GH is to HB as FE to ED ; and 
 as HB to HL, so is ED to EK (4. G.) ; therefore, ex jrquali (22. 5 ), GH 
 is to HL, as FE to EK : for the same reason, AB is to BL, as CD to DK • 
 and BL is to LH, as (4. 6.) DK to KE, because the triangles BLH, DK K 
 are equiangular : therefore, because the five-sided figures AGHLB 
 CFEKD are equiangular, and have their sides about the equal angles pro- 
 portionals, they are similar to one another ; and in the same manner a rec-
 
 (36 
 
 ELEMENTS 
 
 tilineal figure 01 six, or more, sides may be described upon a given straight 
 line similar to one given, and so on. 
 
 PROP. XIX. THEOR. 
 
 Similar triangles are to one another in the duplicate ratio of the homologous 
 
 sides. 
 
 Let ABC, DEF be simi- 
 lar triangles, having the an- 
 gle B equal to the angle E, 
 and let AB be to BC, as 
 DE to EF, so that the side 
 BC is homologous to EF 
 (def. 13. 5.) : the triangle 
 ABC has to the triangle 
 DEF, the duplicate ratio 
 of that which BC has to 
 EF. 
 
 Take BG a third proportional to BC and EF (11. 6.), or such that 
 BC : EF . : EF : BG, and join GA. Then, because 
 AB : BC : : DE : EF, alternately (16. 5.), 
 AB : DE : : BC : EF ; but 
 BC : EF : : EF : BG ; therefore (11. 5.) 
 AB : DE :: EF : BG ; wherefore the sides of the triangles 
 ABG, DEF, which are about the equal angles, are reciprocally propor- 
 tional ; but triangles, which have the sides about two equal angles recipro- 
 cally proportional, are equal to 
 
 one another (15. 6.) : therefore A 
 
 the triangle ABG is equal to 
 the triangle DEF; and because 
 
 that BC is to EF, as EF to / / \ 
 
 BG ; and that if three straight 
 lines be proportionals, the first 
 has to the third the duplicate 
 ratio of that which it has to the 
 
 second ; BC therefore has to B G C ID F 
 
 BG the duplicate ratio of that which BC has to EF. But as BC to BG 
 so is (1. 6.) the triangle ABC to the triangle ABG : therefore the triangle 
 ABC has to the triangle ABG the duplicate ratio of that which BC has to 
 EF : and the triangle ABG is equal to the triangle DEF ; wherefore also 
 the triangle ABC has to the triangle DEF the duplicate ratio of that which 
 BC has to EF. 
 
 Cor. From this, it is manifest, that if three straight lines be propor- 
 tionals, as the first is to the third, so is any trianglo upon the first to a 
 similar, and similarly described triangle upon the second.
 
 OF GEOMETRY. BOOK VI. 13? 
 
 PROP. XX. THEOR. 
 
 Similar polygons may be divided into the same number of similar triangles, hav- 
 ing the same ratio to one another that the polygons have ; and the polygons 
 have to one another the duplicate ratio of that which their homologous sides 
 have. 
 
 Let ABCDE, FGHKL, be similar polygons, and let AB be the homo- 
 logous side to FG: the polygons ABCDE, FGHKL, may be divided into 
 the same number of similar triangles, whereof each has to each the same 
 ratio which the polygons have ; and the polygon ABCDE has to the poly- 
 gon FGHKL a ratio duplicate of that which the side AB has to the sid« 
 FG. 
 
 Join BE, EC, GL, LH : and because the polygon ABCDE is similar 
 to the polygon FGHKL, the angle BAE is equal to the angle GFL (def. 
 1. 6.), and BA : AE : : GF : FL (def. 1.6.): wherefore, because the tri- 
 angles ABE, FGL have an angle in one equal to an angle in the other 
 and their sides about these equal angles proportionals, the triangle ABE is 
 equiangular (6. 6.), and therefore similar, to the triangle FGL (4. 6.) : 
 wherefore the angle ABE is equal to the angle FGL : and, because the 
 polygons are similar, the whole angle ABC is equal (def. 1. 6.) to the whole 
 angle FGH ; therefore the remaining angle EBC is equal to the remain- 
 ing angle LGH : now because the triangles ABE, FGL are similar, 
 
 EB : BA : : LG : GF; and also because the 
 polygons are similar, AB : BC : : FG : GH (def. 1.6.); therefore, ex 
 eequali (22. 5.) EB : BC : : LG : GH, that is, the sides about the equal 
 angles EBC, LGH are proportionals ; therefore (6. 6.) the triangle EBC 
 
 is equiangular to the triangle LGH, and similar to it (4. 6.). For the 
 »ame reason, the triangle ECD is likewise similar to the triangle LHK ; 
 therefore the similar polygons ABCDE, FGHKL are divided into the same 
 number of similar triangles. 
 
 Also these triangles have, each to each, the same ratio which the poly- 
 gons have to one another, the antecedents being ABE, EBC, ECD, and 
 the consequents FGL, LGH, LHK : and the polygon ABCDE has to the 
 polygon FGHKL the duplicate ratio of that which the side AB has to the 
 homologous side FG. 
 
 Because the triangle ABE is similar to ihe triangle FGL, ABE has to 
 FGL the duplicate ratio (19. 6.) of that wl ich the side BE has to the side 
 
 18
 
 138 ELEMENTS 
 
 GL • foi the same reason, the triangle BEC has to GLH the duplicate 
 ratio of that which BE has to GL : therefore, as the triangle ABE to the 
 triangle FGL, so (1 1. 5.) is the triangle BEC to the triangle GLH. Again 
 because the triangle EBC is similar to the triangle LGH, EBC has to 
 LGH the duplicate ratio of that which the side EC has to the side LH : 
 for the same reason, the triangle ECD has to the triangle LHK, the du- 
 plicate ratio of that which EC has to LH : therefore, as the triangle EBC 
 to the triangle LGH, so is (11. 5.) the triangle ECD to the triangle LHK : 
 but it has been proved, that the triangle EBC is likewise to the trianglo 
 LGH, as the triangle ABE to the triangle FGL. Therefore, as the trian- 
 gle ABE is to the triangle FGL, so is the triangle EBC to the triangle 
 LGH, and the triangle ECD to the triangle LHK : and therefore, as one 
 of the antecedents to one of the consequents, so are all the antecedents tc 
 all the consequents (12. 5.). Wherefore, as the triangle ABE to the tri 
 
 angle FGL, so is the polygon ABCDE to the polygon FGHKL: but the 
 triangle ABE has to the triangle FGL, the duplicate ratio of that which 
 the side AB has to the homologous side FG. Therefore also the polygon 
 ABCDE has to the polygon FGHKL the duplicate ratio of that which 
 AB has to the homologous side FG. 
 
 Cor. 1. In like manner it may be proved, that similar figures of four 
 sides, or of any number of sides, are one to another in the duplicate ratio of 
 their homologous sides, and the same has already been proved of triangles : 
 therefore, universally, similar rectilineal figures are to one another in the 
 duplicate ratio of their homologous sides. 
 
 Cor. 2. And if to AB, FG, two of the homologous sides, a third pro- 
 portional M be taken, AB has (def. 11. 5.) to M the duplicate ratio of that 
 which AB has to FG : but the four-sided figure, or polygon, upon AB has 
 to th3 four-sided figure, or polygon, upon FG likewise the duplicate ratio 
 of that which AB has to FG : therefore, as AB is to M, so is the figure 
 upon AB to the figure upon FG, which was also proved in triangles (Cor. 
 19. 6.). Therefore, universally, it is manifest, that if three straight lines 
 be proportionals, as the first to the third, so is any rectilineal figure upon 
 the first, to a similar, and similarly described rectilineal figure upon the se- 
 cond. 
 
 Cor. 3. Because all squares are similar figures, the ratio of any two 
 squares to one another is the same with the duplicate ratio of their sides. ; 
 and hence, also, any two similar rectilineal figures are to one another as ihe 
 squares of their homologous sides.
 
 OF GEOMETRY. BOOK VI. 139 
 
 SCHOLIUM. 
 
 If t#o polygons are composed of the same number of triangles similar, 
 and similarly situated, those two polygons will be similar. 
 
 For the similarity of the two triangles will give the angles EAB=LFG 
 ABE=FGL,EBC=LGH: hence, ABC=FGH, likewise BCD=GHK 
 &c. Moreover, we shall have, EA : LF : : AB : FG : : EB : LG : : BC 
 : GH, &c. ; hence the two polygons have their angles equal and their sides 
 proportional ; consequently they are similar. 
 
 PROP. XXI THEOR. 
 
 Rectilineal figures which are similar to the same rectilineal figure, are also 
 similar to one another. 
 
 Let each of the rectilineal figures A, B be similar to the rectilineal figure 
 C : The figure A is similar to the figure B. 
 
 Because A is similar to C, they are equiangular, and also have their 
 sides about the equal angles proportionals (def. 1. 6.). Again, because B 
 is similar to C, th^.y are equiangular, and have their sides about the equal 
 angles proportionals (def. 1.6.): therefore the figures A, B, are each of 
 
 them equiangular to C, and have the sides about the equal angles of each 
 of them, and of C, proportionals. Wherefore the rectilineal figures A and 
 B are equiangular (1. Ax. 1.), and have their sides about the equal angles 
 proportionals (11. 5.). Therefore A is similar (def. 1. 6.) to B. 
 
 PROP. XXII THEOR. 
 
 If four straight lines be proportionals, the similar rectilineal figures similarly 
 described upon them shall also be proportionals ; and if the similar rectilineal 
 figures similarly described upon four straight lines be proportionals, those 
 straight lines shall be proportionals. 
 
 Let the four straight lines, AB, CD, EF, GH be proportionals, viz. AB 
 to CD, as EF to GH, and upon AB, CD let the similar rectilineal figures 
 KAB, LCD be similarly described ; and upon EF, GH the similar recti- 
 lineal figures MF, NH, in like manner : the rectilineal figure KAB is to 
 LCD, as MF to NH. 
 
 m To AB, CD take a third proportional (11. 6.) X ; and to EF, GH.s 
 tiiird proportional ; and because
 
 140 
 
 ELEMENTS 
 
 AB : CD : : EF : GH, and 
 
 CD : X : : GH : (1 1. 5.) 0, ex sequali (22. 5.) 
 
 AB : X : : EF : O. But 
 
 AB : X (2. Cor. 20. 6.) : : KAB : LCD ; and 
 
 EF : O : : (2. Cor. 20. 6.) MF : NH ; therefore 
 KAB : LCD (2. Cor. 20. 6.) : : MF : NH. 
 
 And if the figure KAB be to the figure LCD, as the figure MF to tho 
 figure NH, AB is to CD, as EF to GH. 
 
 Make (12. 6.) as AB to CD, so EFto PR, and upon PR describe (18. 
 6.) the rectilineal figure SR similar, and similarly situated to either of the 
 
 E T G H O 'JF~ H 
 
 figures MF, NH : then, because that as AB to CD, so is EF to PR, 
 
 and 
 
 upon AB, CD are described the similar and similarly situated rectilineals 
 KAB, LCD, and upon EF, PR, in like manner, the similar rectilineals 
 MF, SR ; KAB is to LCD, as MF to SR ; but by the hypothesis, KAB 
 is to LCD, as MF to NH ; and therefore the rectilineal MF having the 
 same ratio to each of the two NH, SR, these two are equal (9. 5.) to one 
 another ; they are also similar, and similarly situated ; therefore GH is 
 equal to PR : and because as AB to CD, so is EF to PR, and because PR 
 is equal to GH, AB is to CD, as EF to GH. 
 
 PROP. XXIII. THEOR. 
 
 Equiangular parallelograms have to one another the ratio which is compounded 
 of the ratios of their sides. 
 
 Let AC, CP be equiangular parallelograms having the angle BCD 
 equal to the angle ECG ; the ratio of the parallelogram AC to the paral 
 lelogram CF, is the same with the ratio which is compounded of the ratiof 
 of their sides. 
 
 Let BC, CG be placed in a straight line ; therefore DC and CE are also 
 in a straight line (14. 1.); complete the parallelogram DG ; and, taking 
 any straight line K, make (12. 6.) as BC to CG, so K to L ; and as DC 
 to CE, so make (12. 6.) L to M : therefore the ratios of K to L, and L to 
 M, are the same with the ratios of the sides, viz. of BC to CG, and of DC 
 to CE- But the ratio of K to M, is that which is said to be compounded 
 (def. 10. 5.) of the ratios of K to L, and L to M ; wherefore also K has to
 
 OF GEOMETRY. BOOK "VI. 
 
 141 
 
 M the ratio compounded of the ratios of 
 the sides of the parallelograms. Now, 
 because as BC to CG, so is the parallel- 
 ogram AC to the parallelogram CH (1. 
 6.) ; and as BC to CG, so is K to L ; 
 therefore K is ( 1 1 . 5.) to L, as the paral- 
 lelogram AC to the parallelogram CH : 
 again, because as DC to CE, so is the 
 parallelogram CH to the parallelogram 
 CF : and as DC to CE, so is L to M ; 
 therefore L is (1 1. 5.) to M, as the paral- 
 lelogram CH to the parallelogram CF : 
 therefore, since it has been proved, that 
 as K to L, so is the parallelogram AC 
 to the parallelogram CH ; and as L to M, so the parallelogram CH to the 
 parallelogram CF ; ex aequali (22. 5.), K is to M, as the parallelogram 
 AC to the parallelogram CF ; but K has to M the ratio which is com- 
 pounded of the ratios of the sides ; therefore also the parallelogram AC 
 has to the parallelogram CF the ratio which is compounded of the ratios 
 of the sides. 
 
 Cor. Hence, any two rectangles are to each other as the products of 
 their bases multiplied by their altitudes. 
 
 SCHOLIUM. 
 
 Hence the product of the base by the altitude may be assumed as the 
 neasure of a rectangle, provided we understand by this product the pro- 
 duct of two numbers, one of which is the number of linear units contained 
 in the base, the other the number of linear units contained in the altitude. 
 
 Still this measure is not absolute but relative : it supposes that the area 
 of any other rectangle is computed in a similar manner, by measuring its 
 sides with the same linear unit ; a second product is thus obtained, and 
 the ratio of the two products is the same as that of the two rectangles, 
 agreeably to the proposition just demonstrated. 
 
 For example, if the base of the rectangle A contained three units, and its 
 altitude ten, that rectangle will be represented by the number 3x10, or 
 30, a number which signifies nothing while thus isolated ; but if there is a 
 second rectangle B, the base of which contains twelve units, and the alti- 
 tude seven, this rectangle would be represented by the number 12 x 7=84 ; 
 and we shall hence be entitled to conclude that the two rectangles are to 
 each other as 30 is to 84 ; and therefore, if the rectangle A were to be as- 
 sumed as the unit of measurement in surfaces, the rectangle B would then 
 have |^ for its absolute measure ; or, which amounts to the same thing, it 
 would be equal to |^ of a superficial unit. 
 
 It is more common and more simple to assume the squares as the unit of 
 surface ; and to select that square whose side is the unit of length. In 
 this case, the measurement which we have regarded merely as relative, 
 becomes absolute : tho number 30, for instance, by which the rectangle A 
 M as measured, now represents 30 superficial units, or ?>0 of those squares 
 which have each of their sides equal to unity.
 
 142 ELEMENTS 
 
 Cop 1. Hence, the area of any parallelogram is equal to the product of 
 its bas ; by its altitude. 
 
 Cor. 2. It likewise follows, that the area of any triangle is equal to the 
 product of its base by half its altitude. 
 
 . PROP. XXIV. THEOR. 
 
 The parallelograms about the diameter of any parallelogram, are similar to the 
 whole, and to one another. 
 
 Let ABCD be a parallelogram, of which the diameter is AC ; and EG, 
 HK the parallelograms about the diameter: the parallelograms EG, HK 
 are similar, both to the whole parallelogram ABCD, and to one another. 
 
 Because DC, GF are parallels, the angle ADC is equal (29. 1.) to the 
 angle AGF : for the same reason, because BC, EF are parallels, the an- 
 gle ABC is equal to the angle AEF : and each of the angles BCD, EFG 
 is equal to the opposite angle DAB (34. 1.), and therefore are equal to one 
 another, wherefore the parallelograms ABCD, AEFG are equiangular 
 And because the angle ABC is equal to the angle AEF, and the angle 
 BAC common to the two triangles BAC, 
 EAF, they are equiangular to one another ; 
 therefore (4. 6.) as AB to BC, so is AE to 
 EF ; and because the opposite sides of paral- 
 lelograms are equal to one another (34. 1.), 
 AB is (7. 5.) to AD, as AE to AG ; and DC 
 to CB, as GF to FE ; and also CD to DA. 
 as FG to G A : therefore the sides of the pa- 
 rallelograms ABCD, AEFG about the equal 
 angles are proportionals ; and they are 
 therefore similar to one another (def. 1.6.); for the same reason, the pa- 
 rallelogram ABCD is similar to the parallelogram FHCK. Wherefore 
 each of the parallelograms, GE, KH is similar to DB : but rectilinea. 
 figures which are similar to the same rectilineal figure, are also similar »c 
 one another (21. 6.) ; therefore the parallelogram GE is similar to KH. 
 
 PROP. XXV. PROB. 
 
 To describe a rectilineal figure which shall be similar to one, and equal to 
 another given rectilineal figure. 
 
 Let ABC be the given rectilineal figure, to which the figure to be de- 
 scribed is required to be similar, and D that to which it must be equal. It 
 is required to describe a rectilineal figure similar to ABC, and equal to D. 
 
 Upon the straight line BC describe (Cor. Prop. 45. 1.) the parallelogram 
 BE equal to the figure ABC ; also upon CE describe (Cor. Prop. 45 1.) 
 the parallelogram CM equal to D, and having the angle FCE equal to the 
 angle CBL : therefore BC and CF are in a straight line (29.1. or 14.1.), as 
 also LE and EM ; between BC and CF find (13. 6.) a mean proportional 
 GH, and upon GH describe (18. 6.) the rectilineal figure KGH similar, 
 and similarly situated, to the figure ABC. And because BC is to GH as
 
 OF GEOMETRY. BOOK VL 
 
 143 
 
 GH to CF, and if three straight lines be proportionals, as the first is to th*> 
 third, so is (2. Cor. 20. 6.) the figure upon the first to the similar and simi 
 larly described figure upon the second ; therefore as BC to CF, so is the 
 
 figure ABC to the figure KGH : but as BC to CF, so is (1. 6.) the paral 
 lelogram BE to the parallelogram EF : therefore as the figure ABC s to 
 the figure KGH, so is the parallelogram BE to the parallelogram EF (11 
 5.): but the rectilineal figure ABC is equal to the parallelogram BE ; there 
 fore the rectilineal figure KGH is equal (14. 5.) to the parallelogram EF : 
 but EF is equal to the figure D ; wherefore also KGH is equal to D ; and 
 it is similar to ABC. Therefore the rectilineal figure KGH has been de- 
 scribed similar to the figure ABC, and equal to D. 
 
 PROP. XXVI. THEOR. 
 
 If two similar parallelograms have a common angle, and be similarly situated, 
 they are about the same diameter. 
 
 Let the parallelograms ABCD, AEFG be similar and similarly situated, 
 and have the angle DAB common; ABCD and AEFG are about the 
 same diameter. 
 
 For, if not, let, if possible, the parallelogram 
 BD have its diameter AHC in a different 
 straight line from AF, the diameter of the pa- 
 rallelogram EG, and let GF meet AHC in H ; 
 and through H draw HK parallel to AD or 
 BC ; therefore the parallelograms ABCD, 
 AKHG being about the same diameter, are 
 similar to one another (24. 6.) : wherefore, as 
 DA to AB, so is (def. 1, 6.) GA to AK; but 
 because ABCD and AEFG are similar paral- 
 lelograms, as DA is to AB, so is GA to AE ; therefore (11. 5.) as GA to 
 AE, so GA to AK ; wherefore GA has the same ratio to each of the straight 
 lines AE, AK ; and consequently AK is equal (9. 5.) to AE, the less to 
 the greater, which is impossible ; therefore ABCD and AKHG are not 
 about the same diameter ; wherefore ABCD and AEFG must be about 
 the same diameter.
 
 144 ELEMENTS 
 
 PROP. XXVII. THEOR. 
 
 Of all the rectangles contained by the segments of a given straight line, the 
 greatest is the square which is described on half the line. 
 
 Let AB be a given straight line, which is bisected in C ; and let D be 
 
 any point in it, the square on AC is greater 
 
 than the rectangle AD, DB. A C D B 
 
 For, since the straight line AB is divided into two equal parts in C, and 
 into two unequal parts in D, the rectangle contained by AD and DB, to- 
 gether with the square of CD, is equal to the square of AC (5. 2.). The 
 square of AC is therefore greater than the rectangle AD.DB. 
 
 PROP. XXVIII. PROB. 
 
 To divide a given straight line, so that the rectangle contained by its segments 
 may be equal to a given space ; but that space must not be greater than the 
 square of half the given line. 
 
 Let AB be. the given straight line, and let the square upon the given 
 straight line C be the space to which the rectangle contained by the seg- 
 ments of AB must be equal, and this square, by the determination, is not 
 greater than that upon half the straight line AB. 
 
 Bisect AB in D, and if the square upon AD be equal to the square upon 
 C, the thing required is done : But if it be not equal to it, AD must be 
 greater than C, according to the deter- 
 mination : Draw DE at right angles to 
 AB, and make it equal to C : produce 
 ED to F, so that EF be equal to AD 
 or DB, and from the centre E, at the 
 distance EF, describe a circle meeting 
 AB in G. Join EG ; and because AB -A- x s v ^ ^^tjr B 
 
 is divided equally in D, and unequally ~~^ 
 
 in G, AG.GB + DG 2 =(5. 2.) DB 2 = ■ 
 
 EG 2 . But (47. 1.) ED 2 +DG 2 =EG 2 ; therefore, AG.GB + DG 2 =ED 2 
 4-DG 2 , and taking away DG 2 , AG.GB=ED 2 . Now ED=C, therefore 
 the rectangle AG.GB is equal to the square of C : and the given line AB 
 is divided in G, so that the rectangle contained by the segments AG, GB 
 is equal to the square upon the given straight line C. 
 
 PROP. XXIX. PROB. 
 
 To produce a given straight line, so that the rectangle contained by the segmenti 
 between the extremities of the given line, and the points to which it is pro- 
 duced, may be equal to a given space. 
 
 Let AB be the given straight line, and let the square upon the given 
 straight line C be the space to which the rectangle under the segments of 
 \B produced, must be equal.
 
 OF GEOMETRY. BOOK VI. 
 
 145 
 
 Bisect AB in D, and draw BE at right angles to it, so that BE be equ». 
 to C ; and having joined DE, from the centre D at the distance DE de 
 scribe a circle meeting AB produced in G. 
 And because AB is bisected in D, and 
 produced to G, (6. 2.) AG.GB+DB 2 = 
 DG 2 =DE 2 . 
 
 But (47. 1.) DE 2 =DB 2 +BE 2 , there- 
 fore AG.GB + DB 2 = DB 2 + BE 2 , and 
 AG.GB=BE 2 . Now, BE = C ; where- 
 fore the straight line AB is produced to 
 G, so that the rectangle contained by the 
 segments AG, GB of the line produced, 
 is equal to the square of C. 
 
 PROP. XXX. PROB. 
 
 R 
 
 To cut a given straight line in extreme and mean ratio. 
 
 Let AB be the given straight line ; it is required to cut it in extreme zaui 
 mean ratio. 
 
 Upon AB describe (Prop. 46. l.)the square BC, and produce CA to D, 
 so that the rectangle CD.DA may be equal to the square CB (29. 6.). 
 Take AE equal to AD, and complete the rectangle DF under DC and 
 AE, or under DC and DA. Then, because the 
 rectangle CD.DA is equal to the square CB, the 
 rectangle DF is equal to CB. Take away the 
 common part CE from each, and the remainder 
 FB is equal to the remainder DE. But FB is 
 the rectangle contained by FE and EB, that is, 
 by AB and BE ; and DE is the square upon AE ; 
 therefore AE is a mean proportional between 
 AB and BE (17. 6.), or AB is to AE as AE to EB. 
 But AB is greater than AE ; wherefore AE is 
 greater than EB (14. 5.): Therefore the straight 
 line AB is cut in extreme and mean ratio in E (def. 
 3. 6.). 
 
 Otherwise. 
 
 Let AB be the given straight line; it is required to cut it in exhume 
 and mean ratio. 
 
 Divide AB in the point C, so that the rectangle contained by AB, BC 
 be equal to the square of AC (11. 2.): Then be- 
 cause the rectangle AB.BC is equal to the square j£~~ q 5 
 of AC, as BA to AC, so is AC to CB (17. 6.) ; 
 Therefore AB is cut in extreme and mean ratio in C (def. 3. 6.). 
 
 19
 
 146 
 
 ELEMENTS 
 
 PROP. XXXI. THEOR. 
 
 »'n right angled triangles, the rectilineal figure described upon the side oppo- 
 site to the right angle, is equal to the similar, and similarly described 
 figures upon the sides containing the right angle. 
 
 Let ABC bo a right angled triangle, having the right angle BAC : The 
 rectilineal figure described upon BC is equal to the similar, and similarly 
 described figures upon BA, AC. 
 
 Draw the perpendicular AD ; therefore, because in the right angled tri- 
 angle ABC, AD is drawn from the right angle at A perpendicular to the 
 base BC, the triangles A.BD, ADC are similar to the whole triangle ABC, 
 and to one another (8. 6.), and because the triangle ABC is similar to 
 ADB, as CB to BA, so is BA to BD (4. 6.) ; and because these three 
 straight lines are proportionals, as the first to the third, so is the figure upon 
 the first to the similar, and similarly described figure upon the second (2. 
 Cor. 20. 6.) : Therefore, as CB to BD, 
 so is the figure upon CB to the similar 
 and similarly described figure upon 
 BA : and inversely (B. 5.), as DB to 
 BC, so is the figure upon BA to that 
 upon BC ; for the same reason as DC 
 to CB, so is the figure upon CA to that 
 upon CB. Wherefore, as BD and DC 
 together to BC, so are the figures upon 
 BA and on AC, together, to the figure 
 upon BC (24. 5.) ; therefore the figures on BA, and on AC, are together 
 equal to that on BC ; and they are similar figures. 
 
 PROP. XXXII. THEOR. 
 
 If two triangles, which have two sides of the one proportional to two sides of 
 the other, be joined at one angle, so as to have their homologous sides pa- 
 rallel to one another ; their remaining sides shall be in a straight line. 
 
 Let ABC, DCE be two triangles which have two sides BA, AC propor- 
 tional to the two CD, DE, viz. BA to AC, as CD to DE ; and let AB be 
 parallel to DC, and AC to DE ; BC and CE are in a straight line. 
 
 Because AB is parallel to DC, and the straight line AC meets them, the 
 alternate angles BAC, ACD are equal (29 1.) ; for the same reason, the 
 angle CDE is equal to the angle 
 ACD ; wherefore also BAC is equal 
 o CDE : And because the triangles 
 ABC, DCE have one angle at A 
 equal to one at D, and the sides about 
 these angles proportionals, viz. BA to 
 AC, as CD to DE, the triangle ABC 
 is equiangular (6. 6.) to DCE : 
 Therefore the angle ABC is equal to
 
 OF GEOMETRY. BOOK VI. 
 
 147 
 
 the angle DCE : And the angle BAC was proved to be equal to ACD . 
 Therefore the whole angle ACE is equal to the two angles ABC, BAC ; 
 add the common angle ACB, then the angles ACE, ACB are equal to the 
 angles ABC, BAC, ACB . But ABC, BAC, ACB are equal to two right 
 angles (32. 1.) ; therefore also the angles ACE, ACB are equal to two 
 right angles : And since at the point C, in the straight line AC, the two 
 straight lines BC. CE, which are on the opposite sides of it, make the ad- 
 jacent angles ACE, ACB equal to two right angles ; therefore (14. 1.) BC 
 tnd CE are in a straight line. 
 
 PROP. XXXIII. THEOR. 
 
 In equal circles, angles, whether at the centres or circumferences, have the same 
 ratio which the arcs, on which they stand, have to one another : So also have 
 the sectors. 
 
 Let ABC, DEF be equal circles ; and at their centres the angles BGC, 
 EHF, and the angles BAC, EDF at their circumferences ; as the arc BC 
 to the arc EF, so is the angle BGC to the angle EHF, and the angle BAC 
 to the angle EDF : and also the sector BGC to the sector EHF. 
 
 Take any number of arcs CK, KL, each equal to BC, and any number 
 whatever FM, MN each equal to EF ; and join GK, GL, HM, HN. Be- 
 cause the arcs BC, CK, KL are all equal, the angles BGC, CGK, KGL 
 are also all equal (27. 3.) : Therefore, what multiple soever the arc BL is 
 of the arc BC, the same multiple is the angle BGL of the angle BGC : For 
 the same reason, whatever multiple the arc EN is of the arc EF the same 
 multiple is the angle EHNof the angle EHF. But if the arc BL, be equal 
 to the arc EN, the angle BGL is also equal (27. 3.) to the angle EHN ; 
 or if the arc BL be greater than EN, likewise the angle BGL is greater 
 than EHN : and if less, less : There being then four magnitudes, the two 
 arcs, BC, EF, and the two angles BGC, EHF, and of the arc BC, and of 
 the angle BGC, have been taken any equimultiples whatever, viz. the arc 
 BL, and the angle BGL ; and of the arc EF, and of the angle EHF, any 
 equimultiples whatever, viz. the arc EN, and the angle EHN : And it 
 han been proved, that if the arc BL be greater than EN, the angle BGL 
 8 greater than EHN ; and if equal, equal ; and if less, less ; As therefore, 
 «he arc BC to the arc EF, so (def. 5.5.) is the angle BGC to the angle
 
 U8 ELEMENTS 
 
 EHF But as the angle BGC is to the angle EHF, so is (15 5.) the an 
 gle B AC to the angle EDF, for each is double of each (20. 3.) : Therefore, 
 as the circumference BC is to EF, so is the angle BGC to the angle EHF, 
 and the angle BAC to the angle EDF. 
 
 Also, as the arc BC to EF, so is the sector BGC to the sector EHF. 
 Join BC, CK, and in the arcs BC, CK take any points X, 0, and join BX, 
 XC, CO, OK : Then, because in the triangles GBC, GCK, the two sides 
 BG, GC are equal to the two CG, GK, and also contain equal angles ; the 
 base BC is equal (4. 1.) to the base CK, and the triangle GBC to the tri- 
 angle GCK : And because the arc BC is equal to the arc CK, the remain- 
 ing part of the whole circumference of the circle ABC is equal to the re- 
 maining part of the whole circumference of the same circle : Wherefore 
 the angle BXC is equal to the angle COK (27. 3.) ; and the segment 
 BXC is therefore similar to the segment COK (def. 9. 3.) ; and they are 
 upon equal straight lines BC, CK : But similar segments of circles upon 
 equal straight lines are equal (24. 3.) to one another : Therefore the seg 
 ment BXC is equal to the segment COK : And the triangle BGC is equal 
 to the triangle CGK ; therefore the whole, the sector BGC is equal to the 
 whole, the sector CGK : For the same reason, the sector KGL is equal to 
 each of the sectors BGC, CGK ; and in the same manner, the sectors 
 EHF, FHM, MHN, may be proved equal to one another : Therefore, what 
 multiple soever the arc BL is of the arc BC, the same multiple is the sec- 
 tor BGL of the sector BGC. For the same reason, whatever multiple the 
 arc EN is of EF, the same multiple is the sector EHN of the sector EHF ; 
 Now if the arc BL be equal to EN, the sector BGL is equal to the sector 
 
 EHN ; and if the arc BL be greater than EN, the sector BGL is greater 
 than the sector EHN ; and if less, less : Since, then, there are four mag- 
 nitudes, the two arcs BC, EF, and the two sectors BGC, EHF, and of the 
 arc BC, and sector BGC, the arc BL and the sector BGL are any equi- 
 multiples whatever ; and of the arc EF, and sector EHF, the arc EN and 
 sector EHN, are any equimultiples whatever ; and it has been proved, that 
 if the arc BL be greater than EN, the sector BGL is greaterthan the sec 
 'or EHN ; if equal, equal; and if less, less ; therefore (def. 5. 5.) as the 
 arc BC is to the arc EF, so is the sector BGC to the sector EHF
 
 OF GEOMETRY. BOOK VI. 
 
 l49 
 
 PROP. B. THEOR. 
 
 If an angle tf a triangle be bisected by a straight line, which likewise cuts th* 
 base; the rectangle contained by the sides of the triangle is equal to the 
 rectangle contained by the segments of the base, together with the square of 
 the straight line bisecting the angle. 
 
 Let ABC be a triangle, and let the angle BAC be bisected by the 
 straight line AD ; the rectangle BA.AC is equal to the rectangle BD.DC, 
 together with the square of AD. 
 
 Describe the circle (Prop. 5. 4.) ACB about 
 the triangle, and produce AD to the circum- 
 ference in E. and join EC Then, because 
 the angle BAD is equal to the angle CAE, 
 and the angle ABD to the angle (21. 3.) 
 AEC, for they are in the same segment ; the 
 triangles ABD, AEC are equiangular to one 
 another : Therefore BA : AD : : E A : (4. 6.) 
 AC, and consequently, BA.AC= (16. 6.) 
 AD.AE=ED.DA (3. 2.) +DA 2 . But ED. 
 DA=BD.DC, therefore BA.AC =■• BD.DC 
 +DA 2 . 
 
 PROP. C THEOR. 
 
 If from any anglt of a triangle a straight line be drawn perpendicular to the 
 base ; the rectangle contained by the sides of the triangle is equal to the 
 rectangle contained by the perpendicular, and the diameter of the circle de- 
 scribed about the triangle. 
 
 Let ABC be a triangle, and AD the perpendicular from the angle A to 
 the base BC ; the rectangle BA.AC is equal to the rectangle contained by 
 AD and the diameter of the circle described about the triangle. , 
 
 Describe (Prop 5. 4.) the circle ACB 
 about the triangle, and draw its diameter 
 AE, and join EC ; Because the right 
 angle BDA is equal to the angle EC A in 
 a semicircle, and the angle ABD to the 
 angle AEC, in the same segment (21. 
 3 ); the triangles ABD, AEC are equi- 
 angular : Therefore, as (4. 6.) BA to 
 AD, so is EA to AC : and consequently 
 the rectangle BA.AC is equal (16. 6.) to 
 die rectangle EA.AD.
 
 (50 
 
 ELEMENTS 
 
 PROP. D. THEOR. 
 
 The itctangl*. contained by the diagonals of a quadrilateral inscribed in a 
 circle, is equal to both the rectangles, contained by Us opposite sides. 
 
 Let ABCD be any quadrilateral inscribed in a circle, and let AC, BD be 
 drawn ; the rectangle AC.BD is equal to the two rectangles AB.CD, and 
 AD.BC. 
 
 Make the angle ABE equal to the angle DBC ; add to each of these 
 the common angle EBD, then the angle ABD is equal to the angle EBC : 
 Anu the angle BDA is equal to (21. 3.) the angle BCE, because they are 
 in the same segment ; therefore the triangle 
 ABD is equiangular to the triangle BCE. 
 Wherefore (4. 6.), BC : CE : : BD : DA, 
 and consequently (16. 6.) BC.DA=BD.CE. 
 Again, because the angle ABE is equal to 
 the angle DBC, and the angle (21. 3.) BAE 
 to the angle BDC, the triangle ABE is equi- 
 angular to the triangle BCD ; therefore BA 
 : AE :: BD : DC, and BA.DC=BD.AE: 
 But it was shewn that BC.DA=BD.CE ; 
 wherefore BC.DA + BA.DC = BD.CE + 
 BD.AE=BD.AC(1. 2.). That is, the rect- 
 angle contained by BD and AC, is equal to the rectangles contained by 
 AB, CD, and AD, BC. 
 
 PROP. E. THEOR. 
 
 If an arc of a circle be bisected, and from the extremities of the arc, and from 
 the point of bisection, straight lines be drawn to any point in the circum- 
 ference, the sum of the two lines drawn from the extremities of the arc will 
 have to the line drawn from the point of bisection, the same ratio which the 
 straight line subtending the arc has to the straight line subtending half the 
 arc. 
 
 Let ABD be a circle, of which AB is an arc bisected in C, and from A, 
 C, and B to D, any point whatever in the circumference, let AD, CD, BD 
 be drawn ; the sum of the two lines AD 
 and DB has to DC the same ratio that 
 BA has to AC. 
 
 For since ACBD is a quadrilateral in- 
 scribed in a circle, of which the diagonals 
 are AB and CD, AD.CB + DB.AC (D 
 6 ) = AB.CD : but AD.CB + DB.AC = 
 AD.AC + DB.AC, because CB = AC. 
 Therefore AD.AC+DB.AC, that is (1. 
 2.),(AD+DB) AC=AB.CD. And be- 
 cause the sides of equal rectangles are re- 
 ciprocally proportional (14. 6.), AD-fDB 
 . DC : • AB : AC.
 
 OF GEOMETRY. BOOK VI. iSl 
 
 PROP. F. THEOR 
 
 iftwopoints be taken in the diameter of a circle, such that the rectangle contained 
 by the segments intercepted between them and the centre of the circle be equal to 
 the square of the radius: andiffrom these points two straight lines be drawn 
 to any point whatsoever in the circumference of the circle, the ratio of these 
 lines will be the same with the ratio of Oie segments intercepted between the 
 two first mentioned points and the circumference of the circle. 
 
 Let ABC be a circle, of which the centre is D, and in DA produced, let 
 the points E and F be such that the rectangle ED, DF is equal to the 
 square of AD ; from E and F to any point B in the circumference, let EB, 
 FB be drawn ; FB : BE : : FA : AE. 
 
 Join BD, and because the rectangle FD, DE is equal to the square of 
 \D, that is, of DB, FD : DB : : DB : DE (17. 6.). 
 
 The two triangles, FDB, BDE have therefore the sides proportional 
 that are about the common angle D ; the^fore they are equiangular (6. 
 6.), the angle DEB being equal to the angle DBF, and DBE to DFB 
 
 Now, since the sides about these equal angles are also proportional (4. 6.), 
 FB : BD : : BE : ED, and alternately (16. 5.), FB : BE : : BD : ED, or 
 FB : BE : : AD : DE. But because FD : DA : : DA : DE, by division 
 (17. 5.), FA : DA : : AE : ED, and alternately (11. 5.), FA : AE : : DA 
 : ED. Now it has been shewn that FB : BE : : AD : DE, therefore FB 
 • BE : : FA : AE. 
 
 Cor. If AB be drawn, because FB : BE : : FA : AE, the anglo FBE 
 is bisected (3. 6.) by AB. Also, since FD : DC : : DC : DE, by compo- 
 sition (18. 5.), FC : DC : : CE : ED, and since it has been shewn that 
 FA : AD (DC) : : AE : ED, therefore, ex »quo, FA : AE : : FC : CE. 
 ButFB : BE :: FA : AE, therefore FB : BE : : FC : CE (11.5 ), so that 
 if FB be produced to G.and if BC be drawn, the angle EBG is bisected 
 by the line BC (A. 6.).
 
 152 
 
 ELEMENTS 
 
 PROP. G. THEOR. 
 
 If from the extremity of the diameter of a circle a straight line be drawn in the 
 circle, and if either within the circle or produced without it, it meet a line per- 
 pendicular to the same diameter, the rectangle contained by the straight lint 
 drawn in the circle, and the segment of it, intercepted between the extremity 
 of the diameter and the perpendicular, is equal to the rectangle contained b$ 
 the diameter and the segment of it cut off by the perpendicular. 
 
 Let ABC be a circle, of which AC is a diameter, let DE be perpendicu- 
 lar to the diameter AC, and let AB meet DE in F ; the rectangle BA.AF 
 is equa? to the rectangle CA.AD. Join BC, and because ABC is an an- 
 
 gle in a semicircle, it is a right angle (31. 3.): Now, the angle ADF is 
 also a right angle (Hyp.) ; and the angle BAC is either the same with 
 DAF, or vertical to it ; therefore the triangles ABC, ADF are equiangular, 
 and BA : AC : : AD : AF (4. 6.) ; therefore also the rectangle BA.AF, 
 contained by the extremes, is equal to the rectangle ACAD contained by 
 the means (16. 6.). 
 
 PROP. H. THEOR. 
 
 The perpendiculars drawn from the three angles of any triangle to the opposite 
 sides intersect one another in the same point. 
 
 Let ABC be a triangle, BD and CE two perpendiculars intersecting one 
 another in F ; Let AF be joined, and produced if necessary, let it meet BC 
 in G, AG is perpendicular to BC. 
 
 Join DE, and about the triangleAEF let a circle be described, AEP 
 then, because AEF is a right angle, the circle described about the triangle 
 AEF will have AF . >r its diameter (31. 3.). In the same manner, the 
 circle described about the triangle ADF has AF for its diameter ; there- 
 fore the points A, E, F and D, are in the circumference of the fame circle
 
 OF GEOMETRY. BOOK VI. 
 
 '5? 
 
 But because the angle EFB is equal 
 to the angle DFC(15. 1.), and also 
 the angle BEF to the angle CPF, 
 being both right angles, the triangles 
 BEF, and CDF are equiangular, and 
 therefore BF : EF : : CF : FD (4. 6.), 
 or alternately (16.5.) BF : FC : : EF 
 • FD. Since, then, the sides about 
 the equal angles BFC, EFD are pro- 
 portionals, the triangles BFC, EFD 
 are also equiangular (6. 6.) ; where- 
 fore the angle FCB is equal to the an- 
 gle EDF. But EDF is equal to EAF, 
 because they are angles in the same 
 segment (21. 3.); therefore the angle 
 EAF is equal to the angle FCG : Now, the angles AFE, CFG are also 
 equal, because they are vertical angles ; therefore the remaining angles 
 AEF, FGC are also equal (4. Cor. 32. 1.) : But AEF is a right angle, 
 therefore FGC is a right angle, and AG is perpendicular to BC. 
 
 Cor. The triangle ADE is similar to the triangle ABC. For the two 
 triangles BAD, CAE having the angles at D and E right angles, and the 
 angle at A common, are equiangular, and therefore BA : AD : : CA : AE, 
 and alternately BA : CA : : AD : AE ; therefore the two triangles BAC, 
 DAE, have the angle at A common, and the sides about that angle pro- 
 portionals, therefore they are equiangular (6. 6.) and similar. 
 
 Hence the rectangles BA.AE, CA.AD are equal. 
 
 PROP. K. THEOR. 
 
 If from any angle of a triangle a perpendicular be drawn to the opposite side 
 or base : the rectangle contained by the sum and difference of the other two 
 sides, is equal to the rectangle contained by the sum and difference of the 
 segments, into which the. base is divided by the perpendicular. 
 
 Let ABC be a triangle, AD a perpendicular drawn from the angle A on 
 tie base BC, so that BD, DC are the segments of the base ; (AC+AB) 
 *C-AB)=(CD+ DB) (CD-DB.)
 
 /54 
 
 ELEMENTS 
 
 Ficm A as a centre with the radius AC, the greater of the wo sides, 
 describe the circle CFG : produce AB to meet the circumference in E and 
 F, and CB to meet it in G. Then because AF=AC, BF=AB *-AC, 
 the sum of the sides; and since AE=AC, BE=AC — AB=the diffe- 
 rence of the sides. Also, because AD drawn from the centre cuts GC at 
 right angles, it bisects it ; therefore, when the perpendicular falls within 
 the triangle, BG=DG — DB = DC— DB= the difference of the segments 
 of the base, and BC=BD-f-DC= the sum of the segments. But when 
 AD falls without the triangle, BG = DG+DB=CD-f-DB= the sum of 
 the segments of the base, and BC=CD — DB= the difference of the seg- 
 ments of the base. Now, in both cases, because B is the intersection of 
 the two lines FE, GC, drawn in the circle, FB.BE=CB.BG ; that is, as 
 has been shewn, (AC-|-AB) (AC-AB)=(CD+DB) (CD-DB) 
 
 PROBLEMS 
 
 RELATING TO THE SIXTH BOOK. 
 
 PROP. L. PROBLEM. 
 To construct a square that shall be equivalent to a given rectilineal figure. 
 
 Let A be the given rectilineal figure ; it is required to describe a square 
 that shall be equivalent to A. 
 
 Describe (Prop. 45.1.) the 
 rectangular parallelogram 
 BCDE equivalent to the rec- 
 tilineal figure A ; produce 
 one of the sides BE, of this 
 rectangle, and make EF = 
 ED; bisect BF in G, and 
 from the centre G, at the 
 distance GB, or GF, de- 
 scribe the semicircle BHF, 
 and produce DE to H. 
 
 HE 2 =BE x EF, (13. 6.) ; therefore the square described upon HE will 
 be equivalent to the rectilineal figure A. 
 
 SCHOLIUM. 
 
 This problem may be considered as relating to the second Book : Thus, 
 join GH, the rest of the construction being the same, as above ; because 
 the straight line BF is divided into two equal parts in the point G, and into 
 two unequal in the point E, the rectangle BE.EF, together with the square 
 of EG, is equal (5. 2.) to the square of GF : but GF is equal to GH ,
 
 OF GEOMETRY. BOOK VI. 
 
 155 
 
 ihei afore the rectangle BE, EF, together with the square of EG, is equal 
 to the square of GH : But the squares of HE and EG, are equal (47. 1 ^ 
 to the square of GH : Therefore also the rectarfgle BE.EF, together witL 
 the square of EG, is equal to the squares of HE and EG. Take away 
 the square of EG, which is common to both, and the remaining rectangle 
 BE.EFis equal to the square of EH : But BD is the rectangle containeJ 
 by BE and EF, because EF is equal to ED ; therefore BD is equal to the 
 square of EH ; and BD is also equal to the rectilineal figure A ; therefore 
 the rectilineal figure A is equal to the square of EH : Wherefore a square 
 has been made equal to the given rectilineal figure A, viz. the square de- 
 scribed upon EH. 
 
 Note. This operation is called squaring the rectilineal figure, or finding 
 the quadrature of it. 
 
 PROP. M. PROB. 
 
 To construct a rectangle that shall be equivalent to a given square, and the 
 difference of whose adjacent sides shall be equal to a given line. 
 
 Suppose C equal to the given square, and 
 AB the difference of the sides. 
 
 Upon the given line AB as a diameter, de- 
 scribe a circle ; at the extremity of the diam- 
 eter draw the tangent AD equal to the side 
 of the square C ; through the point D, and the 
 centre O, draw the secant DF ; then will DE 
 and DF be the adjacent sides of the rectangle 
 required. 
 
 First, the difference of their sides is equal 
 to the diameter EF or AB ; secondly, the rect- 
 angle DE.DF is equal to AD 2 (36. 3.) ; hence 
 that rectangle is equivalent to the given square C. 
 
 ?ROP. N. PROB. 
 
 To construct a rectangle equivalent to a given square, and having the sum 
 of its adjacent sides equal to a given line. 
 
 Let C be the given square, and AB equal to the sum of the sides of the 
 required rectangle 
 
 Upon AB as a diameter, 
 describe a semicircle ; draw 
 the line DE parallel to the 
 diameter, at a distance AD 
 from it, equal to the side of 
 die given square C ; from the 
 
 point E, where the parallel "^ F B 
 
 cuts the circumference, draw EF perpendicular to the diameter ; aF 
 and FB will be the sides of the rectangle required.
 
 156 
 
 ELEMENTS 
 
 For their sum is equal to AB ; and their rectangle AF.FB is equal to th« 
 •quare EF, or to the square AD ; hence that rectangle is equivalent to the 
 given square C. * 
 
 SCHOLIUM. 
 
 To render the problem possible, the distance AD must not exceed the 
 radius ; that is, the side of the square C must not exceed the half of the 
 line AB. 
 
 PROP. 0. PROB. 
 
 To construct a square that skall be to a given square as a given line to a given 
 
 line. 
 
 Upon the indefinite straight line GH take GK=E, and KH=F ; de- 
 scribe on GH a semicircle, and draw the perpendicular KL. Through 
 
 
 D 
 
 C 
 
 the points G, H, draw the 
 straight lines LM, LN, mak- 
 ing the former equal AB, the 
 side of the given square, and 
 through the point M, draw 
 MN parallel to GH, then will 
 LN be the side of the square 
 sought. 
 
 For, since MN is parallel 
 to GH, LM : LN : : LG : 
 LH ; consequently, LM 2 : LN 2 : : LG 2 : LH 2 (22. 6.) ; but, since the trian- 
 gle LGH is right angled, we have LG 2 : LH 2 : : GK : KH ; hence LM 2 : 
 LN 2 : : GK : KH ; but, by construction GK=E, and KH=F, also LM 
 =AB ; therefore, the square described on AB is to that described on LN, 
 as the line E is to the line F. 
 
 PROP. P. PROB. 
 
 To divide a triangle into two parts by a line from the vertex of one of its angles^ 
 so that the parts may be to each other as a straight line M to another straight 
 line N. 
 
 Divide BC into parts BD, DC propor- 
 tional to M, N; draw the line AD, and 
 the triangle ABC will be divided as re- 
 quired. 
 
 For, since the triangles of the same 
 altitude are to each other as their bases, 
 we have ABD : ADC : : BD : DC : : 
 M- N. 
 
 SCHOLIUM. 
 
 A triangle may evidently be divided into any number of parts propoi 
 tional to given lines, by dividing the base in the same proportion
 
 OF GEOMETRY. BOOK VI. 
 
 157 
 
 PROP. Q. PROB. 
 
 To divide a triangle into ttvo parts by a line drawn parallel to one of its sides, 
 so that these parts may be to each other as two straight lines M, N 
 
 As M+N : N, so make AB' to AD 2 
 (Prob. 4.) ; Draw DE parallel to BC, 
 *nd the triangle is divided as required. 
 
 For the triangles ABC, ADE being 
 similar, ABC : ADE : : AB 3 : AD 2 ; but 
 M+N : N : : AB 3 : AD 3 ; therefore ABC 
 : ADE : : M+N : N ; consequently 
 BDEC : ADE : : M : N. 
 
 PROP R. PROB. 
 
 To divide a triangle into two parts, by a line drawn from a given point in 
 one of its sides, so that the parts may be to each other as two given lines 
 
 M, N. 
 
 Let ABC be the given triangle, and P the given point ; draw PC, and 
 divide AB in D, so that AD is to DB as M is to N ; draw DE parallel to 
 PC, join PE, and the triangle will be divid- 
 ed by the line PE into the proposed parts. 
 
 For join DC ; then because PC, DE are 
 parallel, the triangles PDE, CDE are equal ; 
 to each add the triangle DEB, then PEB = 
 DCB ; and consequently, by taking each from 
 the triangle ABC, there results the quadri- 
 lateral ACEP equivalent to the triangle 
 ACD. 
 
 Now, ACD : DCB : : AD : DB : : M : N ; 
 ACEP : PEB : : M : N 
 
 B 
 
 consequently, 
 
 SCHOLIUM. 
 
 The above operation suggests the method of dividing a triangle into any 
 number of equal parts by lines drawn from a given point in one of its sides ; 
 for if AB be divided into equal parts, and lines be drawn from the points of 
 equal division, parallel to PC, they will intersect BC, and AC ; and from 
 these several points of intersection if lines be drawn to P, they will divide 
 the trianglo into equal parts.
 
 158 
 
 ELEMENTS 
 
 PROP. S. 
 
 PROB. 
 
 To divide a triat^ i into three equivalent parts by lines drawn from the ver- 
 tices ojf he angles to the same point within the triangle. 
 
 Make BD equal to a third part of BC, and draw DE parallel to BA, Ine 
 side to which BD is adjacent. From F, tho middle of DE, draw the 
 straight lines FA, FB, FC, and they will 
 divide the triangle as required. 
 
 For, draw DA ; then since BD is one 
 third of BC, the triangle ABD is one 
 third of the triangle ABC ; but ABD = 
 ABF (37. 1.) ; therefore ABF is one 
 hird of ABC ; also, since DF=FE, 
 BDF = AFE ; likewise CFD = CFE , 
 consequently the whole triangle FBC 
 is equal to the whole triangle FCA ; and 
 FBA has been shown to be equal to a third part of the whole triangle 
 ABC ; consequently the triangles FBA, FBC, FCA, are each equal to a 
 third part of ABC. 
 
 PROP. T. PROB. 
 
 To divide a triangle into three equivalent parts, by lines drawn Jrom a given 
 
 point within it. 
 
 Divide BC into three equal parts in the points D, E, and draw PD, PE ; 
 draw also AF parallel to PD, and AG parallel to PE; then if the lines 
 PF, PG, PA be drawn, the trian- 
 gle ABC will be divided by them 
 into three equivalent parts. 
 
 For, join AD, AE ; then because 
 AF, PD are parallel, the triangle 
 AFP is equivalent to the triangle 
 AFD ; consequently, if to each of 
 these there be added the triangle 
 ABF, there will result the quadri- 
 lateral ABFP equivalent to the 
 triangle ABD ; but since BD is a 
 third part of BC, the triangle ABD 
 is a third part of the triangle ABC ; 
 consequently the quadrilateral ABFP is a third part of the triangle ABC. 
 Again, because AG, PE are parallel, the triangle AGP is equivalent to 
 the triangle AGE and if to each of these there be added the triangle A CO 
 the quadrilateral ACGP will be equivalent to the triangle ACF ; but this 
 triangle is one third of ABC ; hence the quadrilateral ACGP is one thirJ 
 of the triangle ABC : consequently, the spaces ABFP, ACPG, PFO are 
 each equal to a third part of the triangle ABC.
 
 OF GEOMETRY. BOOK VI 
 
 159 
 
 PROP. U. PROB. 
 
 To hvide a quadrilateral into two parts by a straight line drawn from the vertex 
 of one of its angles, so that the parts may be to each other as a line M to an- 
 other line N. 
 
 Draw CE perpendicular to AB, and construct a rectangle equivalent tc 
 the given quadrilateral, of which one side may be CE ; let the other side 
 be EF ; and divide EF in G, so that 
 M : N : : GF : EG ; take BP equal 
 to twice EG, and join PC, then the 
 quadrilateral will be divided as re- 
 quired. 
 
 For, by construction, the triangle 
 CPB is equivalent to the rectangle 
 CE.EG ; therefore the rectangle CE, 
 GF is to the triangle CPB as GF is 
 to EG. Now CE.GF is equivalent 
 to the quadrilateral DP, and GF is to EG as M is to N ; therefore, 
 
 DP : CPB : : M : N ; 
 that is, the quadrilateral is divided, as required. 
 
 PROP. W. PROB. 
 
 To divide a quadrilateral into two parts by a line parallel to one oj its sides 
 so that these parts may be to each other as the line M is to the line N. 
 
 Produce AD, BC till they meet in E ; draw the perpendicular EF and 
 bisect it in G. Upon the side GF construct a rectangle equivalent to tha 
 triangle EDC, and let HB be equal 
 to the other side of this rectangle. 
 Divide AH in K, so that AK : KH 
 : : M : N, and as AB is to KB, so 
 make EA 2 to Ea 2 ; draw ab paral- 
 lel to AB, and it will divide the quad- 
 rilateral into the required parts. 
 
 For since the triangles EAB, Eab 
 are similar, we have the proportion 
 EAB : Eab : : EA 2 : Ea a ; but by 
 conotTuction, EA 2 : Ea 2 : : AB : 
 KB ; so that EAB : Eab : : AB : KB 
 quently, since by construction EAB=AB.GF, it follows that En6 = KB 
 GF, and therefore AK.GF=sA5, and since by construction AH.GF=AC 
 it follows that KH.GF=aC. Now AK.GF : KH.GF : : AK : KH ; b*n 
 \K : KH : : M : N ; consequently, 
 
 A* : aC : : M : N ; 
 that is, the quadrilateral is divided, as required. 
 
 A. K E H B 
 
 : : AB.GF : KB.GF ; and const,
 
 60 
 
 ELEMENTS, &c. 
 
 PROP. X. PROB. 
 
 To divide a quadrilateral into two parts by a line dranm from a point in one oj 
 'ts sides, so that the parts may be to each other as a line M is to a line N. 
 
 Draw PD, upon which construct a rectangle equivalent to the given 
 quadrilateral, and let DK be the other 
 side of this rectangle ; divide DK in 
 L, so that DL : LK : : M : N ; make 
 DF=2DL, and FG equal to the per- 
 pendicular Aa ; draw Gp parallel to 
 DP ; join the points P, p, and the 
 quadrilateral figure will be divided, 
 as required. 
 
 For draw the perpendicular pb ; 
 then by construction, PD.DK = AC, 
 and PD.DF = PD.Aa + PDpb, that 
 is, PD.DF is equivalent to twice the 
 sum of the triangles APD, pFD , 
 consequently, since DL is half DF, 
 PD.DL=APjdD ; and therefore PD. 
 LK=PBC;> ; but PD.DL : PD.LK : : DL : LK : : M : N ; consequently, 
 
 AP^D : FBCp : : M : N ; 
 hence the quadrilateral is divided, as required. 
 
 PROP. Y. 
 
 PROB 
 
 To divide a quadrilateral by a line perpendicular to one of its sides, so that the 
 two parts may be to each other as a line M is to a line N. 
 
 Let ABCD be the given quadrilateral, which is to be divided in the ratio 
 of M to N by a perpendicular to the side AB. 
 
 Construct on DE perpendicular 
 to AB*, a rectangle DE.EF, equi- 
 valent to the quadrilateral AC, 
 and divide FE in G, so that FG : 
 GE : : M : N. Bisect AE in H, 
 and divide the quadrilateral EC 
 into two parts by a line PQ, paral- 
 lel to DE, so that those parts may 
 be to each other as FG is to GH, 
 then PQ will also divide the quadri- 
 lateral AC as required. 
 
 For, by construction DE.LF=AC, and DE.EH=DAE ; hence DE. 
 HF=EC, and consequently, since the quadrilateral EC is divided in the 
 same proportion as the base FH of its equivalent rectangle, it follows that 
 QC=DE.FG, and EP=DE.GH, also AP=DE.GE ; consequently, 
 
 QC : AP : : FG : GE : : M : N ; 
 tnat is, the quadrilateral is divided, as required. 
 
 RlHE
 
 SUPPLEMENT 
 
 TO THE 
 
 ELEMENTS 
 
 OP 
 
 GEOMETRY. 
 
 31
 
 ELEMENTS 
 
 OF 
 
 GEOMETRY. 
 
 SUPPLEMENT. 
 
 BOOK I* 
 
 OF THE QUADRATURE OF THE CIRCLE. 
 
 LEMMA 
 
 A v curve line, or any polygonal line, which envelopes a convex line from one 
 end to the other, is longer than the enveloped line. 
 
 Let AMB be the enveloped line ; then will it be less than the line 
 APBB which envelopes it. 
 
 We have already said that by the 
 tenn convex line we understand a line, 
 pol/gonal, or curve, or partly curve and 
 partly polygonal, such that a straight 
 line cannot cut it in more than two 
 points. If in the line AMB there were 
 any sinuosities or re-entrant portions, it 
 would cease to be convex, because a 
 straight line might cut it in more than 
 
 two points. The arcs of a circle are essentially convex ; but the present 
 proposition extends to any line which fulfils the required conditions. 
 
 This being premised, if the line AMB is not shorter than any of those 
 which envelope it, there will be found among the latter, a line shorter than 
 all the rest, which is shorter than AMB, or, at most, equal to it. Let 
 ACDEB be this enveloping line : any where between those two lines, 
 draw the straight line PQ, not meeting, or ?>t least only touching, the line 
 AMB. The straight line PQ is shorter than PCUEQ ; hence, if instead 
 *f the part PCDEQ, we substitute the straight line PQ, the enveloping line 
 APQB will be shorter than APDQB. But, by hypothesis, this tatter was 
 shorter tha* any other ; hence that hypothesis was false : hence all nt »he 
 enveloping lines are longer than AMB
 
 -F 
 
 -G 
 
 (64 SUPPLEMENT TO THE ELEMENTS 
 
 C<»K 1. Mence the perimeter of any polygon inscribed in a circle 1* 
 Ifess than the circumference of the circle. 
 
 Cor. 2. If from a point two straight lines be drawn, touch'ng a circle, 
 these two lines are together greater than the arc intercepted between 
 them ; and hence the perimeter of any polygon described about a circle i* 
 greater than the circumference of the circle. 
 
 PROP. I. THEOR. 
 
 If jrom the greater of two unequal magnitudes there be taken away its half 
 and from the remainder its half ; and so on ; There will at length remain 
 a magnitude less than the least of the proposed magnitudes. 
 
 Let AB and C be two unequal magnitudes, of which AB is the greater. 
 If from AB there be taken away its half, and from the -^ 
 
 remainder its half, and so on ; there shall at length 
 remain a magnitude less than C. 
 
 For C may be multiplied so as, at length, to be- 
 come greater than AB. Let DE, therefore, be a -rr 
 multiple of C, which is greater than AB, and let it 
 contain the parts DF, FG, GE, each equal to C. 
 From AB take BH equal to its half; and from the -rr 
 remainder AH, take HK equal to its half, and so on, 
 until there be as many divisions in AB as there are 
 la DE ; And let the divisions in AB be AK, KH, 
 HB. And because DE is greater than AB, and EG 
 taken from DE is not greater than its half, but BH 
 taken from AB is equal to its half; therefore the re- 
 mainder GD is greater than the rem tinder HA. B C 
 Again, because GD is greater than HA, and GF is 
 not greater than the half of GD, but HK is equal to the half of HA ; there- 
 fore the remainder FD is greater than the remainder AK. And FD is 
 equal to C, therefore C is greater than AK ; that is, AK is less than C. 
 
 PROP. II. THEOR. 
 
 Equilateral polygons, of the same number of sides, inscribed in circles, art 
 similar, and are to one another as the squares of the diameters of the 
 circles. 
 
 Let ABCDEF and GHIKLM be two equilateral polygons of the same 
 number of sides inscribed in the circles ABD and GHK ; ABCDEF and 
 GHIKLM are similar, and are to one another as the squares of the diame- 
 ters of the circles ABD, GHK. 
 
 Find N and O the centres of the circles, join AN and BN, as also GO 
 and HO, and produce AN and GO till they meet the circumferences in D 
 and K- 
 
 Because the straight lines AB, BC, CD, DE, EF, FA, are all equal 
 the arcs AB, BC, CD, DE, EF, FA are also equal (28. 3.). For the 
 same, reason, the arcs GH, HI, IK, KL, LM, MG are aP equal, and they
 
 OF GEOMETRY. BOOK I. 
 
 165 
 
 are equal in number to the others ; therefore, whatever part the arc AB :s 
 of the whole circumference ABD, the same is the arc GH of the circum- 
 ference GHK. But the angle ANB is the same part of four right angles, 
 *hat the arc AB is of the circumference ABD (33. 6.) ; and the angle 
 GOH is the same part of four right angles, that the arc GH is of the cir- 
 cumference GHK (33. 6.), therefore the angles ANB, GOH are each of 
 them the same part of four right angles, and therefore they are equal to 
 one another. The isosceles triangles ANB, GOH are therefore equian- 
 gular, and the angle ABN equal to the angle GHO ; in the same manner, 
 by joining NC, 01, it may be proved that the angles NBC, OHI are equal 
 to one another, and to the angle ABN. Therefore the whole angle ABO 
 
 is equal to the whole GHI ; and the same may be proved of the angles 
 BCD, HIK, and of the rest. Therefore, the polygons ABCDEF and 
 GHIKLM are equiangular to one another ; and since they are equilateral, 
 the sides about the equal angles are proportionals ; the polygon ABCDEF 
 is therefore similar to the polygon GHIKLM (def. 1.6.). And because simi- 
 lai polygons are as the squares of their homologous sides (20. 6.), the po- 
 lygon ABCDEF is to the polygon GHIKLM as the square of AB to the 
 Bquare of GH ; but because the triangles ANB, GOH are equiangular, 
 tlie square of AB is to the square of GH as the square of AN to the square 
 of GO (4. 6.), or as four times the square of AN to four times the square 
 15. 5.) of GO, that is, as the square of AD to the square of GK, (2. Cor. 
 8. 2.). Therefore also, the polygon ABCDEF is to tho polygon GHIKLM
 
 166 SUPPLEMENT TO THE ELEMENTS 
 
 as thr square of AD to the square of GK ; and they have also been shewn 
 to be similar. 
 
 Cor. Every equilateral polygon inscribed in a circle is also equiangu 
 lar : For the isosceles triangles, which have their common vertex in the 
 centre, are all equal and similar ; therefore, the angles at their bases are 
 all equal, and the angles of the polygon are therefore also equal. 
 
 PROP. III. PROB. 
 
 The side of any equilateral polygon inscribed in a circle being given, to find the 
 side of a polygon of the same number of sides described about the circle. 
 
 Lei ABCDEF be an equilateral polygon inscribed in the circle ABD ; 
 it is required to find the side of an equilateral polygon of the same number 
 of sides described about the circle. 
 
 Find G the centre of the circle ; join GA, GB, bisect the arc AB in H ; 
 and through H draw KHL touching the circle in H, and meeting GA and 
 GB produced in K and L ; KL is the side of the polygon required. 
 
 Produce GF to N, so that GN maybe equal to GL ; join KN, and from 
 G draw GM at right angles to KN, join also HG. 
 
 Because the arc AB is bisected in H, the angle AGH is equal to the 
 angle BGH (27. 3.) ; and because 
 KL touches the circle in H, the 
 angles LHG, KHG are right an- 
 gles (18. 3.); therefore, there are 
 two angles of the triangle HGK, 
 equal to two angles of the triangle 
 HGL, each to each. But the side 
 GH is common to these triangles ; 
 therefore they are equal (26. l.),and 
 GL is equal to GK. Again, in 
 the triangles KGL, KGN, because 
 GN is equal to GL ; and GK com- 
 mon, and also the angle LGK equal 
 to the angle KGN ; therefore the 
 base K L is equal to the base KN 
 (4. 1.). But because the triangle KGN is isosceles, the angle GKN is 
 equal to the angle GNK, and the angles GMK, GMN are both right an 
 gles by construction ; wherefore, the triangles GMK, GMN have two aw 
 gles of the one equal to two angles of the other, and they have also the 
 side GM common, therefore they are equal (26. l.),and the side KM is equal 
 to the side MN, so that KN is bisected in M. But KN is equal to KL, 
 and therefore their halves KM and KH are also equal. Wherefore, in the 
 triangles GKH, GKM, the two sides GK and KH are equal to the two 
 GK and KM, each to each ; and the angles GKH, GKM, are also equal, 
 therefore GM is equal to GH (4. 1.) ; wherefore, the point M is in the cir- 
 cumference of the circle ; and because KMG is a right angle, KM touches 
 the circle. And in the same manner, by joining the centre and the other 
 angular points of the inscribed polygon, an equilateral polygon may be
 
 OF GEOMETRY. BOOK I. .67 
 
 described about the circle, the sides of which will each be equal t« KL, and 
 wid be equal in number to the sides of the inscribed polygon. Therefore. 
 KL is the side of an equilateral polygon, described about the circle, of the 
 Bame number of sides with the inscribed polygon ABCDEF. 
 
 Cor. 1. Because GL, GK, GN, and the other straight lines drawc 
 from the centre G to the angular points of the polygon described about th© 
 circle ABD are all equal ; if a circle be described from the centre G, with 
 the distance GK, the polygon will be inscribed in that circle ; and there- 
 fore it is similar to the polygon ABCDEF. 
 
 Cor. 2. It is evident that AB, a side of the inscribed polygon, is to KL, 
 a side of the circumscribed, as the perpendicular from G upon AB, to the 
 perpendicular from G upon KL, that is, to the radius of the circle ; there- 
 fore also, because magnitudes have the same ratio with their equimultiples 
 (15. 5.), the perimeter of the inscribed polygon is to the perimeter of the 
 circumscribed, as the perpendicular from the centre, on a side of the in- 
 scribed polygon, to the radius of the circle. 
 
 PROP. IV. THEOR. 
 
 A circle being given, two similar polygons may be found, the one described about 
 the circle, and the other inscribed in it, which shall differ from one another by 
 a space less than any given space. 
 
 Let ABC be the given circle, and the square of D any given space ; a 
 polygon may be inscribed in the circle ABC, and a similar polygon describ- 
 ed about it, so that the difference between them shall be less than the 
 square of D. 
 
 In the circle ABC apply the straight line AE equal to D, and let AB be 
 a fourth part of the circumference of the circle. From the circumference 
 AB take away its half, and from the remainder its half, and so on till the 
 circumference AF is found less than the circumference AE (I. 1. Sup.). 
 Find the centre G ; draw the diameter AC, as also the straight lines AF 
 and FG ; and having bisected the circumference AF in K, join KG, and 
 draw HL touching the circle in K, and meeting GA and GF produced in 
 H and L ; join CF. 
 
 Because the isosceles triangles HGL and AGF have the common an- 
 gle AGF, they are equiangular (6. 6.) and the angles GHK, GAF arc 
 therefore equal to one another. But the angle GKH, CFA are also equal, 
 for they are right angles ; therefore the triangles HGK, ACF, are like- 
 wise equiangular (4. Cor. 32. 1.). 
 
 And because the arc AF was found by taking from the arc AB its half, 
 and from that romainder its half, and so on, AF will be contained a certain 
 number of times, exactly, in the arc AB, and therefore it will also be con- 
 tained a certain number of times, exactly, in the whole circumference 
 ABC ; and the straight line AF is therefore the side of an equilateral poly- 
 gon inscribed in the circle ABC. Wherefore also, HL is the side of an 
 equilateral polygon, of the same number of sides, described about ABC (3. 
 1. Sup.). Let the polygon described about the circle be called M, and the 
 polygon inscribed be called N ; then, because these polygons are similar
 
 68 SUPPLEMENT TO THE ELEMENTS 
 
 B 
 
 tltey are as the squares of the homologous sides HL and AF (3. Oorol. 
 20. 6.), that is, because the triangles HLG, AFG are similar, as the square 
 of HG to the square of AG, that is of GK. But the triangles HGK, ACF 
 have been proved to be similar, and therefore the square of AC is to the 
 square of CF as the polygon M to the polygon N ; and, by conversion, 
 the square of AC is to its excess above the squares of CF, that is, to the 
 square of AF (47. 1.), as the polygon M to its excess above the polygon 
 N. But the square of AC, that is, the square described about the circle 
 ABC is greater than the equilateral polygon of eight sides described about 
 the circle, because it contains that polygon ; and, for the same reason, the 
 polygon of eight sides is greater than the polygon of sixteen, and so on ; 
 therefore, the square of AC is greater than any polygon described about 
 the circle by the continual bisection of the arc AB ; it is therefore greater 
 than the polygon M. Now, it has been demonstrated, that the square of 
 AC is to the square of AF as the polygon M to the difference of the poly- 
 gons ; therefore, since the square of AC is greater than M, the square of 
 AF is greater than the difference of the polygons (14. 5.). The difference 
 of the polygons is therefore less than the square of AF ; but AF is less 
 than D ; therefore the difference of the polygons is less than the square of 
 D ; that is, than the given space. 
 
 Cor. 1. Because the polygons M and N differ from one another more 
 than either of them differs from the circle, the difference between each of 
 them and the circle is less than the given space, viz. the square of D. And 
 therefore, however small any given space may be, a polygon may be in- 
 scribed in the circle, and another described about it, each of which shall 
 differ from the circle by a space less than the given space. 
 
 Cor. 2. The space B, which is greater than any polygon that can be 
 inscribed in the circle A, and less than any polygon that can be described 
 about it, is equal to the circle A. If not, let them be unequal ; and first, 
 let B exceed A by the space C. Then, because the polygons described 
 about the circle A are all greater than D, by hypothesis ; and because B 
 is greater than A by the space C therefore no polygon can bo described
 
 OF GEOMETRY. BOOK I. 
 
 169 
 
 about the circle A, but what must exceed it by a space greater than C, 
 which is absurd. In the same manner, if B be less than A by the space 
 C, it is shewn that no polygon can be inscribed in the circle A, but what 
 is less than A by a space greater than C, which is also absurd. Therefore, 
 A and B are not unequal ; that is, they are equal to one another. 
 
 PROP. V. THEOR. 
 
 The area of any circle is equal to the rectangle contained by the semi-diameter, 
 and a straight line equal to half the circumference. 
 
 Let ABC be a circle of which the centre is D, and the diameter AC ; if 
 in AC produced there be taken AH equal to half the circumference, the 
 area of the circle is equal to the rectangle contained by DA and AH. 
 
 Let AB be the side of any equilateral polygon inscribed in the circle 
 ABC ; bisect the circumference AB in G, and through G draw EGF 
 KUching the circle, and meeting DA produced in E, and DB produced in 
 
 k n l 
 
 N-r-0 
 
 F ; EF will be the side of an equilateral polygon described about the cit 
 cle ABC (3. 1. Sup.). In AC produced take AK equal to half the peri- 
 meter of the polygon whose side is AB ; and AL equal to half the perime- 
 ter of the polygon whose side is EF. Then AK will be less, and AL 
 greater than the straight line AH (Lem. Sup.). Now, because in the 
 triangle EOF, DG is drawn perpendicular to the base, the triangle EDF 
 
 22
 
 170 SUPPLEMENT TO THE ELEMENTS 
 
 is equal to the rectangle contained by DG and the half of EF (41. 1.) ana 
 as the .same is true of all the other equal triangles having their vertices ii 
 D, which make up the polygon described about the circle ; therefore, the 
 whole polygon is equal to the rectangle contained by DG and AL, half the 
 perimeter of the polygon (1. 2.), or by DA and AL. But AL is 
 greater than AH, therefore the rectangle DA.AL is greater than the reel- 
 angle DA.AH ; the rectangle DA.AH is therefore less than the rectangle 
 DA.AL, that is, than any polygon described about the circle ABC. 
 
 Again, the triangle ADB is equal to the rectangle contained by DM the 
 perpendicular, and one half of the base AB, and it is therefore less than the 
 rectangle contained by DG, or DA, and the half of AB And as the same 
 
 is true of all the other triangles having their vertices in D, which make 
 up the inscribed polygon, therefore the whole of the inscribed polygon is 
 less than the rectangle contained by DA, and AK half the perimeter of the 
 polygon. Now, the rectangle DA.AK is less than DA.AH ; much more, 
 therefore, is the polygon whose side is AB less than DA.AH ; and the 
 rectangle DA.AH is therefore greater than any polygon inscribed in th«» 
 circle ABC. But the same rectangle DA.AH has been proved to be lesb 
 than any polygon described about the circle ABC ; therefore the rectangle 
 DA.AH is equal to the circle ABC (2. Cor. 4. 1. Sup.). Now DA is the 
 8emidiameter of the circle ABC, and AH the half of its circumference. 
 
 Cor. 1. Because DA : AH : : DA 2 : DA.AH (1. 6.), and because by 
 this proposition, DA.AH = the area of the circle, of which DA is the ra- 
 dius : therefore, as the radius of any circle to the semicircumference, or as 
 the diameter to the whole circumference, so is the square of the radius to 
 the area of the circle. 
 
 Cor. 2. Hence a polygon may be described about a circle, the perime- 
 ter of which shall exceed the circumference of the circle by a line that is 
 less than any given line. Let NO be the given line. Take in NO the 
 part NP less than its half, and also than AD, and let a polygon be describ- 
 ed about the circle ABC, so that its excess above ABC may be less than 
 the square of NP (1. Cor. 4. 1. Sup.). Let the side of this polygon be EF. 
 \nd since, as has been proved, the circle is equal to the rectangle DA.AH, 
 and the polygon to the rectangle DA.AL, the excess of the polygon above 
 the circle is equal to the rectangle DA.HL ; therefore the rectangle DA
 
 OF GEOMETRY. BOOK I. 
 
 171 
 
 HL is l.js» ihan the square of NP ; and therefore, since D A 'a greater than 
 NP, HL rs less than NP, and twice HL less than twice NP, wherefore 
 much more is twice HL less than NO. But HL is the difference between 
 half the perimeter of the polygon whose side is EF, and half the circum- 
 ference of the circle ; therefore, twice HL is the difference between the 
 whole perimeter of the polygon and the whole circumference of the circle 
 (5. 5.). The difference, therefore, between the perimeter of the polygon 
 and the circumference of the circle is less than the given line NO. 
 
 Cor. 3. Hence, also, a polygon may be inscribed in a circle, such 
 that the excess of the circumference above the perimeter of the polygon 
 may be less than any given line. This is proved like the preceding. 
 
 PROP. VI. THEOR. 
 
 The areas of circles are to one another in the duplicate ratio, or as the squares 
 of their diameters. 
 
 Let ABD and GHL be two circles, of which the diameters are AD and 
 GL; the circle ABD is to the circle GHL as the square of AD to the 
 square of GL. 
 
 Let ABCDEF and GHKLMN be two equilateral polygons of the same 
 number of sides inscribed in the circles ABD, GHL ; and let Q be such a 
 
 DG 
 
 •pace that the square of AD is to the square of GL as the circle ABP to 
 t;..> spec* Q. Because the polygons ABCDEF and GHKLMN are equi 
 lateral an J of the same number of sides, they are similar (2. I. Sup.\ and
 
 172 SUPPLEMENT TO THE ELEMENTS 
 
 men ared:-. are as the squares of the diameters of the circles in which they 
 are inscribed. Therefore AD 2 : GL 2 : : polygon ABCDEF : polygon 
 GHKLMN; but AD 3 : GL 2 : : circle ABD : Q ; and therefore, ABCDEF 
 : GHKLMN :: circle ABD : Q. Now, circle ABD 7 ABCDEF; there- 
 fore Q 7 GHKLMN (14. 5.), that is, Q is greater than any polygon in- 
 scribed in the circle GHL. 
 
 In the same manner it is demonstrated, that Q is less than any polygon 
 described about the circle GHL ; wherefore the space Q is equal to tho 
 circle GHL (2. Cor. 4. 1. Sup.). Now, by hypothesis, the circle ABD is 
 to the space Q as the square of AD to the square of GL ; therefore the 
 circle ABD is to the circle GHL as the square of AD to the square of GL. 
 
 Cor. 1. Hence the circumferences of* circles are to one another as 
 their diameters. 
 
 Let the straight line X be equal to half the circumference of the circle 
 \BD and the straight line Y to half the circumference of the circle GHL • 
 
 And because the rectangles AO.X and GP.Y are equal to the circles ABD 
 and GHL (5. 1. Sup.), therefore AO.X : GP.Y : : AD 2 : GL 2 : : AO 2 : 
 GP 2 ; and alternately, AO.X : AO 2 : : GP.Y : GP 2 ; whence, because 
 rectangles that have equal altitudes are as their bases (1. G.). X : AO : : 
 Y : GP,and again alternately, X : Y : : AO : GP: wherefore, taking the 
 doubles of each, the circumference ABD is to the circumference GHL a* 
 the diameter AD to the diameter GL. 
 
 Cor. 2. The circle that is described upon the side of a right angled 
 triangle opposite to the right angle, is equal to the two circles described on 
 the other two sides. For the circle described upon SR is to the circle de- 
 scribed upon RT as the square of SR to the square of RT ; and the circl« 
 described upon TS is to the circle described upon RT as the square of ST 
 to the square of RT. Wherefore, 
 the circles described on SR and on 
 ST are to the circle described on RT 
 as the squares of SR and of ST to 
 the square of RT (24. 5.). But the 
 squares of RS and of ST are equal 
 to the square of RT (47. 1.) ; there- 
 fore the circles described on RS and 
 ST are equal to the circle described 
 on RT 
 
 PROP. VII. THEOR. 
 
 Equiangular parallelograms are to one another as the products of the nnm 
 bers proportional to their sides. 
 
 Let AC and DF be two equiangular parallelograms, and let M, N, P 
 *nd Q be four numbers, such that AB : BC : : M . N ; AB : DE : ■ \t
 
 UF GEOMETRY. BOOK I. 173 
 
 P , and AB : EF : : M : Q, and therefore ex aequali, BC : EF : . N . Q 
 The parallelogram AC is to the parallelogram DF as MN to PQ. 
 
 Let NP he the product of N into P, and the ratio of MN to PQ will be 
 compounded of the ratios (def. 10. 5.) of MN to NP, and NP to PQ 
 But the ratio of MN to NP is the same with that of M to P (15. 5.), be 
 
 E 
 
 A. B D E 
 
 cause MN and NP are equimultiples of M and P ; and for the same reason, 
 the ratio of NP to PQ is the same with that of N to Q ; therefore the ratio 
 of MN to PQ is compounded of the ratios of M to P, and of N to Q. Now, 
 the ratio of M to P is the same with that of the side AB to the side DE (by 
 Hyp.) ; and the ratio of N to Q the same with that of the side BC to the 
 side EF. Therefore, the ratio of MNtoPQ is compounded of the ratios 
 of AB to DE, and of BC to EF. And the ratio of the parallelogram AC 
 to the parallelogram DF is compounded of the same ratios (23. 6.) ; there- 
 fore, the parallelogram AC is to the parallelogram DFas MN, the product 
 of the numbers M and N, to PQ, the product of the numbers P and Q. 
 
 Cor. 1. Hence, if GH be to KL as the number M to the number N ; 
 the square described on GH will be to 
 
 the square described on KL as MM. the G H K L 
 
 square of the number M to NN, the 
 square of the number N. 
 
 Cor. 2. If A, B, C, D, &c. are any lines, and m, n, r, s, &c. numbers 
 proportional to them ; viz. A : B : : m : n, A : C : : m : r, A : D : : m : s, 
 &c. ; and if the rectangle contained by any two of the lines be equal to the 
 square of a third line, the product of the numbers proportional to the first 
 two, will be equal to the square of the number proportional to the third , 
 that is, if A.C = B 2 , mXr=nX/«, or=// 2 . 
 
 For by this Prop. A.C : B 2 : : mXr : n 2 ; but A.C=B 2 , therefore mXt 
 =n 2 . Nearly in the same way it may be demonstrated, that whatever ia 
 the relation between the rectangles contained by these lines, there is the 
 same between the products of the numbers proportional to them. 
 
 So also conversely if m and r be numbers proportional to the lines A and 
 C ; if also A.C=B 2 , and if a number n be found such, that n 2 =mr, then 
 A : B : : m : n. For let A : B : : m : q, then since m, q, r are proportional 
 to A, B, and C, and A.C=B 2 ; therefore, as has just been proved. q 2 =m 
 Xr; but n 2 =qXr, by hypothesis, therefore n 2 =y 2 , and n=^ ; wherefore 
 A : B : : m : n 
 
 SCHOLIUM. 
 In order to have numbers proportional to any set of magnitudes ol the
 
 174 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 same kind, suppose one of them to be divided into any number m, of eqaal 
 parts, and let H be one of those parts. Let H be found n times in the mag- 
 nitude B, r times in C, s times in D, &c, then it is evident that the num- 
 bers m, n, r, s are proportional to the magnitudes A, B, C and D. When 
 therefore it is said in any of the following propositions, that a line as A=s 
 a number m, it is understood that A=mxH, or that A is equal to the given 
 magnitude H multiplied by m, and the same is understood of the other 
 magnitudes, B, C, D, and their proportional numbers, H being the common 
 measure of all the magnitudes. This common measure is omitted for the 
 sake of brevity in the arithmetical expression ; but is always implied, when 
 a line, or other geometrical magnitude, is said to be equal to a number 
 Also, when there are fractions in the number to which the magnitude is 
 called equal, it is meant that the common measure H is farther subdivided 
 into such parts as tho numerical fraction indicates. Thus, if A=360.375, 
 it is meant that there is a certain magnitude H, such that A=360 X H+ 
 
 375 
 
 X H, or that A is ?qual to 360 times H, together with 375 of the 
 
 thousandth parts of H. And the same is true in all other cases, where 
 numbers are used to express the relations of geometrical magnitudes. 
 
 PROP. VIII. THEOR. 
 
 The perpendicular drawn from the centre of a circle on the chord of any arc is a 
 mean proportional between half the radius and the line made up of the radius 
 and the perpendicular drawn from the centre on the chord of double that arc : 
 And the chord of the arc is a mean proportional between the diameter and a line 
 which is the difference between the radius and the aforesaid perpendicular from 
 the centre 
 
 Let ADB be a circle, of which the centre is C ; DBE any arc, and DB 
 the half of it ; let the chords DE, DB be drawn : as also CF and CG at 
 right angles to DE and DB ; if CF be produced it will meet the circum 
 ference in B : let it meet it again in A, and let AC be bisected in H ; CG
 
 OF GEOMETRY. BOOK I. 
 
 175 
 
 18 a mean proportional between AH and AF ; and BD a mean proportional 
 between AB and BF, the excess of the radius above CF. 
 
 Join AD ; and because ADB is a right angle, being an angle in a semi 
 circle ; and because CGB is also a right angle, the triangles ABD, CBQ 
 are equiangular, and, AB : AD : : BC : CG (4. 6.), or alternately, AB : 
 BC : : AD : CG ; and therefore, because AB is double of BC, AD is dou- 
 ble of CG, and the square of AD therefore equal to four times the square 
 of CG. 
 
 But, because ADB is a right angled triangle, and DF a perpendicular 
 on AB, AD is a mean proportional between AB and AF (8. 6.), and AD' 
 =AB.AF (17. 6.), or since AB is =4AH, AD 2 =4AH.AF. Therefore 
 also, because 4CG 2 =AD 2 , 4CG 2 =4AH.AF, and CG^AH.AF ; where 
 fore CG is a mean proportional between AH and AF, that is, between half 
 the radius and the line made up of the radius, and the perpendicular on the 
 chord of twice the arc BD. 
 
 Again, it is evident that BD is a mean proportional between AB and BF 
 (8. 6.), that is, between the diameter and the excess of the radius above 
 the perpendicular, on the chord of twice the arc DB. 
 
 PROP. IX. THEOR.* 
 
 The circumference of a circle exceeds three times the diameter, by a line less 
 than ten of the parts, of which the diameter contains seventy, but greatet 
 than ten of the parts whereof the diameter contains seventy-one. 
 
 Let ABD be a circle, of which the centre is C, and the diameter AB ; 
 the circumference is greater than three times AB, by a line less than =rr, ot 
 
 70' 
 
 -, of AB, but greater than — of AB. 
 
 * In thia proposition, the character -{- placed after a numb-r, si^nh.^s that romething n tt 
 to added to it ; and th» character - ,on the other hand, sanities that something is to be takai 
 •v f from it
 
 176 SUPPLEMENT TO THE ELEMENTS 
 
 In the circle ABD apply the straight line BD equal to the radius BC . 
 Draw DF perpendicular to BC, and let it meet the circumference again in 
 E ; draw also CG perpendicular to BD : produce BC to A, bisect AC in 
 H, and join CD. 
 
 It is evident, that the arcs BD, BE are each of them one-sixth of the 
 circumference (Cor. 15. 4.), and that therefore the arc DBE is one third of 
 the circumference. Wherefore, the line (8. 1. Sup.) CG is a mean pro- 
 portional between AH, half the radius, and the line AF. Now because the 
 sides BD, DC, of the triangle BDC are equal, the angles DCF, DBF are 
 also equal ; and the angles DFC, DFB being equal, and the side DF com- 
 mon to the triangles DBF, DCF, the base BF is equal to the base CF, and 
 BC is bisected in F. 
 
 Therefore, if AC or BC = 1000, AH=500, CF=500, AF=1500, and 
 CG being a mean proportional between AH and AF, CG 2 =(17. 6.) AH. 
 AF=500x 1500=750000; wherefore CG=866.0254+, because (866. 
 0254) 2 is less than 750000. Hence also, AC+CG=1866.0254+. 
 
 Now, as CG is the perpendicular drawn from the centre C, on the chord 
 of one-sixth of the circumference, if P = the perpendicular from C on the 
 chord of one-twelfth of the circumference, P will be a mean proportional 
 between AH (8. 1. Sup.) and AC+CG, and P 2 =AH (AC+CG)= 
 .500 X (1866.0254+) = 933012.7+. Therefore, P = 965.9258+ , be- 
 cause (965. 9258) 2 is less than 933012.7. Hence also, AC + P= 1965. 
 9258+. 
 
 Again, if Q = the perpendicular drawn from C on the chord of one 
 twenty-fourth of the circumference, Q will be a mean proportional between 
 AH and AC+P, and Q 2 =AH (AC+P)=500(1965.9258+)=982962. 
 9+ ; and therefore Q=991.4449+, because (991.4449) 2 is less than 
 982962.9. Therefore also AC+Q=1991.4449+. 
 
 In like manner, if S be the perpendicular from C on the chord of one 
 forty-eighth of the circumference, S 2 =AH (AC+Q)=500 (1991.4449+) 
 =995722.45+ ; and S=997.8589+, because (997.8589) 2 is less than 
 995722.45. Hence also, AC+S = 1997.8589+. 
 
 Lastly, if T be the perpendicular from C on the chord of one ninety-sixth 
 of the circumference, T 2 =AH (AC + S)=500 (1997.8589+)=998929. 
 45+, and T=999.46458+. Thus T, the perpendicular on the chord of 
 one ninety-sixth of the circumference, is greater than 999.46458 of those 
 parts of which the radius contains 1000. 
 
 But by the last proposition, the chord of one ninety-sixth part of the cir- 
 cumference is a mean proportional between the diameter and the excess of 
 the radius above S, the perpendicular from the centre on the chord of one 
 forty-eighth of the circumference. Therefore, the square of the chord of 
 one ninety-sixth of the circumference =AB (AC— S)=2000x(2.1411— ,) 
 =4282.2--; and therefore the chord itself =65.4386 — , because (65. 
 4386) 2 is greater than 4282.2. Now, the chord of one ninety-sixth of the 
 circumference, or the side of an equilateral polygon of ninety-six sides in- 
 scribed in the circle, being 65.4386 — , the perimeter of that polygon will be 
 =(65.4386—) 96=6282.1056—. 
 
 Let the perimeter of the circumscribed polygon of the same number of 
 aides, be M,then (2. Cor. 2. 1. Sup.) T : AC : : 6282.1056— : M, that is, 
 (since T=999.46458+, as already shewn),
 
 OF GEOMETRY. BOOK I. 
 
 177 
 
 999.46458+ : 1000 : : 6282.1056— : M ; if then N be such 
 -hat 999.46458 : 1000 : : 6282.1056— : N ; ex aequo perturb. 999.46458 
 + : 999.46458 : : N : M ; and, since the first is greater than the second, 
 the third is greater than the fourth, or N is greater than M. 
 
 Now, if a fourth proportional be found to 999.46458, 1000 and 6282. 
 1056 vi2 6285.461— , then, 
 
 because, 999.46458 : 1000 ; : 6282.1056 : 6285.461—, 
 and as before, 999.46458 : 1000 : : 6282.1056— : N ; 
 therefore, 6282.1056 : 6282.1056— : : 6285.461— N, and as the first o. 
 ihese proportionals is greater than the second, the third, viz. 6285 46 1 
 
 is greater than N, the fourth. But N was proved to be greater than M ; 
 much more, therefore, is 6285.461 greater than M, the perimeter of a poly- 
 gon of ninety-six sides circumscribed about the circle ; that is, the perime- 
 ter of that polygon is less than 6285.461 ; now, the circumference of the 
 circle is less than the perimeter of the polygon ; much more, therefore, is it 
 less than 6285.461 ; wherefore the circumference of a circle is less than 
 6285.461 of thoso parts of which the radius contains 1000. The circum- 
 ference, therefore has to the diameter a less ratio (8. 5.) than 6285.461 has 
 to 2000, or than 3142.7305 has to 1000 : but the ratio of 22 to 7 is greater 
 than the ratio of 3142.7305 to 1000, therefore the circumference has a less 
 ratio to the diameter than 22 has to 7, or the circumference is less than 22 
 of the parts of which the diameter contains 7. 
 
 It remains to demonstrate, that the part by which the circumference ex- 
 ceeds the diameter is greater than — of the diameter. 
 
 It was before shewn, that CG 2 =750000 ; wherefore CG=866.02545— . 
 because (866.02545) 2 is greater than 750000 ; therefore AC + CG = 1866 
 02545—. 
 
 Now, P being, as before, the perpendicular from the centre on the chord 
 of one twelfth of the circumference, P 2 =AH (AC+CG) =^500x(1866 
 02545) =933012.73—; and P = 965.92585— , because (965.02585)* 
 is greater than 633012.73. Henefl also, AC + P = 1965.92585 — 
 
 23
 
 178 SUPPLEMENT TO THE ELEMENTS 
 
 Next, as Q= the perpendicular drawn from the centre on the chord o( 
 one twenty-fourth of the circumference, Q 2 =AH (AC+P)=500x (1965. 
 92585- ) =982962.93— ; and Q = 991.44495 — .because (991.44496) 2 
 is greater than 982962.93. Hence also, AC + Q = 1991.44495 — . 
 
 In like manner, as S is the perpendicular from C on the chord of one 
 forty-eighth of the circumference, S 2 = AH (AC + Q)=500(1991.44495 — ) 
 =995722.475 — , and S = (997.85895 — ) because (997.85895) 2 is greater 
 than 995722.475. 
 
 But the square of the chord of the ninety-sixth part of the circumference 
 = AB (AC— S)=2000 (2.14105 + )=4282.1 + , and the chord itself =-. 
 65.4377+ because (65.4377) 2 is less than 4282.1 : Now the chord ofoiw 
 ninety-sixth part of the circumference being =65.4377 + , the perimele. 
 of a polygon of ninety-six sides inscribed in the circle =(65.4377+ )96=- 
 6282.019 + . But the circumference of the circle is greater than he pe 
 rimeter of the inscribed polygon ; therefore the circumference is greater 
 than 6282.019, of those parts of which the radius contains 1000 ; or thu.y 
 3141.009 of the parts of which the radius contains 500, or the diameter 
 
 contains 1000. Now, 3141.009 has to 1000 a greater ratio than 3+ -- 
 
 to 1 ; therefore the circumference of the circle has a greater ratio to the 
 
 diameter than 3+ ^-r nas t0 * 5 ^ at * s > ^ e excess of the circumference 
 
 / 1 
 
 above three times the diameter is greater than ten of those parts of which 
 the diameter contains 71 ; and it has already been shewn to be less than 
 ten of those of which the diameter contains 70. 
 
 Cor. 1. Hence the diameter of a circle being given, the circumference 
 may be found nearly, by making as 7 to 22, so the given diameter to a 
 fourth proportional, which will be greater than the circumference. Ani 
 
 if as 1 to 3 + — , or as 71 or 223, so the given diameter to a fourth pro 
 
 portional, this will be nearly equal to the circumference, but will be less 
 than it. 
 
 Cor. 2. Because the difference between - and — - is — — , therefore the 
 
 7 71 497 
 
 lines found by these proportionals differ by — — of the diameter. There- 
 fore the difference of either of them from the circumference must be less 
 than the 497th part of the diameter. 
 
 Cor. 3. As 7 to 22, so the square of the radius to the area of the circle 
 nearly. 
 
 For it has been shewn, that (1. Cor. 5. 1. Sup.) the diameter of a cir 
 cle is to its circumference as the square of the radius to the area of the 
 circle ; but the diameter is to the circumference nearly as 7 to 22, there- 
 fore the square of the radius is to the area of the circle nearly in that same 
 ratio
 
 OF GEOMETRY. BOOK I. 
 
 179 
 
 SCHOLIUM. 
 
 It is evident that the method employed in this proposition, for finding 
 the limits of the ratio of the circumference of the diameter, may be carrie? 
 to a greater degree of exactness, by finding the perimeter of an inscribed 
 And of a circumscribed polygon of a greater number of sides than 96. The 
 manner in which the perimeters of such polygons approach nearer to one 
 another, as the number of their sides increases, may be seen from the fol- 
 lowing Table, which is constructed on the principles explained in the fore 
 going Proposition, and in which the radius is supposed = 1 . 
 
 NO. of Sides 
 
 Perimeter of the 
 
 Perimeter of the 
 
 of the Poly- 
 
 inscribed Poly- 
 
 circumscribed 
 
 gon. 
 
 gon. 
 
 Polygon. 
 
 6 
 
 6.000000 
 
 6.822033 — 
 
 12 
 
 6.211657+ 
 
 6.430781 — 
 
 24 
 
 6.265257+ 
 
 6.319320— 
 
 48 
 
 6.278700+ 
 
 6.292173 — 
 
 96 
 
 6.282063 + 
 
 6.285430 — 
 
 192 
 
 6.282904+ 
 
 6.283747— 
 
 384 
 
 6.283115+ 
 
 6.283327— 
 
 768 
 
 6.283167+ 
 
 6.283221 — 
 
 1536 
 
 6.283180+ 
 
 6.283195 — 
 
 3072 
 
 6.283184+ 
 
 6.283188— 
 
 6144 
 
 6.283185 + 
 
 6.283186 — 
 
 The part that is wanting in the numbers of the second column, to make 
 up the entire perimeter of any of the inscribed polygons, is less than unit 
 in the sixth decimal place ; and in like manner, the part by which the 
 numbers in the last column exceed the perimeter of any of the circumscrib- 
 ed polygons is less than a unit in the sixth decimal place, that is, than 
 
 of the radius. Also, as the numbers in the second column are 
 
 less than the perimeters of the inscribed polygons, they are each of them 
 less than the circumference of the circle ; and for the same reason, each of 
 those in the third column is greater than the circumference. But when 
 
 the arc of - of the circumference is bisected ten times, the number of sides 
 o 
 
 in the polygon is 6144, and the numbers in the Table differ from one an- 
 other only by part of the radius, and therefore the perimeters o. 
 
 the polygons differ by less than that quantity ; and consequently the cir- 
 cumference of the circle, which is greater than the least, and less than the 
 greatest of these numbers, is determined within less than the millionth 
 pan of the radius. 
 
 Hence also, if R be the radius of any circle, the circumference is gTeatei 
 than Rx 6.283 185. or than 2Rx 3.1 41592, but less than 2Rx3.141593 .
 
 180 SUPPLEMENT TO THE ELEMENTS, &c. 
 
 and these numbers differ from one another only by a millionth part of the 
 radius. So also R 2 +3.141592 is less, and R 2 X 3.141593 greater than the 
 area of the circle ; and these numbers differ from One another only by a 
 millionth part of the square of the radius. 
 
 In this way, also, the circumference and the area of the circle may be 
 found still nearer to the truth ; but neither by this, nor by any other me- 
 thod yet known to geometers, can they be exactly determined, though the 
 errors of both may be reduced to a less quantity than any that can be a»» 
 tinned
 
 ELEMENTS 
 
 OF 
 
 GEOMETRY. 
 
 SUPPLEMENT. 
 
 BOOK II. 
 
 OF THE INTERSECTION OF PLANES. 
 
 DEFINITIONS. 
 
 * A straight line is perpendicular or at right angles to a plane, when 
 it makes right angles with every straight line which it meets in that 
 plane. 
 
 2. A plane is perpendicular to a plane, when the straight lines drawn in 
 one of the planes perpendicular to the common section of the two planes 
 are perpendicular to the other plane. 
 
 3. The inclination of a straight line to a piano is the acute angle contained 
 by that straight line, and another drawn from the point in which the 
 first line meets the plane, to the point in which a perpendicular to the 
 plane, drawn from any point of the first line, meets the same plane. 
 
 4. The angle made by two planes which cut one another, is the angle con 
 tained by two straight lines drawn from any, the same point in the line 
 of their common section, at right angles to that line, the one, in the one 
 plane, and the other, in the other. Of the two adjacent angles made by 
 two lines d/awn in this manner, that which is acute is also called the In- 
 clination of the planes to one another. 
 
 5. Two planes are said to have the same, or a like inclination to one an- 
 other, which two other planes have, when the angles of inclination above 
 defined are equal to one another. 
 
 •. A straight line is said to be parallel to a plane, when it does not meet 
 the plane, though produced ever so far.
 
 182 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 7. Planes are said to be parallel to one another, which do not meet, though 
 produced ever so far. 
 
 8. A solid angle is an angle made by the meeting of more than two plane 
 angles, which are not in the same plane in one point. 
 
 PRCP. I. THEOR. 
 
 One part of a straight line cannot be in a plane and another part above it. 
 
 If it be possible let AB, part of the straight line ABC, be in the plane, 
 and the part BC above it : and since the 
 straight line AB is in the plane, it can be 
 produced in that plane (2. Post. 1.) ; let 
 it be produced to D : Then ABC and 
 ABD are two straight lines, and they 
 have the common segment AB, which is 
 impossible (Cor. def. 3. 1.). Therefore 
 ABC is not a straight line. 
 
 PROP. II. THEOR 
 
 Any three straight lines which meet one another, not in the same point, are m 
 
 one plane. 
 
 Let the three straight lines AB, CD, CB meet one another in the points 
 B, C and E ; AB, CD, CB are in one plane. 
 
 Let any plane pass through the straight line 
 EB, and let the plane be turned about EB, pro- 
 duced, if necessary, until it pass through the 
 point C : Then, because the points E, C are in 
 this plane, the straight line EC is in it (def. 5. 1.) : 
 for the same reason, the straight line BC is in 
 the same ; and, by the hypothesis, EB is in it ; 
 therefore the three straight lines EC, CB, BE 
 are in one plane : but the whole of the lines DC, 
 AB, and BC produced, are in the same plane 
 with the parts of them EC, EB, BC (1. 2. 
 Sup.) Therefore AB, CD, CB, are all in one 
 plane. 
 
 Cor. It is manifest, that any two straight lines which cut one anothex 
 are in one plane . Also, that any three points whatever are in one plane
 
 OF GEOMETRY. BOOK II. 
 
 183 
 
 PROP. III. THEOR. 
 
 [fntni planes cut one another, their common section ts a straight Une. 
 
 Let two planes AB, BC cut one another, 
 and let B and D be two points in the line of 
 their common section. From B to D draw the 
 straight line BD ; and because the points B 
 and D are in the plane AB, the straight line 
 BD is in that plane (def. 5. 1.): for the same 
 reason it is in the plane CB ; the straight line 
 13 D is therefore common to the planes AB 
 and BC, or it is the common section of these 
 olanes. 
 
 PROP. IV. THEOR. 
 
 If a straight line stand at right angles to each of two straight lines in the 
 point of their intersection, it will also be at right angles to the plane in 
 which these lines are. 
 
 Let the straight line AB stand at right angles to each of the straight 
 lines EF, CD in A, the point of their intersection : AB is also at right an- 
 gles to the plane passing through EF, CD. 
 
 Through A draw any line AG in the 
 plane in which are EF and CD ; let G be 
 any point in that lino ; draw GH parallel 
 to AD; and make HF=HA, join FG ; and 
 when produced let it meet CA in D ; join 
 BD, BG, BF. Because GH is parallel to 
 AD, and FH = HA : therefore FG=GD, 
 so that the line DF is bisected in G. And 
 because BAD is a right angle, BD 2 =AB 2 
 -f-AD 2 (47. 1.); and for the same reason, 
 BF 2 = AB 2 +AF 2 , therefore BD 2 +BF 2 = 
 2AB 2 + AD 2 + AF 2 ; and because DF is 
 bisected in G (A. 2.), Al) 2 +AF 2 =2AG 2 -f- 
 2GF 2 , therefore BD 2 +BF 2 =2AB 2 +2AG 2 
 +2GF 2 . But BD 2 + BF 2 = (A. 2.) 2BG 2 +2GF», therefore 2BG 2 -j- 
 2GF 2 =2AB 2 +2AG 2 +2GF 2 ; and taking 2GF 2 from both, 2BG 2 =2AB» 
 4-2AG 2 , or BG 2 =AB 2 +AG 2 ; whence BAG (48. 1.) is a right angle. 
 Now AG is any straight line drawn in the plane of the lines AD, AF ; and 
 "hen a straight line is at right angles to any straight line which it meet* 
 with in a plane, it is at right angles to the plane itself (def. 1. 2. Sup.). AB 
 is therefore at right angles to the plane of the lines AF, AD.
 
 84 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 PROP. V. THEOR. 
 
 If three s 'raight lines meet all in one point, and a straight line stand at rtghl 
 angles to each of them in that point ; these three straight lines are in one 
 . and the same plane. 
 
 Let the straight line AB stand at right angles to each of the straight 
 lines BC, BD, BE, in B, the point where they meet ; BC, BD, BE are in 
 one and the same plane. 
 
 If not, let BD and BE, if possible, be in one plane, and BC be above it ; 
 and let a plane pass through AB, BC, the common section of which with 
 the plane, in which BD and BE are, shall be a straight (3. 2. Sup.) line ; 
 let this be BF : therefore the three straight lines AB, BC, BF are all in 
 one plane, viz. that which passes through AB, BC ; and because AB 
 stands at right angles to each of the straight lines BD, BE, it is also at 
 right angles (4. 2. Sup.) to the plane passing 
 through them ; and therefore makes right an- 
 gles with every straight line meeting it in that 
 plane ; but BF which is in that plane meets it ; 
 therefore the angle ABF is a right angle ; but 
 the angle ABC, by the hypothesis is also a right 
 angle ; therefore the angle ABF is equal to the 
 angle ABC, and they are both in the same 
 plane, which is impossible : therefore the straight 
 line BC is not above the plane in which are BD 
 and BE : Wherefore the three straight lines 
 BC, BD, BE are in one and the same plane. 
 
 PROP. VI. THEOR. 
 
 Two straight lines which are at right angles to the same plane, are parallel to 
 
 one another. 
 
 Let the straight lines AB, CD be at right angles to the same plane BDE ; 
 AB is parallel to CD. 
 
 Let them meet the plane in the points B, D. 
 Draw DE at right angles to DB, in the plane BDE, 
 and let E be any point in it: Join AE, AD, EB. 
 Because ABE is a right angle, AB 2 +BE 2 = (47. 1 .) 
 AE 2 , and because BDE is a right angle, BE 2 =BD 2 
 + DE 2 ; therefore AB 2 +BD 2 -f DE 2 =AE 2 ; now, 
 AB 2 +BD 2 =AD 2 , because ABD is a right angle, 
 therefore AD 2 +DE 2 =AE 2 , and ADE is therefore 
 a (48. 1.) right angle. Therefore ED is perpendi- 
 cular to the three lines BD, DA, DC, whence these 
 lines are in one plane (5. 2. Sup.). But AB is in the 
 plane in which are BD, DA, because any three 
 straight lines, which meet one another, are in one
 
 OF GEOMETRY. BOOK II. 185 
 
 plane (2. 2. Sup.) : therefore AB, BD, DC are in one plane ; and each ol 
 the angles ABD, BDC is a right angle ; therefore AB is parallel (Cor. 28 
 l.)toCD. 
 
 PROP. VII. THEOR. 
 
 If two straight lines be parallel, and one of them at right angles to a vlane 
 the other is also at right angles to the same plane. * 
 
 c& 
 
 Let AB, CD be two parallel straight 
 lines, and let one of them AB be at 
 right angles to a plane ; the other CD 
 is at right angles to the same plane. 
 
 For, if CD be not perpendicular to 
 the plane to which AB is perpendicular, 
 let DG be perpendicular to it. Then 
 (6. 2. Sup.) DG is parallel to AB : DG E 
 and DC therefore are both parallel to ^ 
 AB, and are drawn through the same 
 point D, which is impossible (11. Ax. \ **" \tx 
 
 PROP. VIII. THEOR. 
 
 Two straight lines which are each of them parallel to the same straight line, 
 though not both in the same plane with it, are parallel to one another. 
 
 Let AB, CD be each of them parallel to EF, and not in the same plane 
 with it ; AB shall be parallel to CD. 
 
 In EF take any point G, from which draw, in the plane passing through 
 EF, AB, the straight line GH at right angles to EF; and in the plane 
 passing through EF, CD, draw GK at right angles to the same EF. 
 And because EF is perpendicular both to GH and GK, it is perpendicular 
 (4. 2. Sup.) to the plane HGK passing through them : and EF is parallel 
 to AB ; therefore AB is at right 
 angles (7. 2. Sup.) to the plane 
 HGK. For the same reason, CD 
 is likewise at right angles to the 
 plane HGK. Therefore AB, CD 
 are each of them at right angles 
 to the plane HGK. But if two 
 straight lines are at right angles 
 to the same plane, they are paral- 
 
 K 
 
 'el (6. 2. Sup.) to one another. Therefore AB is parallel to CD. 
 
 PROP. IX. THEOR. 
 
 If two straight lines meeting one another be parallel to two others that meet one 
 another, though not in the same plane with the first two ; the first two and the 
 other two shall contain equal angles. 
 
 Let the two straight lines AB, BC which meet one another, be parallel 
 
 24
 
 186 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 to the two straight lines DE, EF that meet one another, anil are not in the 
 tame plane with Ali, BC. The angle ABC is equal to the angle DEF 
 
 Take BA, BC, ED, EF all equal to one an- 
 other ; and join AD, CF, BE, AC, DF : Because 
 BA is equal and parallel to ED, therefore AD is 
 (33. 1.) both equal and parallel to BE. For the 
 same reason, CF is equal and parallel to BE. 
 Therefore AD and CF are each of them equal and 
 parallel to BE. But straight lines that are paral- 
 lel to the same straight line, though not in the 
 same plane with it, are parallel (8. 2. Sup.) to one 
 another. Therefore AD is parallel to CF ; and it 
 is equal to it, and AC, DF join them towards the 
 same parts ; and therefore (33. 1.) AC is equal 
 and parallel to DF. And because AB, BC are 
 <qual to DE, EF, and the base AC to the base 
 DF ; the angle ABC is equal (8. 1.) to the angle 
 DEF. 
 
 PROP. X. PROB. 
 
 To draw a straight line perpendicular to a plane, from a given point above it 
 
 Let A be the given point above the plane BH, it is required to draw from 
 (he point A a straight line perpendicular to the plane BH. 
 
 In the plane draw any straight line BC, and from the point A draw (Prop 
 12. 1.) ADpeipendicular to BC. If then AD be also perpendicular to the 
 plane BH, the thing required is already done ; but if it be not, from the 
 point D draw (Prop. 11. 1), in the 
 plane BH, the straight line DE at 
 right angles to BC ; and from the 
 point A draw AF perpendicular to 
 DE ; and through F draw (Prop. 31 
 1.) GH parallel to BC : and because 
 BC is at right angles to ED, and DA, 
 BC is at right angles (4. 2. Sup.) to 
 the plane passing through ED, DA. 
 And GH is parallel to BC ; but if two 
 straight lines be parallel, one of which is at right angles to a plane, the 
 other shall be at right (7. 2. Sup.) angles to the same plane ; wherefore 
 GH is at right angles to the plane through ED, DA, and is perpendicular 
 (def. 1. 2. Sup.) to every straight line meeting it in that plane. But AF, 
 which is m the plane through ED, DA, meets it : Therefore GH is per- 
 pendicular to AF, and consequently AF is perpendicular to GH ; and AF 
 is also perpendicular to DE : Therefore AF is perpendicular to each of ihe 
 straight lines GH, DE. But if a straight line stands at right angles a 
 each of two straight lines in the point of their intersection, it. is also at right 
 angles to the plane passing through them (4. 2. Sup.). And the plane 
 passing through ED, GH is the plane BH ; therefore A v is perpendiculai
 
 OF GEOMETRY. BOOK II. 
 
 187 
 
 to the plane BH ; so that, from the given point A, above the plane BH, 
 the straight line AF is drawn perpendicular to that plane. 
 
 Cor. If it be required from a point C in a plane to erect a perpen 
 dicular to that plane, take a point A above the plane, and draw AF per 
 pendicular to the plane ; then, if from C a line be drawn parallel to AF 
 it will be the perpendicular required ; for being parallel to A F it will be 
 perpendicular to the same plane to which AF is perpendicular (7. 2. Sup.) 
 
 PROP. XI. THEOR. 
 
 From the same point in a plane, there cannot be two straight lines at right 
 angles to the plane, upon the same side of it; And there can be but one 
 perpendicular to a plane from a point above it. 
 
 For if it be possible, let the two straight lines AC, AB be at right angles 
 to a given plane from the same point A in the plane, and upon the same 
 side of it ; and let a plane pass through BA, AC ; the common section o( 
 this plane with the given plane is a straight (3. 2. Sup.) line passing through 
 A : Let DAE be their common section : Therefore the straight lines AB, 
 AC, DAE are in one plane: And because CA is at right angles to the 
 given plane, it makes right angles with every 
 straight line meeting it in that plane. But 
 DAE, which is in that plane, meets CA: there- 
 fore CAE is a right angle. For the same rea- 
 son BAE is a right angle. Wherefore the an- 
 gle CAE is equal to the angle BAE ; and 
 they are in one plane, which is impossible. 
 Also, from a point above a plane, there can bo 
 but one perpendicular to that plane ; for if there 
 could be two, they would be parallel (6. 2. Sup.) to one another, which is 
 absurd. 
 
 PROP. XII. THEOR. 
 
 Planes to which the same straight line is perpendicular, are parallel to one 
 
 another. 
 
 Let the straight line AB be perpendicular to 
 each of the planes CD, EF : these planes are pa- 
 rallel to one another. 
 
 If not, they must meet one another when pro- 
 duced, and their common section must be a straight 
 line GH, in which take any point K, and join AK, 
 BK : Then, because AB is perpendicular to the 
 plate EF, it is perpendicular (def. 1. 2. Sup.) to 
 the straight line BK which is in that plane, and 
 therefore ABK is a right angle. For the same 
 reason, BAK is • right angle ; wherefore the two 
 anglei ABK, BAK of the triangle ABK are 
 fual to two right angles, which is impossible,
 
 98 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 (17 1.): Therefore the planes CD, EF, though produced, do not meet 
 one another ; that is, they are parallel (def. 7. 2. Sup.) 
 
 PROP. XIII. THEOR. 
 
 If two straight lines meeting one another, be parallel to two straight hnet 
 which also meet one another, but are not in the same plane with the first 
 two : the plane which passes through the first two is parallel to the plane 
 passing through the others. 
 
 Let AB, BC, two straight lines meeting one another, be parallel to DE, 
 EF that meet, one another, but are not in the same plane with AB, BC : 
 The planes through AB, BC, and DE, EF shall not meet, though pro- 
 duced. 
 
 From the point B draw BG perpendicular (10. 2. Sup.) to the plane 
 which passes through DE, EF, and let it meet that plane in G ; and 
 through G draw GH parallel to ED (Prop. 31. 1.), and GK parallel to EF : 
 And because BG is perpendicular to the plane through DE, EF, it rata 
 make right angles with every 
 straight line meeting it in that 
 plane (1. def. 2. Sup.). But 
 the straight lines GH, GK in 
 that plane meet it : Therefore 
 each of the angles BGH, BGK 
 is a right angle : And because 
 BA is parallel (8. 2. Sup.) to 
 GH (for each of them is paral- 
 lel to DE), the angles GBA, 
 BGH are together equal (29 
 1.) to two right angles: And 
 
 BGH is aright angle ; therefore also GBA is a right angle, and GB per- 
 pendicular to BA : For the same reason, GB is perpendicular to BC : 
 Since, therefore, the straight line GB stands at right angles to the two 
 straight lines BA, BC, that cut one another in B ; GB is perpendicular 
 (4. 2. Sup.) to the plane through BA, BC : And it is perpendicular to the 
 plane through DE, EF ; therefore BG is perpendicular to each of the 
 planes through AB, BC, and DE, EF : But planes to which the same 
 straight line is perpendicular, are parallel (12. 2. Sup.) to one another: 
 Therefore the plane through AB, BC, is parallel to the plane through 
 DE, EF. 
 
 Cor. It follows from this demonstration, that if a straight line meet 
 two parallel planes, and be perpendicular to one of them, it must be per* 
 pendicular to the other also.
 
 OF GEOMETRY. BOOK II. 
 
 189 
 
 PROP. XIV. THEOR. 
 
 lj two parallel planes be cut by another plane, their common sections with U 
 
 are parallels. 
 
 Let the parallel planes AB, 
 CD, be cut by the plane EFHG, 
 and let their common sections with 
 it be EF, GH ; EF is parallel to 
 GH. 
 
 For the straight lines EF and 
 GH are in the same plane, viz. 
 EFHG which cuts the planes 
 AB and CD ; and they do not 
 meet though produced ; for the 
 planes in which they are do not 
 meet; therefore EF and GH are parallel (def. 30. I.). 
 
 ^\ 
 
 PROP. XV. THEOR. 
 
 If two parallel planes be cut by a third plane, they have the same inclination 
 
 to that plane. 
 
 Let AB and CD be two parallel planes, and EH a third plane cutting 
 them ; The planes AB and CD are equally inclined to EH. 
 
 Let the straight lines EF and GH be the common section of the plane 
 EH with the two planes AB and CD ; and from K, any point in EF, draw 
 in the plane EH the straight line KM at right angles to EF, and let it 
 meet GH in L ; draw also KN at right angles to EF in the plane AB : 
 and through the straight lines KM, KN, let a plane be made to pass, cut- 
 ting the plane CD in the line LO. And because EF and GH are the 
 common sections of the plane EH with the two parallel planes AB and 
 CD, EF is parallel to GH (14. 2. Sup.). But EF is at right angles to 
 the plane that passes through KN and KM (4. 2. Sup.), because it is at 
 right angles to the lines KM and KN : therefore GH is also at right an- 
 gles to the same plane (7. 2. Sup.), and it is therefore at right angles to 
 
 ~K J) 
 
 A 
 
 No 
 
 =N 
 
 o 
 
 M
 
 190 SUPPLEMENT TO THE ELEMENTS 
 
 the lines LM, LO which it meets in that plane. Therefore, since LM and 
 LO are at right angles to LG, the common section of the two planes CD 
 and EH, the angle OLM is the inclination of the plane CD to the plane 
 EH (1. def. 2. Sup.)- For the same reason the angle MKN is the inclina- 
 tion of the plane AB to the plane EH. But because KN and LO are pa- 
 rallel, beina the common sections of the parallel planes AB and CD with 
 a third plane, the interior angle NKM is equal to the exterior angle OLM 
 (29. 1.) ; that is, the inclination of the plane AB to the plane EH, is equal 
 to the inclination of the plane CD to the same plane EH. 
 
 PROP. XVI. THEOR. 
 If two straight lines be cut by parallel planes, they must be cut in the same ratio 
 
 Let the straight lines AB, CD be cut by the parallel planes GH, KL, 
 MN, in the points A, E, B ; C, F, D : 
 As AE is to EB, so is CF to FD. 
 
 Join AC, BD, AD, and let AD meet 
 the plane KL in the point X ; and join 
 EX, XF : Because the two parallel 
 planes KL, MN are cut by the plane 
 EBDX, the common sections EX, BD, 
 are parallel (14. 2. Sup.). For the same 
 reason, because the two parallel planes 
 GH, KL are cut by the plane AXFC, 
 the common sections AC, XF are paral- 
 lel : And because EX is parallel to BD, 
 a side of the triangle ABD, as AE to 
 EB, so is (2. 6.) AX to XD. Again, be- 
 cause XF is parallel to AC, aside of the 
 triangle ADC, AX to XD, so is CF to 
 FD : and it was proved that AX is to XD, 
 as AE to EB : Therefore (11. 5.), as AE 
 to EB, so is CF to FD. 
 
 PROP. XVII. THEOR. 
 
 If a straight line be at right angles to a plane, every plane which passes through 
 that line is at right angles to the Jirst mentioned plane. 
 
 Let the straight line AB be at right angles to the plane CK ; every plane 
 which passes through AB is at right angles to the plane CK. 
 
 Let any plane DE pass through AB, and let CE be the common section 
 of the planes DE, CK ; take any point F in CE, from which draw FG in 
 the plane DE at right angles to CE : And because AB is perpendicular 
 to -he plane CK, therefore it is also perpendicular to every straight line 
 meeting it in that plane (1. def. 2. Sup.); and consequently it is perpen- 
 dicular to CE : Wherefore ABF is a right angle ; But GFB is likewise a 
 right angle ; therefore AB is parallel (28. 1.) to FG. And AB is at right 
 angles to the plane CK : therefore FG is also at right angles to the same
 
 O* GEOMETRY. BOOK II. 
 
 «9J 
 
 A U 
 
 B E 
 
 plane (7. 2. Sup.). But one plane is 
 at right angles to another plane when 
 the straight lines drawn in one of the 
 planes, at right angles to their com- 
 mon section, are also at right angles 
 to the other plane (def. 2. 2. Sup.) ; an;l 
 any straight line FG in the plane DE, 
 which is at right angles to CE, the 
 common section of the planes, has been 
 proved to be perpendicular to the other 
 plane CK ; therefore the plane DE 
 
 is at right angles to the plane CK. In like manner, it may be proved 
 that all the planes which pass through AB are at right angles to the plane 
 CK. 
 
 PROP. XVIII. THEOR. 
 
 If two planes cutting one another be each of them perpendicular to a third plane 
 their common section is perpendicular to the same plant. 
 
 Let the two planes AB, BC be each of them perpendicular to a third 
 plane, and Bl) be the common section of the first two ; BD is perpendicular 
 to the plane ADC. 
 
 From 1) in the plane ADC, draw DE perpen- 
 dicular to AD, and DF to DC. Because DE is 
 perpendicular to AD, the common section of the 
 planes AB and ADC ; and because the plane 
 AB is at right angles to ADC, DE is at right 
 angles to the plane A B (def. 2.2. Sup.), and there- 
 fore also to the straight line BD in that plane 
 (def. 1. 2. Sup.). For the same reason, DF is at 
 right angles to DB. Since BD is therefore at 
 right angles to both the lines DE and DF, it is 
 at right angles to the plane in which DE and 
 DF are, that is, to the plane ADC (4. 2. Sup.). 
 
 A. E 
 
 PROP. XIX. PROB. 
 
 Two straight lines not in the same plane being given in position, to draw a 
 straight line perpendicular to them both. 
 
 Let AB and CD be the given lines, which are not in the same plane ; it 
 is required to draw a straight line which shall be perpendicular both to AB 
 and CD. 
 
 In AB take any point E, and through E draw EF parallel to CD, and 
 let EG be drawn perpendicular to the plane which passes through EB, 
 EF (10. 2. Sup.). Through AB and EG let a plane pass, viz. GK, and let 
 this plane meet CD in H; from H draw HK perpendicular to AB ; and 
 HK is the line required. Through H, draw HG parallel to AB.
 
 92 SUPPLEMENT TO THE ELEMENTS 
 
 Then, since HK and GE, which are in the same plane, are both at right 
 angles to the straight line AB, they are parallel to one another. And be- 
 cause the lines HG, HD are parallel to the lines EB, EF, each to each, 
 ..he plane GHD is parallel to the plane (13. 2. Sup.) BEF ; and therefore 
 EG, which is perpendicular to the plane BEF, is perpendicular also to the 
 plane (Cor. 13. 2. Sup.) GHD. Therefore HK, which is parallel to GE, 
 is also perpendicular to the plane GHD (7. 2. Sup.), and it is therefore per- 
 pendicular to HD (def. 1. 2. Sup.), which is in that plane, and it is also 
 perpendicular to AB ; therefore HK is drawn perpendicular to the two 
 given lines, AB and CD. 
 
 PROP. XX. THEOR. 
 
 If a solid angle be contained by three plane angles, any two of these angles are 
 greater than the third. 
 
 Let the solid angle at A be contained by the three plane angles BAC, 
 CAD, DAB. Any two of them are greater than the third. 
 
 If the angles BAC, CAD, DAB be all equal, it is evident that any two 
 of them are greater than the third. But if they are not, let BAC be that 
 angle which is not less than either of the other two, and is greater than 
 one of them, DAB ; and at the point A in the 
 straight line AB, make in the plane which 
 passes through BA, AC, the angle BAE equal 
 (Prop. 23. 1.) to the angle DAB; and make 
 AE equal to AD, and through E draw BEC 
 cutting AB, AC in the points B, C, and join 
 DB, DC. And because DA is equal to AE, 
 and AB is common to the two triangles ABD, 
 ABE. and also the angle DAB equal to the 
 angle EAB ; therefore the base DB is equal (4. 1.) to the base BE. And 
 because BD, DC are greater (20. 1.) than CB, and one of them BD has 
 been proved equal to BE, a part of CB, therefore the other DC is greater 
 than the remaining part EC. And because DA is equal to AE, and AC 
 common, but the base DC greater than the base EC ; therefore the angle 
 DAC is greater (25. 1.) than the angle EAC ; and, by the construction,
 
 OF GEOMETRY. BOOK II 193 
 
 the angle DAB is equal to the angle BAE ; wherefore the angles DAB, 
 DAC are together greater than BAE, EAC, that is, than the angle BAC 
 But BAC is not less than either of the angles DAB, DAC ; therefore 
 BAC, with either of mem, is greater than the other. 
 
 PROP. XXI. THEOR. 
 
 The plane angles which contain any solid angle are together less than four 
 
 right angles. 
 
 Let A be a solid ang e contained by any number of plane angles BAC, 
 CAD, DAE, EAF, FAB ; these together are less than four right angles. 
 
 Let the planes which contain the solid angle at A be cut by another 
 plane, and let the section of them by that plane be the rectilineal figure 
 BCDEF. And because the solid angle at B is contained by three plant 
 angles CBA, ABF, FBC, of which any two 
 are greater (20. 2. Sup) than the third, the 
 angles CBA, ABF are greater than the an- 
 gle FBC : For the same reason, the two 
 plane angles at each of the points C, D, E, 
 F, viz. the angles which are at the bases of 
 the triangles having the common vertex A, 
 are greater than the third angle at the same 
 point, which is one of the angles of the figure 
 BCDEF : therefore all the angles at the 
 bases of the triangles are together greater 
 than all the angles of the figure : and be- 
 cause all the angles of the triangles are to- 
 gether equal to twice as many right angles as there are triangles (32. 1.) J 
 that is, as there are sides in the figure BCDEF ; and because all the an- 
 gles of the figure, together with four right angles, are likewise equal to 
 twice as many right angles as there are sides in the figure(lcr. 32. l.);there- 
 fore all the angles of the triangles are equal to all the angles of the rectili 
 neal figure, together with four right angles. But all the angles at the bases 
 of the triangles are greater than all the angles of the rectilineal, as has 
 been proved. Wherefore, the remaining angles of the triangles, viz. those 
 at the vertex, which contain the solid angle at A, are less than four right 
 »ngl««. 
 
 Otherwise.. 
 
 Let the sum of all the angles at the bases of the triangles =S; the 
 «um of all the angles of the rectilineal figure BCDEF=^ ; the sum of the 
 plane angles at A=X, and let R= a right angle. 
 
 Then, because S+X= twice (32 1.) as many right angles as there are 
 triangles, or as there are sides of the rectilineal figure BCDEF, and as 
 .Z+4R is also equal to twice as many right angles as there are sides of the 
 lame figure ; therefore S-f-X = 2"-j- iR. But because of the three plane 
 angles which contain a solid angle, anv two are greater than the third, 
 
 25
 
 194 SUPPLEMENT TO THE ELEMENTS, &c. 
 
 872 ; and therefore X/4R ; that is, the sum of the plane angles which 
 contain the solid angle at A is less than four right angles 
 
 SCHOLIUM. 
 
 It is evident, that when any of the angles of the figure BCDEF ia ex- 
 terior, like the angle at D, in the an- 
 nexed figure, the reasoning in the _^_ 
 above proposition does not hold, be- 
 cause the solid angles at the base 
 are not all contained by plane an- 
 gles, of which two belong to the tri- 
 angular planes, having their com- 
 mon vertex in A, and the third is an 
 interior angle of the rectilineal figure, 
 
 or base. Therefore, it cannot be ^, 
 
 concluded that Sis necessarily great- U v^ 
 
 er than 2. This proposition, therefore, is subject to a limitation, which .'* 
 farther explained in the notes on this Book.
 
 ELEMENTS 
 
 GEOMETRY, 
 
 SUPPLEMENT. 
 
 BOOK III. 
 
 OF THE COMPARISON OF SOLIDS. 
 
 DEFINITIONS. 
 
 1. A Solid is that which has length, breadth, and thickness. 
 
 2. Sii.iilar solid figures are such as are contained by the same number ol 
 sim.Iar planes similarly situated, and having like inclinations to one an- 
 other. 
 
 3. A pyramid is a solid figure contained by planes that are constituted be- 
 twixt one plane and a point above it in which they meet. 
 
 4. A prism is a solid figure contained by plane figures, of which two that 
 are opposite are equal, similar, and parallel to one another ; and the 
 others are parallelograms. 
 
 5. A parallelopiped is a solid figure contained by six quadrilateral figures, 
 whereof every opposite two are parallel. 
 
 6. A cube ia a solid figure contained by six equal squares. 
 
 7. A sphere is a solid figure described by the revolution of a semicircle 
 about a dismeter, which remains unmoved. 
 
 8. The axis of a sphere is the fixed straight line about which the semi 
 circle revoLes. 
 
 9. The centre of a sphere is the same with that of the semicircle. 
 
 J 0. The diameter of a sphere is any straight line which passes through 
 the centre, an I is terminated both ways by the superficies of the sphere
 
 196 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 11. A cone is a solid figure described by the revolution of a right angled 
 triangle about one of the sides containing the right angle, which side 
 remains fixed. 
 
 12. The axis of a cone is the fixed straight line about which the triangle 
 revolves. 
 
 3. The base of a cone is the circle described by that side, containing th« 
 right angle, which revolves. 
 
 14. A cylinder is a solid figure described by the revolution of a rignt an- 
 gled parallelogram about one of its sides, which remains fixed. 
 
 15. The axis of a cylinder is the fixed straight line about which the paral- 
 lelogram revolves. 
 
 16. The bases of a cylinder are the circles described by the two revolving 
 opposite sides of the parallelogram. 
 
 17. Similar cones and cylinders are those which have their axes, and the 
 diameters of their bases proportionals. 
 
 PROP. I. THEOR. 
 
 If two solids be contained by the same number of equal and similar planes 
 similarly situated, and if the inclination of any two contiguous planes in the 
 one solid be the same with the inclination of the two equal, and similarly 
 situated planes in the other, the solids themselves are equal and similar. 
 
 Let AG and KQ be two solids contained by the same number of equal 
 and similar planes, similarly situated so that the plane AC is similar and 
 equal to the plane KM, the plane AF to the plane KP ; BG to LQ, GD 
 to QN, DE to NO, and FH to PR. Let also the inclination of the plane 
 AF to the plane AC be the same with that of the plane KP to the plane 
 KM, and so of the rest ; the solid KQ is equal and similar to the solid AG. 
 
 Let the solid KQ be applied to the solid AG, so that the bases KM and 
 
 n 
 
 D 
 
 J2 
 
 S 
 
 L\ 
 
 _Q 
 
 sO 
 
 M 
 
 35 
 
 TL 
 
 AC which are equal and similar, may coincide (8. Ax. 1.), the point N 
 coinciding with the point D, K with A L with B, and so on. And be- 
 cause the plane KM coincides with the plane AC, and, by hypothesis, the
 
 OF GEOMETRY. BOOK III. 197 
 
 inclination of KR to KM is the same with the inclination of AH to j*C 
 the plane KR will be upon the plane AH, and will coincide with it, because 
 they are similar and equal (8. Ax. 1 .), and because their equal sides KN 
 and AD coincide. And in the same manner it is shewn that the olhet 
 planes of the solid KQ coincide with the other planes of the solid AG, 
 each with each : wherefore the solids KQ and AG do wholly coincide, 
 and are equal and similar to one another. 
 
 PROP. II. THEOR 
 
 If a solid be contained by six planes, two and two of which are parallel, the op- 
 posite planes are similar and equal parallelograms. 
 
 Let the solid CDGH be contained by the parallel planes AC, GF ; BG, 
 CE ; FB, AE : its opposite planes are similar and equal parallelograms. 
 
 Because the two parallel planes BG, CE,are cut by the plane AC, their 
 common sections AB, CD are parallel (14. 2. Sup.). Again, because the 
 two parallel planes BF, AE are cut by the plane AC, their common sec- 
 tions AD, BC are parallel (14. 2. Sup.) : and AB is parallel to CD ; there- 
 fore AC is a parallelogram. In like manner, it may be proved that each 
 of the figures CE, FG, GB, BE, AE is a pa- 
 rallelogram; join AH, DF; and because AB 
 is parallel to DC, and BH to CF ; the two 
 straight lines AB, BH, which meet one an- 
 other, are parallel to DC and CF, which meet 
 one another ; wherefore, though the first two 
 are not in the same plane with the other two, 
 they contain equal angles (9. 2. Sup.) ; the 
 angle ABH is therefore equal to the angle -^ 
 
 DCF. And because AB, BH, are equal to DC, CF, and the angle ABH 
 equal to the angle DCF ; therefore the base AH is equal (4. 1.) to the base 
 DF, and the triangle ABH to the triangle DCF: For the same reason, 
 the trian«le AGH is equal to the triangle DEF : and therefore the paral- 
 lelogram BG is equal and similar to the parallelogram CE. In the same 
 manner', it may be proved, that the parallelogram AC is equal and similar 
 to the parallelogram GF, and the parallelogram AE to BF. 
 
 PROP. III. THEOR. 
 
 If a solid parallelopipcd be cut by a plane parallel to two of its opposite planes, 
 it will be divided into two solids, which will be to one another as the bases. 
 
 Let the solid parallelopipcd A BCD be cut by the plane EV, which is 
 parallel to the opposite planes A R, HD, and divides the whole into the 
 solids ABFV, EGCD : as the base AEFY to the base EHCF, so is the 
 solid ABFV to the solid EGCD. 
 
 Produce AH both ways, ami take any number of straight lines II M, 
 MN, each equal to EH, and any number AK. K Leach equal to EA, and 
 complete the parallelograms LO, KY, EfQ, MS, and the solids LP KR
 
 98 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 FIU, MT then, because the straight lines LK, KA, AE are all equal, and 
 also tue straight lines KO, AY, EF which make equal angles with LK, 
 KA, AE, the parallelograms LO, KY, AF are equal and similar (36. 1. 
 & def. 1. 6.) : and likewise the parallelograms KX, KB, AG ; as also 
 
 X 
 
 K 
 
 
 E 
 
 H 
 
 m In 
 
 o 
 
 3? C Q S 
 
 (2. 3. Sup.) the parallelograms LZ, KP, AR, because they are opposite 
 planes. For the same reason, the parallelograms EC, HQ, MS are equal 
 (36. 1. & def. 1. 6.); and the parallelograms HG, HI, IN, as also (2. 3. 
 Sup.) HD, MU, NT ; therefore three planes of the solid LP, are equal and 
 similar to three planes of the solid KR, as also to three planes of the solid 
 AV : but the three planes opposite to these three are equal and similar to 
 them (2. 3. Sup.) in the several solids ; therefore the solids LP, KR, AV 
 are contained by equal and similar planes. And because the planes LZ, 
 KP, AR are parallel, and are cut by the plane XV, the inclination of LZ 
 to XP is equal to that of KP to PB ; or of AR to BV (15. 2. Sup.) and 
 the same is true of the other contiguous planes, therefore the solids LP 
 KR, and AV, are equal to one another (1. 3. Sup.). For the same rea- 
 son, the three solids, ED, HU, MT are equal to one another; therefore 
 what multiple soever the base LF is of the base AF, the same multiple is 
 the solid L V of the solid AV ; for the same reason, whatever multiple the 
 base NF is of the base HF, the same multiple is the solid NV of the solid 
 ED : And if the base LF be equal to the base NF, the solid LV is equal 
 (1.3. Sup.) to the solid NV ; and if the base LF be greater than the base 
 NF, the solid LV is greater than the solid NV : and if less, less. Since 
 then there are four magnitudes, viz. the two bases AF, FH, and the two 
 solids AV, ED, and of the base AF and solid AV, the base LF and solid 
 LV are any equimultiples whatever ; and of the base FH and solid ED, 
 the base FN and solid NV are any equimultiples whatever ; and it has 
 been proved, that if the base LF is greater than the base FN, the solid LV 
 is greater than the solid NV ; and if equal, equal : and if less, less : There 
 fore (def. 5. 5.) as the base AF is to the base FH, so is the solid AV to 
 the solid ED. 
 
 Cor. Because the parallelogram AF is to the parallelogram FH a» YF 
 to FC (1 6.), therefore the solid AV is to the solid ED as YF to FC
 
 OF GEOMETRY. BOOK III. 
 
 199 
 
 PROP. IV. THEOR. 
 
 f a solid parallelopiped be cut by a plane passing through the diagonals o, 
 two of the opposite planes, it will be cut into two equal prisms. 
 
 Let AB be a solid parallelopiped, and DE, CF the diagonals of the op- 
 posite parallelograms AH, GB, viz. those which are drawn betwixt the 
 equal angles in each; and because CD, FE are each of them parallel to 
 GA, though not in the same plane with it, CD, FE are parallel (8. 2. Sup.) 
 wherefore the diagonals CF, DE are in the plane in which the parallel 
 are, and are themselves parallels (14. 2. Sup.) ; 
 the plane CDEF cuts the solid AB into two 
 equal parts. 
 
 Because the triangle CGF is equal (34. 1.) 
 to the triangle CBF, and the triangle DAE to 
 DHE ; and since the parallelogram CA is equal 
 (2. 3. Sup.) and similar to the opposite one BE ; 
 and the parallelogram GE to CH : therefore the 
 planes which contain the prisms CAE, CBE, 
 are equal and similar, each to each ; and they 
 are also equally inclined to one another, because 
 the planes AC, EB are parallel, as also AF and 
 BD, and they are cut by the plane CE (15. 2. Sup.). Therefore the prism 
 CAE is equal to the prism CBE (1. 3. Sup.), and the solid AB is cut into 
 two equal prisms by the plane CDEF. 
 
 N. B. The insisting straight lines of a parallelopiped, mentioned in 
 the followrng propositions, are the sides of the parallelograms betwixt the 
 base and the plane parallel to it. 
 
 PROP. V. THEOR. 
 
 Solid parallelopipeds upon the same base, and of the same altitude, the in 
 sisting straight lines of which are terminated in the same straight lines in 
 the plane opposite to the base are equal to one another. 
 
 Let the solid parallelopipeds AH, AK be upon the same base AB, and 
 of the same altitude, and let their insisting straight lines AF, AG, LM, LN 
 be terminated in the same straight line FN, and let the insisting lines CD 
 CE, BH, BK be terminated in the same straight line DK ; the solid AH 
 is equal to the solid AK. 
 
 Because CH, CK are parallelograms, CB is equal (34. 1.) to each oi 
 the opposite sides DH, EK : wherefore DH is equal to EK : add, or take 
 away the common part HE; then DE is equal to HK : Wherefore also 
 th( triangle CDE is equal (38. 1.) to the triangle BHK : and the parallel- 
 ogram DG is equal (36. 1.) to the parallelogram HN. For the same rea- 
 son, the triangle AFG is equal to the triangle LMN, and the parallelogram 
 CF is equal (2. 3. Sup.) to the parallelogram BM, and CG to BN ; for 
 they are opposite. Therefore the planes which contain the prism DAG 
 are similar and equal to those which contain the prism HLN, each to each
 
 200 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 and the contiguous planes are also equally inclined to one another (15. 2 
 Sup ), » ecause that the parallel planes AD and LH, as also AE and LK 
 
 are cut by the same piane DN : therefore the prisms DAG, HLN are 
 equal (1. 3. Sup.). If therefore the prism LNH be taken from the solid, 
 of which the base is the parallelogram AB, and FDKN the plane opposite 
 to the base ; and if from this same solid there be taken the prism AGD, 
 the remaining solid, viz. the parallelopiped AH is equal to the remaining 
 parallelopiped AK. 
 
 PROP. VI. THEOR. 
 
 Solid parallelepipeds upon the same base, and of tne same altitude, the in- 
 sisting straight lines of which are not terminated in the same straight lines 
 in the plane opposite to the base, are equal to one another. 
 
 Let the parallelopipeds CM, CN, be upon the same base AB, and of the 
 same altitude, but their insisting straight lines AF, AG, LM, LN, CD, 
 CE, BH, BK, not terminated in the same straight lines ; the solids CM, 
 CN are equal to one another. 
 
 Produce FD, MH, and NG, KE, and let them meet one another in the 
 poin's 0, P, Q, R ; and join AO, LP, BQ, CR. Because the planes (def. 
 5. 3. Sup.), LBHM and ACDF are parallel, and because the plane LBHM 
 is that in which are the parallels LB, MHPQ (def. 5. 3. Sup.), and in which
 
 OF GEOMETRY. BOOK III. 
 
 20 
 
 also is the figure BI PQ ; and because the plane ACDF is that in which 
 are the parallels A*", FDOR, and in which also is the figure CAOR. 
 therefore the figure? BLPQ, CAOR,, are in parallel planes. In like man- 
 ner, because the planes ALNG and CBKE are parallel, and the plane 
 ALNG is that in which are the parallels AL, OPGN, and in which also 
 is the figure ALPO ; and the plane CBKE is that in which are the paral- 
 lels CB, RQEK, and in which also is the figure CBQR ; therefore the 
 figures ALPO, CBQR, are in parallel planes. But the planes ACBL 
 ORQP are also parallel ; therefore the solid CP is a parallelopiped. Now 
 the solid parallelopiped CM is equal (5. 2. Sup.) to the solid parallelopiped 
 CP, because they are upon the same base, and their insisting straight lines 
 AF, AO, CD, CP ; LM, LP, BH, BQ are terminated in the same straight 
 lines FR, MP ; and the solid CP is equal (5. 2. Sup.) to the solid CN ; 
 for they are upe<i the same base ACBL, and their insisting straight lines 
 AO, AG, LP, LW ; CR, CE. BQ, BK are terminated in the same straight 
 lines ON, RK ; Therefore the solid CM is equal to the solid CN. 
 
 PROP. VII. THEOR. 
 
 Solid parallelopipeds, which arc upon equal bases, and of the same altitude, 
 are equal to one another. 
 
 Let the solid parallelopipeds, AE, CF, be upon equal bases AB, CD. 
 and be of the same altitude ; the solid AE is equal to the solid CF. 
 
 Case 1. Let the insisting straight lines be at right angles to the bases 
 AB, CD, and let the bases be placed in the same plane, and so as that the 
 Bides CL, LB, be in a straight line; therefore the straight line LM, which 
 is at right angles to the plane in which the bases are, in the point L, is 
 common (11. 2. Sup.) to the two solids AE, CF ; let the other insisting 
 lines of the solids be AG, HK, BE ; DF, OP, CN : and first, let the angle 
 ALB be equal to the angle CLD ; then AL, LD are in a straight line (14. 
 1 .). Produce OD, HB, and let them meet in Q and complete the solid 
 parallelopiped LR, the base of which is the parallelogram LQ, and of 
 which LM is one of its insisting straight lines : therefore, because the pa- 
 rallelogram AB is equal to CD, as the base AB is to the base LQ, so is 
 (7. 5.) the base CD to the same LQ : and because the solid parallelopiped 
 AR is cut by the plane LMEB, which is parallel to the opposite planes 
 AK. DR ; as the base AB is to the base LQ, so is (3. 3. Sup.) the soliJ
 
 202 SUPPLEMENT TO THE ELEMENTS 
 
 AE to the solid LR : for the same reason because the solid parallelopiped 
 CR is cut by the plane LMFD, which is parallel to the opposite planes 
 CP BR ; as the base CD to the base LQ ; so is the solid CF to the solid 
 LR , but as the base AB to the base LQ, so the base CD to the base LQ, 
 as has been proved : therefore as the solid AE to the solid LR, so is the 
 solid CF to the solid LR ; and therefore the solid AE is equal (9. 5.) to 
 the solid CF. 
 
 But let the solid parallelopipeds, SE, CF be upon equal bases SB, CD, 
 and be of the same altitude, and let their insisting straight lines be at right 
 angles to the bases ; and place the bases SB, CD in the same plane, so 
 that CL, LB be in a straight line ; and let the angles SLB, CLD, be un- 
 equal ; the solid SE is also in this case equal to the solid CF. Produce 
 DL, TS until they meet in A, and from B draw BH parallel to DA ; and 
 let HB, OD produced meet in Q, and complete the solids AE, LR : there- 
 fore the solid AE, of which the base is the parallelogram LE, and AK the 
 plane opposite to it, is equal (5. 3. Sup.) to the solid SE, of which the base 
 is LE, and SX the plane opposite ; for they are upon the same base LE, 
 and of the same altitude, and their insisting straight lines, viz. LA, LS, 
 BH, BT ; MG, MU, EK, EX, are in the same straight lines AT, GX : 
 and because the parallelogram AB is equal (35. 1.) to SB, for they are 
 upon the same base LB, and between the same parallels LB, AT ; and 
 because the base SB is equal to the base CD ; therefore the base AB is 
 equal to the base CD : but the angle ALB is equal to the angle CLD : 
 therefore, by the first case, the solid AE is equal to the solid CF ; but the 
 solid AE is equal to the solid SE, as was demonstrated : therefore the 
 solid SE is equal to the solid CF. 
 
 Case 2. If the insisting straight lines AG, HK, BE, LM ; CN, RS, 
 
 l)h , OP, be not at right angles to the bases AB, CD ; in this case likewise 
 the solid AE is equal to the solid CF. Because solid parallelopipeds on 
 the same base, and of the same altitude, are equal (6. 3. Sup.), if two solid 
 parallelopipeds be constituted on the bases AB and CD of the same alti- 
 tude with the solids AE and CF, and with their insisting lines perpendicu- 
 lar to their bases, they will be equal to the solids AE and CF ; and, by the 
 first case of this proposition, they will be equal to one inothei ; wherefore 
 the solids AE and CF are also equal.
 
 OF GEOMETRY. BOOK III. 
 
 203 
 
 PROP. VIII. THEOR. 
 
 Solid parallelopipeds which have the same altitude, are to one another as 'km 
 
 bases. 
 
 Let AB, CD be solid parallelopipeds of the same altitude ; they are to 
 one another as their bases ; that is, as the base AE to the base CF, 80 i* 
 the solid AB to the solid CD. 
 
 To the straight line FG apply the parallelogram FH equal (Cor. Prop 
 45. 1.) to AE, so that the angle FGH be equal to the angle LCG ; and 
 
 complete the solid parallelepiped GK upon the base FH, one of whose in 
 sisting lines is FD, whereby the solids CD, GK must be of the same alti- 
 tude. Therefore the solid AB is equal (7. 3. Sup.) to the solid GK, be- 
 cause they are upon equal bases AE, FH, and are of the same altitude : 
 and because the solid parallelopiped CK is cut by the plane DG which is 
 parallel to its opposite planes, the base HF is (3. 3. Sup.) to the base FC, 
 as the solid HD to the solid DC : But the base HF is equal to the base 
 AE, and the solid GK to the solid AB : therefore, as the base AE to the 
 base CF, so is the solid AB to the solid CD. 
 
 Cor. 1 . From this it is manifest, that prisms upon triangular bases, and 
 of the same altitude, are to one another as their bases. Let the prisms 
 BNM, DPG, the bases of which are the triangles AEM, CFG, have the 
 same altitude : complete the parallelograms AE, CF, and the solid paral 
 lelopipeds AB, CD, in the first of which let AN, and in the other let CP 
 be one of the insisting lines. And because the solid parallelopipeds AB, 
 CD have the same altitude, they are to one another as the base AE is to 
 the base CF ; wherefore the prisms, which are their halves (4. 3. Sup.) 
 are to one another, as the base AE to the base CF ; that is, as the trian- 
 gle AEM to the triangle CFG. 
 
 Cor. 2. Also a prism and a parallelopiped, which have the same alti- 
 tude, are to one another as their bases ; that is, the prism BNM is to the 
 parallelopiped CD as the triangle AEM to the parallelogram LG. For 
 by the last Cor. the prism BNM is to the prism DPG as the triangle AME 
 to the triangle CGF, and therefore the prism BNM is to twice the prism 
 DPG as the triangle AME to twice the triangle CGF (4. 5 ) ; that is, the 
 prism BNM is to the parallelopiped CD as the triangle AME to the paral- 
 lelogram LG.
 
 204 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 PROP. IX. THEOR. 
 
 Solid pa r dlcelopipcds are to one another in the ratio that is compounded oftfu 
 ratios of the areas of their bases, and of their altitudes. 
 
 Let AF and GO be two solid parallelopipeds, of which the bases are the 
 parallelograms AC and GK, and the altitudes, the perpendiculars let fall 
 on the planes of these bases from any point in the opposite planes EF and 
 MO ; the solid AF is to the solid GO in a ratio compounded of the ratios 
 of the base AC to the base GK, and of the perpendicular on AC, to the 
 perpendicular on GK. 
 
 Case 1. When the insisting lines are perpendicular to the bases AC 
 and GK, or when the solids are upright. 
 
 In GM, one of the insisting lines of the solid GO, take GQ equal to AE, 
 one of the insisting lines of the solid AF, and through Q let a plane pass 
 parallel to the plane GK, meeting the other insisting lines of the solid GO 
 
 F 
 
 X 
 
 V 
 
 T2 
 
 
 
 V 
 
 
 
 o 
 
 
 \ 
 
 M 
 
 \ 
 
 "S 
 
 T 
 
 
 
 i 
 
 
 \ 
 
 
 R 
 
 I 
 
 
 X 
 
 \ 
 
 \ 
 
 
 
 T 
 
 
 1L 
 
 in the points R, S and T. It is evident that GS is a solid parallelopiped 
 (def. 5. 3. Sup.) and that it has the same altitude with AF, viz. GQ or 
 AE. Now the solid AF is to the solid GO in a ratio compounded of the 
 ratios of the solid AF to the solid GS (def. 10. 5.), and of the solid GS to 
 the solid GO ; but the ratio of the solid AF to the solid GS, is the same 
 with that of the base AC to the base GK (8. 3. Sup.), because their alti- 
 tudes AE and GQ are equal ; and the ratio of the solid GS to the solid 
 GO, is the same with that of GQ to GM (3. 2. Sup.) ; therefore, the ratio 
 which is compounded of the ratios of the solid AF to the solid GS, and ol 
 the solid GSto the solid GO, is the same with the ratio which is compound- 
 ed of the ratios of the base AC to the base GK, and of the altitude AE to 
 the altitude GM (F. 5.). But the ratio of the solid AF to the solid GO, is 
 that which is compounded of the ratios of AF to GS, and of GS to GO ; 
 therefore, the ratio of the solid AF to the solid GO is compounded of the 
 ratios of the base AC to the base GK, and of the' altitude AE to the alti 
 mde GM. 
 
 Case 2. When the insisting lines are not perpendicular to the basest.
 
 OF GEOMETRY. BOOK III. 205 
 
 Let the parallelograms AC and GK be the bases as before, and let AE 
 and GM be the altitudes of two parallelopipeds Y and Z on these bases. 
 Then, if the upright parallelopipeds AF and GO be constituted on the 
 bases AC and GK, with the altitudes AE and-GM, they will be equal U. 
 the parallelopipeds Y and Z (7. 3. Sup.). Now, the solids AF and GO, 
 by the first case, are in the ratio compounded of the ratios of the bases AC 
 and GK, and of the altitudes AE and GM ; therefore also the solids Y 
 and Z have to one another a ratio that is compounded of the same ratios. 
 
 Cor. 1. Hence, two straight lines may be found having the same ratio 
 with the two parallelopipeds AF and GO. To AB, one of the sides of the 
 parallelogram AC, apply the parallelogram BV equal to GK, having an 
 angle equal to the angle BAD (Prop. 44. 1.) ; and as A E to GM, so let 
 A V be to AX (12. 6.), then AD is to AX as the solid AF to the solid GO. 
 For the ratio of AD to AX is compounded of the ratios (def. 10. 5.) of AD 
 to AV, and of AV to AX ; but the ratio of AD to AV is the same with 
 that of the parallelogram AC to the parallelogram BV (1. 6.) or GK ; 
 and the ratio of A V to AX is the same with that of AE to GM ; therefore 
 the ratio of AD to AX is compounded of the ratios of AC to GK, and of 
 AE to GM (E. 5.). But the ratio of the solid AF to the solid GO is com- 
 pounded of the same ratios ; therefore, as AD to AX, so is the solid AF to 
 the solid GO. 
 
 Cor. 2. If AF and GO are two parallelopipeds, and if to AB, to the 
 perpendicular from A upon DC, and to the altitude of the parallelopiped 
 AF, the numbers L, M, N, be proportional : and if to AB, to GH, to the 
 perpendicular from G on LK, and to the altitude of the parallelopiped GO, 
 the numbers L, /, m, n, be proportional ; the solid AF is to the solid GO 
 as L X M X N lo IXmXn. 
 
 For it may be proved, as in the 7th of the 1st of the Sup. that L X M X 
 Nisto IXmXnin the ratio compounded of the ratio of L X M to I X m, and 
 of the ratio of N to n. Now the ratio of L x M to Ixm is that of the area 
 of the parallelogram AC to that of the parallelogram GK ; and the ratio 
 of N to n is the ratio of the altitudes of the parallelopipeds, by hypothesis, 
 therefore, the ratio of L X M X N to Ixmxn is compounded of the ratio of 
 the areas of the bases, and of the ratio of the altitudes of the parallelopipeds 
 AF and GO ; and the ratio of the parallelopipeds themselves is shewn, in 
 this proposition, to be compounded of the same ratios ; therefore it is the 
 same with that of the product L X M X N to the product lx m X n. 
 
 Cor. 3. Hence all prisms are to one another in the ratio compounded 
 of the ratios of their bases, and of their altitudes. For every prism is 
 equal to a parallelopiped of the same altitude with it, and of an equal base 
 (2. Cor. 8. 3. Sup.). 
 
 PROP. X. THEOR. 
 
 Solid parallelopipeds, which have their bases and altitudes reciprocally propo 
 ttonal, are equal ; and parallelopipeds which are equal, have their bases and 
 altitudes reciprocally proportional. 
 
 Let AG and KQ be two solid parallelopipeds, of which the bases ar»
 
 206 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 AC and KM, and the altitudes AE and KO, and let AC be to KM as KO 
 to AE ; the solids AG and KQ are equal. 
 
 As the base AC to the base KM, so let the straight line KO be to the 
 straight line S. Then, since AC is to KM as KO to S, and also by hypo- 
 thesis, AC to KM as KO to AE, KO has the same ratio to S that it has 
 to A.E (11. 5.) ; wherefore AF is equal to S (9 5.). But the solid AG is 
 
 to the solid KQ, in the ratio compounded of the ratios of AE to KO, and 
 of AC to KM (9. 3. Sup.), that is, in the ratio compounded of the ratios of 
 AE to KO, and of KO to S. And the ratio of AE to S is also compound- 
 ed of the same ratios (def. 10. 5.) ; therefore, the solid AG has to the solid 
 KQ the same ratio that AE has to S. But AE was proved to be equal to 
 S, therefore AG is equal to KQ. 
 
 Again, if the solids AG and KQ be equal, the base AC is to the base 
 KM as the altitude KO to the altitude AE. Take S, so that AC may be 
 to KM as KO to S, and it will be shewn, as was done above, that the solid 
 AG is to the solid KQ as AE to S ; now, the solid AG is, by hypothesis, 
 equal to the solid KQ : therefore, AE is equal to S ; but, by construction, 
 AC is to KM, as KO is to S ; therefore, AC is to KM as KO to AE. 
 
 Cor. In the same manner, it may be demonstrated, that equal prisms 
 have their bases and altitudes reciprocally proportional, and conversely. 
 
 PROP. XI. THEOR. 
 
 Similar solid parallelopipeds are to one another in the triplicate ratio oj their 
 
 homologous sides. 
 
 Let AG, KQ be two similar parallelopipeds, of which AB and KL are 
 two homologous sides ; the ratio of the solid AG to the solid KQ is tripli- 
 cate of the ratio of AB to KL. 
 
 Because the solids are similar, the parallelograms AF, KP are similar 
 (def. 2. 3. Sup.), as also the parallelograms AH, KR ; therefore, the ratios 
 of AB to KL, of AE to KO,and of AD to KN are all equal (def. 1. 6.). 
 But the ratio of the solid AG to the solid KQ is compounded of the ratios 
 o f AC to KM, and of AE to KO. Now, the ratio of AC to KM, because 
 tney are equiangular parallelograms, is compounded (23. 6.) of thcs ratios 
 of AB to KL, and of AD to KN. Wherefore, the ratio of AG to KQ if
 
 OF GEOMETRY. BOOK III. 
 
 207 
 
 IJ 
 
 
 & 
 
 
 
 E 
 
 \ 
 
 J> 
 
 
 
 \ 
 
 \ 
 
 
 A 
 
 E 
 
 H 
 
 Iff , 
 
 a 
 
 \ 
 
 
 
 \ 
 
 
 
 M 
 
 \ 
 
 
 ■ 
 
 compounded of the three ratios of AB to KL, AD to KN, and AE to KO . 
 and the three ratios have already been proved to be equal ; therefore, the 
 ratio that is compounded of them, viz. the ratio of the solid AG to the solid 
 KQ, is triplicate of any of them (def. 12. 5.) : it is therefore triplicate oi 
 the ratio of AB to KL. 
 
 Cor. 1. If as AB to KL, so KL to m, and as KL torn, so is m ton, then 
 AB is to n as the solid AG to the solid KQ. For the ratio of AB to n is 
 triplicate of the ratio of AB to KL (def. 12. 5.), and is therefore equal to 
 that of the solid AG to the solid KQ. 
 
 Cor. 2. As cubes are similar solids, therefore the cube on AB is to the 
 cube on KL in the triplicate ratio of AB to KL, that is in the same ratio 
 with the solid AG, to the solid KQ. Similar solid parallelopipeds are there- 
 fore to one another as the cubes on their homologous sides. 
 
 Cor. 3. In the same manner it is proved, that similar prisms are to one 
 another in the triplicate ratio, or in the ratio of the cubes of their homolo- 
 gous sides. 
 
 PROP. XII. THEOR. 
 
 If two triangular pyramids, which have equal bases and altitudes, be cut by planes 
 that are parallel to the bases, and at equal distances from them, the sections 
 are equal to one another. 
 
 Let ABCDand EFGH be two pyramids, having equal bases BDC and 
 FGH, and equal altitudes, viz. the perpendiculars AQ, and ES drawn from 
 A and E upon the planes BDC and FGH : and let them be cut by planet 
 parallel to BDC and FGH, and at equal altitudes QR and ST above those 
 planes, and let the sections be the triangles KLM, NOP ; KLM and NOP 
 are equal to one another. 
 
 Because the plane ABD cuts the parallel planes BDC, KLM, the com- 
 mon sections BD and KM are parallel (14. 2. Sup.). For the same rea 
 •on, DC and ML are parallel. Since therefore KM and ML are parallel 
 to BD and DC, each to each, though not in the same plane with them, the 
 angle KLM is equal to the angle BDC (9. 2. Sup.). In like manner the 
 other angles of these triangles arc proved to be equal ; therefore, the trian- 
 gles are equiangular, and consequently similar ; and the same is true of the 
 triangles NOP, FGH. 
 
 Now, since the straight lines ARQ, AKB meet the parallel planes BDC
 
 308 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 and KML, they are cut by them proportionally (16. 2. Sup.), or QR : RA 
 : : BK : KA ; and AQ : AR : : AB : AK (18. 5.), for the same reason, 
 ES : ET : : EF : EN ; therefore AB : AK : : EF : EN, because AQ is 
 equal to ES, and AR to ET. Again, because the triangles ABC, AKL 
 are similar, 
 
 AB : AK : : BC : KL ; and for the same reason 
 
 EF : EN : : FG : NO; therefore, 
 
 BC : KL : : FG : NO. And, when four straight lines are propor- 
 tionals, the similar figures described on them are proportionals (22. 6.) ; 
 therefore the triangle BCD is to the triangle KLM as the triangle FGH 
 to the triangle NOP ; but the triangle BDC, FGH are equal ; therefore, 
 the triangle KLM is also equal to the triangle NOP (1. 5.). 
 
 Cor. 1. Because it has been shewn that the triangle KLM is similar 
 to the base BCD ; therefore, any section of a triangular pyramid parallel 
 to the base, is a triangle similar to the base. And in the same manner it is 
 shewn, that the sections parallel to the base of a polygonal pyramid are 
 similar to the base. 
 
 Cor. 2. Hence also, in polygonal pyramids of equal bases and altitudes, 
 the sections parallel to the bases, and at equal distances from them, are 
 equal to one another. 
 
 PROP. XIII. THEOR. 
 
 A series cf prisms of the same altitude may be circumscribed about any pyramxd\ 
 such that the sum of the prisms shall exceed the pyramid by a solid less than 
 any given solid. 
 
 Let ABCD be a pyramid, and Z* a given solid ; a series of prisms hav- 
 ing all the same altitude, may be circumscribed about the pyramid ABCD, 
 so that their sum shall exceed ABCD, by a solid less than Z. 
 
 * The •olid Z is not represented in the figure of this, or me following Proportion
 
 OF GEOMETRY. BOOK III. 
 
 209 
 
 Let Z be equal to a prism standing on the same base with the pyramid, 
 fiz. the triangle BCD, and having for its altitude the perpendicular drawr 
 from a certain point E in the line AC 
 upon the plane BCD. It is evident, that 
 CE multiplied by a certain number m 
 will be greater than AC ; divide CA into 
 as many equal parts as there are units in 
 m, and let these be CF, FG, GH, HA, 
 each of which will be less than CE. 
 Through each of the points F, G, H, let 
 planes be made to pass parallel to the 
 plane BCD, making with the sides of the 
 pyramid the sections FPQ, GRS, HTU, 
 which will be all similar to one another, 
 and to the base BCD (1. cor. 12.3. Sup.). 
 From the point B draw in the plane of 
 the triangle ABC, the straight line BK 
 parallel to CF meeting FP produced in 
 K. In like manner, from D draw DL pa- 
 rallel to CF, meeting FQ in L : Join KL, 
 and it is plain, that the solid KBCDLF 
 is a prism (def. 4. 3. Sup.). By the same 
 construction, let the prisms PM, RO, TV 
 be described. Also, let the straight line IP, which is in the plane of the 
 triangle ABC, be produced till it meet BC in h ; and let the line MQ be 
 produced till it meet DC in g : Join hg ; then hC gQFP is a prism, and is 
 equal to the prism PM (1. Cor. 8. 3. Sup.). In the same manner is describ- 
 ed the prism mS equal to the prism RO, and the prism qU equal to the 
 prism TV. The sum, therefore, of all the inscribed prisms hQ, mS, and 
 qU is equal to the sum of the prisms PM, RO and TV, that is, to the sum 
 of all the circumscribed prisms except the prism BL; wherefore, BL is the 
 excess of the prism circumscribed about the pyramid A BCD above the 
 prisms inscribed within it. But the prism BL is less than the prism which 
 has the triangle BCD for its base, and for its altitude the perpendicular 
 from E upon the plane BCD ; and the prism which has BCD for its base, 
 and the perpendicular from E for its altitude, is by hypothesis equal to the 
 given solid Z ; therefore the excess of the circumscribed, above the inscrib- 
 ed prisms, is less than the given solid Z. But the excess of the circum- 
 scribed prisms above the inscribed is greater than their excess above the 
 pyramid ABCD, because ABCD is greater than the sum of the inscribed 
 prisms. Much more, therefore, is the excess of the circumscribed prisms 
 above the pyramid, less than the solid Z. A series of prisms of the same 
 altitude has therefore been circumscribed about the pyramid ABCD, ex 
 ceeding it by a solid less than the given solid Z. 
 
 PROP. XIV. THEOR 
 
 Pyramids that have equal bases and altitudes are equal to one another 
 
 Let ABCD, EFGH, be two pyramids that have equal bases BCD, FGH 
 
 27
 
 210 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 and also e jual altitudes, viz. the perpendiculars drawn from the vertices A 
 and E upon the planes BCD, FGH : the pyramid ABCD is equal to th« 
 pyramid EFGH. 
 
 If they are not equal, let the pyramid EFGH exceed the pyramid ABCD 
 by the solid Z. Then, a series of prisms of the same altitude may be de 
 scribed about the pyramid ABCD that shall exceed it, by a solid less than 
 Z 03. 3. Sup.) ; let these be the prisms that have for their bases the trian- 
 gles BCD, NQL, ORI, PSM. Divide EH into the same number of equal 
 parts into which AD is divided, viz. HT, TU, UV, VE, and through the 
 
 points T, U and V,let the sections TZW, U.5X, V*Y be made parallel 
 to the base FGH. The section NQL is equal to the section WZT (12. 
 3. Sup.) ; as also ORI to X^U, and PSM to Y*V ; and therefore also the 
 prisms that stand upon the equal sections are equal (1. Cor. 8. 3. Sup.), 
 that is, the prism which stands on the base BCD, and which is between 
 the planes BCD and NQL, is equal to the prism which stands on the base 
 FGH, and which is between the planes FGH and WZT ; and so of the 
 rest, because they have the same altitude : wherefore, the sum of all the 
 prisms described about the pyramid ABCD is equal to the sum of all those 
 described about the pyramid EFGH. But the excess of the prisms de- 
 scribed about the pyramid ABCD above the pyramid ABCD is less than 
 Z (13. 3. Sup.) ; and therefore, the excess of the prism described about 
 the pyramid EFGH above the pyramid ABCD is also less than Z. But 
 the excess of the pyramid EFGH above the pyramid ABCD is equal to 
 Z,by hypothesis, therefore, the pyramid EFGH exceeds the pyramid 
 ABCD, more than the prisms described about EFGH exceeds the same 
 pyramid ABCD. The pyramid EFGH is therefore greater than the sum 
 of the prisms described about it, which is impossible. The pyramids 
 ABCD, EF GH therefore, are not unequal, that is, they are equal to one 
 another.
 
 OF GEOMETRY. BOOK HI. 
 
 211 
 
 PROP. XV. THEOR. 
 
 Every prism having a triangular base may be divided into tnree pyramids that 
 have triangular bases, and that are equal to another. 
 
 Let there be a prism of which the base is the triangle ABC, and lei 
 DEF be the triangle opposite the base : The prism ABC BE F may be 
 divided into three equal pyramids having triangular bases. 
 
 Join AE, EC, CU ; and because ABED is a parallelogram, of which 
 AE is the diameter, the triangle ADE is equal 
 (34. 1.) to the triangle ABE : therefore the py- 
 ramid of which the base is the triangle ADE, 
 and vertex the point C,is equal (14. 3. Sup.) to 
 the pyramid, of which the base is the triangle 
 ABE, and vertex the point C. But the pyra- 
 mid of which the base is the triangle ABE, and 
 vertex the point C, that is, the pyramid ABCE 
 is equal to the pyramid DEFC (14 3. Sup.), 
 for they have equal bases, viz. the triangles 
 ABC, DEF, and the same altitude, viz. the al- 
 titude of the prism ABCDEF. Therefore the 
 three pyramids A DEC, ABEC, DFEC are 
 equal to one another. But the pyramids ADEC, 
 ABEC, DFEC make up the whole prism 
 ABCDEF ; therefore, the prism ABCDEF is 
 divided into three equal pyramids. 
 
 Cor. 1. From this it is manifest, that every pyramid is the third part 
 of a prism which has the same base, and the same altitude with it ; tor if 
 the base of the prism be any other figure than a triangle, it may be divided 
 into prisms having triangular bases. 
 
 Cor. 2. Pyramids of equal altitudes are to one another as their bases ; 
 because the prisms upon the same bases, and of the same altitude, are (1. 
 Cor. 8.3. Sup.) to one another as their bases. 
 
 PROP. XVI. THEOR. 
 
 If from any point in the circumference of the base of a cylinder, a straight 
 line be drawn perpendicular to the plane of the base, it will be wholly in the 
 cylindric superficies. 
 
 Let ABCD be a cylinder of which the base is the circle AEB, DFC 
 the circle opposite to the base, and GH the axis ; from E, any point in the 
 circumference AEB, let EF be drawn perpendicular to the plane of the 
 circle AEB : the straight line EF is in the superficies of the cylinder. 
 
 Let F be the point in which EF meets the plane DFC opposite to the 
 base; join EG and FH ; and let AGHD be the rectangle (It del. 3 
 Sup.) by the revolution of which the cylinder ABCD is described
 
 212 
 
 SUPPLEMENT TO THE ELEMENTS 
 
 Now , because GH is at right angles to GA, 
 ihe straight line, which by its revolution des- 
 cribes the circle AEB, it is at right- angles to 
 all the straight lines in the plane of that circle 
 which meet it in G, and it is therefore at right 
 angles to the plane of the circle AEB. But 
 EF is at right angles to the same plane ; there- 
 fore, EF and GH are parallel (6. 2. Sup.) and 
 in the same plane. And since the plane through 
 GH and EF cuts the parallel planes AEB, 
 DFC, in the straight lines EG and FH, EG is 
 parallel to FH (14. 2. Sup.). The figure 
 EGHF is therefore a parallelogram, and it has 
 the angle EGH a right angle, therefore it is a 
 rectangle, and is equal to the rectangle AH, 
 because EG is equal to AG. Therefore, when 
 in the revolution of the rectangle AH, the straight line AG coincides with 
 EG, the two rectangles AH and EH will coincide, and the straight line 
 AD will coincide with the straight line EF. But AD is always in the 
 superficies of the cylinder, for it describes that superficies ; therefore, EF 
 is also in the superficies of the cylinder. 
 
 PROP. XVII. THEOR. 
 
 A cylinder and a parallelopiped having equal bases and altitudes, are equal to 
 
 one another. 
 
 Let ABCD be a cylinder, and EF a parallelopiped having equal bases, 
 viz. the circle AGB and the parallelogram EH, and having also equal al- 
 titudes ; the cylinder ABGD is equal to the parallelopiped EF. 
 
 QS 
 
 If not, let I'hem be unequal ; and first, let the cylinder be less than the 
 parallelopiped EF ■ and from the parallelopiped EF let there be cut off" »
 
 OF GEOMETRY. BOOK. III. 
 
 2J3 
 
 part EQ by a plane PQ parallel to NF, equal to the cylinder ABCD. In 
 the circle AGB inscribe the polygon AGKBLM that shall differ from the 
 circle by a space less than the parallelogram PH (Cor. 1 4. 1. Sup.), and 
 cut off from the parallelogram EH, a part OR equal to the polygon 
 AGKBLM. The point R will fall between P and N. On the polygon 
 AGKBLM let an upright prism AGBCD be constituted of the same alti 
 tude with the cylinder, which will therefore be less than the cylinder, be- 
 cause it is within it (16. 3. Sup.) ; and if through the point. R a plane RS 
 parallel to NF be made to pass, it will cut off the parallelopiped ES equal 
 (2. Cor. 8. 3. Sup.) to the prism AGBC, because its base is* equal to that 
 of the prism, and its altitude is the same. But the prism AGBC is less 
 than the cylinder ABCD, and the cylinder ABCD is equal to the parallel- 
 opiped EQ, by hypothesis; therefore, ES is less than EQ. and it is also 
 greater, which is impossible. The cylinder ABCD, therefore, is not less 
 than the parallelopiped EF ; and in the same manner, it may be shewn 
 not to be greater than EF. 
 
 PROP. XVIII. THEOR. 
 
 [f a cone and cylinder have the same base and the same altitude, the cone is the 
 third part of the cylinder. 
 
 Let the cone ABCD, and the cylinder BFKG have the same base, viz. 
 the circle BCD, and the same altitude, viz. the perpendicular from the 
 point A upon the plane BCD, the cone ABCD is the third part of the cylin- 
 der BFKG. 
 
 If not, let the cone ABCD be the third part of another cylinder LMNO, 
 having the same altitude with the cylinder BFKG, but let the bases BCD 
 and LIM be unequal ; and first, let BCD be greater than LIM 
 
 Then, because the circle BCD is greater than the circle LIM, a polygon 
 maybe inscribed in BCD, that shall differ from it less than LIM does (A. 
 I. Sup.), and which, therefore, will be greater than LIM. Let this be the 
 polygon BECFD; and upon BECFD, let there be constituted the pyra 
 mid ABECFD, and the prism BCFKHG.
 
 214 SUPPLEMENT TO THE ELEMENTS 
 
 Be '<au&:5 the polygon BECFD is greater than the circle LIM, the prism 
 BCFKHG is greater than the cylinder LMNO, for they have the same 
 altitude, hut the prism has the greater base. But the pyramid ABECFD 
 is ihe third part of the prism (15. 3. Sup.) BCFKHG, therefore it is great- 
 er than the third part of the cylinder LMNO. Now, the cone ABECFD 
 is, by hypothesis, the third part of the cylinder LMNO, therefore the pyra- 
 mid ABECFD is greater than the cone ABCD, and it is also less, because 
 it is inscribed in the cone, which is impossible. Therefore, the cone ABCD 
 »s not less than the third part of the cylinder BFKG : And in the same 
 manner, by circumscribing a polygon about the circle BCD, it may be 
 shewn that the cone ABCD is not greater than the third part of the cylin- 
 der BFKG ; therefore, it is equal to the third part of that cylinder 
 
 PROP. XIX. THEOR. 
 
 If a hemisphere and a cone have equal bases and altitudes, a series of cylinders 
 may be inscribed in the hemisphere, and another series may be described about 
 the cone, having all the same altitudes with one another, and such that their 
 sum shall differ from the sum of the hemisphere, and the cone, by a sAid 
 less than any given solid- 
 
 Let ADB be a semicircle of which the centre is C, and let CD be at right 
 angles to AB ; let DB and DA be squares described on DC, draw CE, 
 and let the figure thus constructed revolve about DC : then, the sector 
 BCD, which is the half of the semicircle ADB, will describe a hemisphere 
 having C for its centre (7 def. 3. Sup.), and the triangle CDE will describe 
 a cone, having its vertex to C, and having for its base the circle (11. def. 
 3. Sup.) described by DE, equal to that described by BC, which is the base 
 of the hemisphere. Let W be any given solid. A series of cylinders may 
 be inscribed in the hemisphere ADB, and another described about the cone 
 ECI, so that their sum shall differ from the sum of the hemisphere and 
 the cone, by a solid less than the solid W. 
 
 Upon the base of the hemisphere let a cylinder be constituted equal to 
 W, and let its altitude be CX. Divide CD into such a number of equal 
 parts, that each of them shall be less than CX ; let these be CH, HG, GF, 
 and FD. Through the points F, G, H, draw FN, GO, HP parallel to 
 CB, meeting the circle in the points K, L and M ; and the straight line 
 CE in the points Q, R and S. From the points K, L, M draw Kf, Lg, 
 Mh, perpendicular to GO, HP and CB ; and from Q, R, and S, draw Qq, 
 Rr, Ss, perpendicular to the same lines. It is evident, that the figure being 
 ♦.hut> constructed, if the whole revolve about CD, the rectangles Ff, Gg, Hh 
 will describe cylinders (14. def. 3. Sup.) that will be circumscribed by the 
 hemispheres BDA ; and the rectangles DN, Fq, Gr, Hs, will also describe 
 cylinders that will circumscribe the cone ICE. Now, it may be demon- 
 strated, as was done of the prisms inscribed in a pyramid (13. 3. Sup.), 
 that the sum of all the cylinders described within the hemisphere, is ex- 
 ceeded by the hemisphere by a solid less than the cylinder generated by 
 the rectangle HB, that is, by a solid less than W, for the cylinder generated 
 by HB is less than W. In the same manner, it may be demonstrated 
 that the sum of the cylinders circumscribing the cone ICE is greater thar
 
 OE GEOMETRY. BOOK III. 
 D 
 
 215 
 
 O £ 
 
 th« cone by a solid less than the cylinder generated by the rectangle DN 
 that is, by a solid less than W. Therefore, since the sum of the cylinder* 
 inscribed in the hemisphere, together with a solid less than W, is equal to 
 the hemisphere ; and, since the sum of the cylinders described about the 
 cone is equal to the cone together with a solid less than W ; adding equals 
 to equals, the sum of all these cylinders, together with a solid less than W, 
 is equal to the sum of the hemisphere and the cone together with a solid 
 less than W. Therefore, the difference between the whole of the cylin- 
 ders and the sum of the hemisphere and the cone, is equal to the difference 
 of two solids, which are each of them less than W ; but this difference 
 must also be less than W, therefore the difference between the two series 
 of cylinders and the sum of the hemisphere and cone is less than the given 
 •olid W. 
 
 PROP. XX. THEOR. 
 
 The same things being supposed as in the last proposition, the sum of all the 
 cylinders inscribed in the hemisphere, and described about the cone, is equal 
 to a cylinder, having the same base and altitude with the hemisphere. 
 
 Let the figure BCD be constructed as before, and supposed to revolve 
 About CD ; the cylinders inscribed in the hemisphere, that is, the cylinders 
 described by the revolution of the rectangles Hh, Gg, Ff, together with 
 those described about the cone, that is, the cylinders described by the revo- 
 lution of the rectangles Hs, Gr, Fq, and DN are equal to the cylinder de 
 scribed by the revolution of the rectangle BD. 
 
 Let L be the point in which GO meets the circle ABD, then, because 
 CGI. is a right angle if CL be joined, the circles described with the dis- 
 tances CG and GL are equal to the circle described with the distance CL 
 (2. Cjt 6. 1 Sup.) or GO; now, CG is equal to OR, because CD is equal 
 to DE, and therefore also, the circles described with the distance GR and 
 GL are together equal to the circle described with the distance GO, that 
 is, the circles described by the revolution of GR and GL about the point 
 G, are together equal to the circle described by the revolution of GO about 
 the same point G; therefore also, the cylinders that stand upon the two 
 first of these circles, having the common altitudes Gil, are enr-i' o *he
 
 216 
 
 SUPPLEMENT TO THE ELEMENTS, &c. 
 
 cylinder which stands on the remaining circle, and which has the same 
 altitude GH. The cylinders described by the revolution of the rectangles 
 Gg, and Gr are therefore equal to the cylinder described by the rectangle 
 GP. And as the same may be shewn of all the rest, therefore the cylin- 
 ders described by the rectangles Hh, Gg, Ff, and by the rectangles Hs, Gr, 
 Fq, DN, are together equal to the cylinder described by BD, that is to the 
 cylinder having the same base and altitude with the hemisphere. 
 
 PROP. XXI. THEOR. 
 
 Every sphere is two-thirds of the circumscribing cylindei . 
 
 Let the figure be constructed as in the two last propositions, and if the 
 hemisphere described by BDC be not equal to two-thirds of the cylinder 
 cbscribed by BD, let it be greater by the solid W. Then, as the cone de- 
 scriued by CDE is one-third of the cylinder (18. 3. Sup.) described by BD, 
 the cone and the hemisphere together will exceed the cylinder by W. But 
 
 that cylinder is equal to the sum of all the cylinders described by the rect- 
 angles Hh, Gg, Ff, Hs, Gr, Fq, DN (20. 3. Sup.) ; therefore the hemisphere 
 and the cone added together exceed the sum of all these cylinders by the 
 given solid W, which is absurd ; for it has been shewn (19. 3. Sup.), that 
 the hemisphere and the cone together differ from the sum of the cylinders 
 by a solid less than W. The hemisphere is therefore equal to two-thirds 
 of the cylinder described by the rectangle BD ; and therefore th« whole 
 sphere is equal to two-thirds of the cylinder described by twice the rectan 
 gle BD, that is, to two-thirds of the circumscribing cylinder. 
 
 END OF THE 8UPPLEME& T TO THE ELEMENTS.
 
 ELEMENTS 
 
 OF 
 
 PLANE TRIGONOMETRY. 
 
 Trigonometry is the application of Arithmetic to Geometry : or, more 
 precisely, it is the application of number to express the relations of the sides 
 and angles of triangles to one another. It therefore necessarily supposes 
 the elementary operations of arithmetic to be understood, and it borrows 
 from that science several ol the signs or characters which peculiarly be- 
 long to it. 
 
 The elements of Plane Trigonometry, as laid down here, are divided into 
 three sections : the first explains the principles ; the second delivers the 
 rules of calculation ; the third contains the construction of trigonometrical 
 tables, together with the investigation of some theorems, useful for extend- 
 ing trigonometry to the solution of the more difficult problems 
 
 SECTION I. 
 
 LEMMA I. 
 
 An angle at the centre of a circle is to four right angles as the arc on which 
 it stands is to the whole circumference. 
 
 Let ABC be an angle at the centre of the circle ACF, standing on the 
 circumference AC : the angle ABC is to four right angles as the arc AC 
 to the whole circumference ACF. 
 
 Produce AB till it meet the circle 
 in E, and draw DBF perpendicular to 
 AE 
 
 Then, because ABC, ABD are two 
 angles at the centre of the circle ACF, 
 the angle ABC is to the angle ABD as 
 the arc AC to the arc AD, (33. 6.) ; 
 and therefore also, the angle ABC is to 
 lour times the angle ABD as the arc 
 AC to four times the arc AD (4. 5.). 
 
 Fut ABD is a right angle, and there- 
 fore four times the arc AD is equa to 
 
 28
 
 218 PLANE TRIGONOMETRY. 
 
 the vl ole circumference ACF ; therefore the angle ABC is to four right 
 angles as the arc AC to the whole circumference ACF. 
 
 Cor. Equal angles at the centres of different circles stand on arcs which 
 have the same ratio to their circumferences. For, if the angle ABC, at 
 the centre of the circles ACE, GHK, stand on the arcs AC, GH, AC is 
 to the whole circumference of the circle ACE, as the angle ABC m fcw 
 right angles ; cUid the arc HG is to the whole circumference of the circle 
 GHK in the same ratio. 
 
 DEFINITIONS. 
 
 1. If two straight lines intersect one another in the centre of a circle, tno 
 re of the circumference intercepted between them is called the Measure 
 
 of the angle which they contain. Thus the arc AC is the measure of 
 the angle ABC. 
 
 2. If the circumference of a circle be divided into 360 equal parts, each of 
 these parts is called a Degree ; and if a degree be divided into 60 equal 
 parts, each of these is called a Minute ; and if a minute be divided into 
 60 equal parts, each of them is called a Second, and so on. And as many 
 degrees, minutes, seconds, &c. as are in any arc, so many degrees, mi- 
 nutes, seconds, &c. are said to be in the angle measured by that arc. 
 
 Cor. 1. Any arc is to the whole circumference of which it is a part, as 
 the number of degrees, and parts of a degree contained in it is to the 
 number 360. And any angle is to four right angles as the number of 
 degrees and parts of a degree in the arc, which is the measure of that 
 angle, is to 360. 
 
 Cor. 2. Hence also, the arcs which measure the same angle, whatever 
 be the radii with which they are described, contain the same number of 
 degrees, and parts of a degree. For the number of degrees and parts of 
 a degree contained in each of these arcs has the same ratio to the num- 
 ber 360, that the angle which they measure has to four right angles 
 (Cor. Lem. 1.). 
 
 The degrees, minutes, seconds, &c. contained in any arc or angle, are 
 usually written as in this example, 49°. 36'. 24". 42'" ; that is, 49 de- 
 grees, 36 minutes, 24 seconds, and 42 thirds. 
 
 3 Two angles, which are together equal to two right angles, or two arcs 
 which are together equal to a semicircle, are called the Supplements of 
 one another. 
 
 4 A straight line CD drawn through C, one of the extremities of the are
 
 PLANE TRIGONOMETRY. 
 
 219 
 
 AC, perpendicular to the diameter 
 passing through the other extremity 
 A, is called the Sine of the arc AC, 
 or of the angle ABC, of which AC 
 is the measure. 
 
 3or 1 . The sine of a quadi ant, or of 
 a right angle, is equal to the radius. 
 
 Cor. 2. The sine of an arc is half the 
 chord of twice that arc : this is evi- 
 dent by producing the sine of any 
 arc till it cut the circumference. 
 
 5. The segment DA of the diameter passing through A, one extremity o' 
 the arc AC, between the sine CD and the point A, is called the Versed 
 sine of the arc AC, or of the angle ABC. 
 
 6. A straight line AE touching the circle at A, one extremity of the arc 
 AC, and meeting the diameter BC, which passes through C the other 
 extremity, is called the Tangent of the arc AC, or of the angle ABC 
 
 Cor. The tangent of half a right angle is equal to the radius. 
 
 7. The straight line BE, between the centre and the extremity of the tan 
 gent AE is called the Secant of the arc AC, or of the angle ABC. 
 
 Cor. to Def. 4, 6,7, the sine, tangent and secant of any angle ABC, are 
 likewise the sine, tangent, and secant of its supplement CBF. 
 
 It is manifest, from Def. 4. that CD is the sine of the angle CBF. Let 
 CB be produced till it meet the circle again in I ; and it is also mani 
 fest, that AE is the tangent, and BE the secant, of the angle ABI, or 
 CBF, from Def. 6. 7. 
 
 Cor. to Def. 4, 5, 6, 7. The sine, versed sine, tangent, and secant of an 
 arc, which is the measure of any gi- 
 ven angle ABC, is to the sine, versed 
 sine, tangent and secant, of any other 
 arc which is the measure of the same 
 angle, as the radius of the first arc is 
 to the radius of the second. 
 
 Let AC, MN be measures of the angle 
 ABC, according to Def. 1 . ; CD the 
 sine, DA the versed Bine. AE the 
 
 OTSID 
 
 tangent, and BE the secant of the arc AC, according to Def. 4, 5, 6, 7 , 
 NO the sine, OM the versed sine, MP the -tangent, and BP the secant 
 of the arc MN. according to the same definitions. Since CD, NO, AE 
 MP are parallel, CD : NO : : rad. CB : rad. NB, and AE : MP : : rmd 
 AB : rad. BM, also BE : BP : : AB : BM ; likewise because BC : BD 
 : : BN : BO, that is, BA : BD : : BM : BO, by conversion and alterna- 
 tion, AD : MO : : AB : MB. Hence the corollary is manifest. And
 
 J20 
 
 PLANE TRIGONOMETRY. 
 
 tnerefore, if tables be constructed, exhibiting in numbers the sines, tan- 
 gents secai.ts, and versed sines of certain angles to a given radius, thev 
 will exhibit the ratios of the sines, tangents, &c. of the same angles to 
 any radius whatsoever. 
 In such tables, which are called Trigonometrical Tables, the radius is 
 either supposed 1, or some in the series 10, 100, 1000, &c. The use 
 and construction of these tables are about to be explained. 
 
 8. The difference between any angle 
 and a right angle, or between any 
 arc and a quadrant, is called the 
 Complement of that angle, or of that 
 arc. Thus, if BH be perpendicular 
 to AB, the angle CBH is the com- 
 plement of the angle ABC, and the 
 arc HC the complement of AC ; 
 also, the complement of the obtuse 
 angle FBC is the angle HBC, its 
 excess above a right angle ; and 
 the complement of the arc FC is 
 HC. 
 
 9. The sine, tangent, or secant of the complement of any angle is called 
 the Cosine, Cotangent, or Cosecant of that angle. Thus, let CL or DB, 
 which is equal to CL, be the sine of the angle CBH ; HK the tangent, 
 and BK the secant of the same angle : CL or BD is the cosine, HK the 
 cotangent, and BK the cosecant of the angle ABC. 
 
 Cor. 1. The radius is a mean proportional between the tangent and the 
 cotangent of any angle ABC; that is, tan. ABC X cot. ABC=R 2 . 
 
 For, since HK, BA are parallel, the angles HKB, ABC are equal, and 
 KHB, BAE are right angles ; therefore the triangles BAE, KHB are 
 similar, and therefore AE is to AB, as BH or BA to HK. 
 
 Cor. 2. The radius is a mean proportional between the cosine and se- 
 cant of any angle ABC ; or 
 cos. ABC x sec. ABC=R 2 . 
 
 Since CD, AE are parallel, BD is to BC or BA, as BA to BE. 
 
 PROP. I. 
 
 In a right angled plane triangle, as the hypotenuse to either of the sides, so 
 the radius to the sine of the angle opposite to that side ; and as either of the 
 sides is to the other side, so is the radius to the tangent of the angle oppo- 
 site to that side. 
 
 Let ABC be aright angled plane triangle, of which BC is the hypote- 
 nuse. From the centre C, with any radius CD, describe the arc DE ■ 
 draw DF at right angles to CE, and from E draw EG touching the circle 
 in E, and meeting CB in G ; DF is the sine, and EG the tanger* of the 
 arc DF or of the angle C.
 
 PLANE TRIGONOMETRY. 
 
 221 
 
 The two triangles DFC, BAC, are equiangular, because the anglei 
 DFC, BAC are right angles, and the 
 angle at C is common. Therefore, 
 CB : BA : : CD : DF ; but CD is 
 the radius, and DF the sine of the 
 angle C, (Def. 4.) ; therefore CB : 
 BA : : R : sin. C. 
 
 Also, because EG touches the cir- 
 cle in E, CEG is a right angle, and 
 therefore equal to the angle BAC ; 
 and since the angle at C is common 
 to the triangles CBA, CGE, these triangles are equiangular, wherefore 
 CA : AB : : CE : EG ; but CE is the radius, and EG the tangent of the 
 angle C ; therefore, CA : AB : : R : tan. C. 
 
 Cor. 1. As the radius to the secant of the angle C, so is the side adja- 
 cent to that angle to the hypotenuse. For CG is the secant of the angle 
 G (def. 7.), and the triangles CGE, CBA being equiangular, CA : CB : : 
 CE : CG, that is, CA : CB : : R : sec. C. 
 
 Cor. 2. If the analogies in this proposition, and in the above corollary 
 be arithmetically expressed, making the radius = 1, they give sin. C = 
 
 tan. C = -r-p^, sec. C = — -^. Also, since sin. C=cos. B, because B 
 
 BC 
 
 AC 
 
 AC 
 
 AB 
 
 of the 
 
 is the complement of C, cos. B =5^, and for the same reason, cos. C 
 
 BC 
 
 AC 
 BC 
 
 Cor. 3. In every triangle, if a perpendicular be drawn from any 
 angles on the opposite side, the segments of 
 that side are to one another as the tangents of 
 the parts into which the opposite angle is di- 
 vided by the perpendicular. For, if in the tri- 
 angle ABC, AD be drawn perpendicular to 
 «he base BC, each of the triangles CAD, ABD 
 being right angled, AD : DC : : R : tan. CAD, 
 and AD : DB : : R : tan. DAB ; therefore, ex 
 »quo, DC : DB : : tan. CAD : tan. BAD. 
 
 . SCHOLIUM. 
 
 The proposition, just demonstrated, is most easily remembered, by stating 
 t. thus : If in a right angled triangle the hypotenuse be made the radius, 
 the sides become the sines of the opposite angles ; and if one of the sides be 
 made the radius, the other side becomes the tangent of the opposite angle, 
 and the hypotenuse the secant of it.
 
 222 
 
 PLANE TRIGONOMETRY 
 
 • PROP. II. THEOR. 
 
 The sides of a plane triangle are to one another as the sines of the opposite 
 
 angles. 
 
 From A any angle in the triangle ABC, 
 let AD be drawn perpendicular to BC. 
 And because the triangle ABD is right 
 angl id at D, AB : AD : : R : sin. B ; and 
 for the same reason, AC : AD : : R : 
 sin. C, and inversely, AD : AC : : sin. 
 C : R • therefore, ex aequo inversely, AB 
 ; AC : sin. C : sin. B. In the same 
 manner it may be demonstrated, that AB 
 i BC : : sin. C : sin. A. 
 
 PROP. III. THEOR. 
 
 The sum of the sines of any two arcs of a circle, is to the difference of their 
 sines, as the tangent of half the sum of the arcs to the tangent of half then 
 difference. 
 
 Let AB, AC be two arcs of a circle ABCD ; let E be the centre, and 
 AEG the diameter which passes through A ; sin. AC+sin. AB : sin. AC 
 —sin. AB : : tan. £ (AC+AB) : tan. \ (AC— AB). 
 
 Draw BF parallel to AG, meeting the circle again in F. Draw BH 
 and CL perpendicular to AE, and they will be the sines of the arcs AB 
 and AC ; produce CL till it meet the circle again in D ; join DF, FC, DE, 
 EB, EC, DB. 
 
 Now, since EL from the centre is perpendicular to CD, it bisects the 
 line CD in L and the arc CAD in A : 
 DL is therefore equal to LC, or to the 
 sine of the arc AC ; and BH or LK 
 being the sine of AB, DK is the sum 
 of the sines of the arcs AC and AB, 
 and CK is the difference of their sines; 
 DAB also is the sum of the arcs AC 
 and AB, becauso AD is equal to AC, 
 and BC is their difference. Now, in 
 the triangle DFC, because FK is per- 
 pendicular to DC, (3. cor. 1.), DK : 
 KC : : tan. DFK : tan. CFK ; but 
 tan. DFK=tan. ^ arc. BD, because 
 the angle DFK (20. 3.) is the half of DEB, and therefore measured by 
 half the arc DB. For the same reason, tan. CFK=tan. \ arc. BC ; and 
 consequently, DK : KC : : tan. £ arc. BD : tan. ^ arc. BC. But DK is 
 the sum of the sines of the arcs AB and AC ; and KC is the difference of 
 the sines ; also BD is the sum of the arcs AB and AC, and BC the diffe- 
 rence of those arcs
 
 PLANE TRIGONOMETRY 223 
 
 Cor. 1. Because EL is the cosine of AC, and EH of AB, FK is the 
 sum of these cosines, and KB their difference ; for FK=£FB + EL=EH 
 + EL, and KB = LH = EH -EL. Now, FK : KB : : tan. FDK : tan. 
 BDK ; and tan. DFK=cotan. FDK, because DFK is the complement 
 of FDK ; therefore, FK : KB : : cotan. DFK : tan. BDK, that is, FK : 
 KB : : cotan. J arc. DB : tan. ^arc. BC. The sum of the cosines of two 
 arcs is therefore to the difference of the same cosines as the cotangent of 
 half the sum of the arcs to the tangent of half their difference. 
 
 Cor. 2. In the right angled triangle FKD, FK : KD : : R : tan. DFK; 
 Now FK=cos. AB+cos. AC, KD= sin. AB + sin. AC, and tan. DFK= 
 tan. I (AB + AC), therefore cos. AB+cos. AC : sin. AB + sin. AC : : R : 
 .an. j (AB+AC). 
 
 In the same manner, by help of the triangle FKC, it may be shewn that 
 cos. AB + cos. AC : sin. AC— sin. AB : : R : tan. j(AC— AB). 
 
 Cor. 3. If the two arcs AB and AC be together equal to 90°, the tan- 
 gent of half their sum, that is, of 45°, is equal to the radius. And the arc 
 BC being the excess of DC above DB, or above 90°, the half of the arc 
 BC will be equal to the excess of the half of DC above the half of DB, that 
 is, to the excess of AC above 45° ; therefore, when the sum of two arcs is 
 90°, the sum of the sines of those arcs is to their difference as the radius to 
 the tangent of the difference between either of them and 45°. 
 
 PROP. IV. THEOR. 
 
 The sum of any two sides of a triangle is to their difference, as the tangent of 
 half the sum of the angles opposite to those sides, to the tangent ofhalft\tw 
 difference. 
 
 Let ABC be any plane triangle ; 
 CA + AB : CA-AB : : tan. \ (B+C) : tan. \ (B-C). 
 For (2.) CA : AB : : sin. B : sin. C ; 
 and therefore (E. 5.) 
 
 CA+AB : CA— AB : : sin. B+sin. C : sin. B— sin. C. 
 But, by the last, sin. B+sin. C : sin. B — sin. C : : 
 tan. \ (B+C) : tan. \ (B— C) ; therefore also, (1 1. 5.) 
 CA f AB : CA-AB : : tan. } (B + C) : tan. } (B-C).
 
 224 PLANE TRIGONOMETRY. 
 
 Otherwise, without the 3d. 
 
 J jet ABC be a triangle ; the sum of AB and AC any two sides, is to the 
 difference of AB and AC as the tangent of half the sum of the angles ACB 
 and ABC, to the tangent ©f half their difference. 
 
 About the centre A wiih the radius AB, the greater of the two sides, de- 
 scribe a circle meeting BC produced in D, and AC produced in E and F 
 Join DA, EB, FB ; and draw FG parallel to CB, meeting EB in G 
 
 Because the exterior angle EAB is equal to the two interior ABC, ACB 
 (32. 1.) : and the angle EFB, at the circumference is equal to half the an- 
 gle EAB at the centre (20. 3.) ; therefore EFB is half the sum of the an- 
 gles opposite to the sides AB and AC. 
 
 Again, the exterior angle ACB is equal to the two interior CAD, ADC, 
 and therefore CAD is the difference of the angles ACB, ADC, that is, of 
 ACB, ABC, for ABC is equal to ADC. Wherefore also DBF, which is 
 the half of CAD, or BFG, which is equal to DBF, is half the difference of 
 the angles opposite to the sides AB, AC. 
 
 Now because the angle FBE,in a semicircle is a right angle, BE is tho 
 tangent of the angle EFB, and BG the tangent of the angle BFG to the 
 radius FB ; and BE is therefore to BG as the tangent of half the sum of 
 the angles ACB, ABC to the tangent of half their difference. Also CE is 
 the sum of the sides of the triangle ABC, and CF their difference ; and be- 
 cause BC is parallel to FG, CE : CF : : BE ; BG, (2. 6.) that is, the sum 
 of the two sides of the triangle ABC is X r j their difference as the tangent of 
 half the sum of the angles opposite tc those sides to the tangent of half 
 their difference.
 
 PLANE TRIGONOMETRY. 
 
 225 
 
 PROP. V. THEOR. 
 
 If a perpendicular be drawn from any angle of a triangle to the opposite sidd, 
 or base ; the sum of the segments of the base is to the sum of the otJier two 
 sides of the triangle as the difference of those sides to the difference of the 
 segments of the base. 
 
 For (K. 6.), the rectangle under the sum and difference of the segments 
 of the base is equal to the rectangle under the sum and difference of the 
 sides, and therefore (16. 6.) the sum of the segments of the base is to the 
 sum of the sides as the difference of the sides to the difference of the seg- 
 ments of the base. 
 
 PROP. VI. THEOR. 
 
 In any triangle, twice the rectangle contained by any two sides ts to the dif- 
 ference between the sum of the squares of those sides, and the square of th» 
 base, as the radius to the cosine of the angle included by the two sides. 
 
 Let ABC be any triangle, 2AB.BC is 
 to the difference between AB 2 +BC 2 and 
 AC 2 as radius to cos. B. 
 
 From A draw AD perpendicular to BC, 
 and (12. and 13. 2.) the difference be- 
 tween the sum of the squares of AB and 
 BC, and the square on AC is equal to 
 2BC.BD. 
 
 But BC.BA : BC.BD : : BA : BD : : 
 R : cos. B, therefore also 2BC.BA : 2BC. B D C 
 
 BD : : R : cos. B. Now 2BC.BD is the difference between AB'+BO* 
 and AC 2 , therefore twice the rectangle 
 
 AB.BC is to the difference between Al 
 
 AB 2 +BC 2 , and AC 2 as radius to the 
 cosine of B. 
 
 Cor. If the radius =1, BD=BA 
 Xcos. B, (1.), and 2BC.BAxcos. B 
 =2BC.BD, and therefore when B is 
 acute, 2BC.BAXcos. B = BC 2 -f BA 3 
 -AC 2 , and adding AC 2 to both; AC 2 
 4- 2 cos. B x BC.BA = BC 2 + BA 2 ; 
 
 and taking 2 cos. Bx BC.BA from both, AC 8 =BC 2 — 2 cos. Bx BC.BA 
 + BA 2 . Wherefore AC= -y/(BC 2 — 2 cos. B X BC.BA+BA 2 ). 
 
 If B is an obtuse angle, it is shewn in the same way that AC« 
 v/(BC 2 +2 cos. Bx BC.BA+BA 2 ). 
 
 29
 
 826 PLANE TRIGONOMETRY. 
 
 PROP. VII. THEOR. 
 
 Four times the rectangle contained by any two sides of a triangle, is tc the 
 rectangle contained by two straight lines, of which one is the base or third 
 side of the triangle increased by the difference of the two sides, and th* 
 other the base diminished by the difference of the same sides, as the square 
 of the radius to the square of the sine of half the angle included between the 
 two sides of the triangle. 
 
 Let ABC be a triangle of which BC is the base, and AB the greater of 
 the two sides ; 4AB.AC : (BC+(AB-AC)) x (BC-(AB-AC)) : : R* 
 • (sin. £ BAC) 2 . 
 
 Produce the side AC to D, so that AD=AB ; join BD, and draw AE 
 
 CF at right angles to it ; from the centre C with the radius CD describe 
 the semicircle GDH, cutting BD in K, BC in G, and meeting BC pro- 
 duced in H. 
 
 It is plain that CD is the difference of the sides, and therefore that BH is 
 the base increased, and BG the base diminished by the difference of the 
 sides ; it is also evident, because the triangle BAD is isosceles, that DE is 
 the half of BD, and DF is the half of DK, wherefore DE— DF=the half 
 of BD— DK (6. 5.), that is, EF=£ BK. And because AE is drawn pa- 
 rallel to CF, a side of the triangle CFD, AC : AD : : EF : ED, (2. 6.) ; 
 and rectangles of the same altitude being as their bases ACAD : AD 2 : : 
 EF.ED : ED 2 (1. 6.), and therefore 4AC.AD : AD 2 : : 4EF.ED : ED 2 , or 
 alternately, 4AC.AD : 4EF.ED : : AD 2 : ED 2 . 
 
 But since 4EF=2BK, 4EF.ED=2BK.ED=2ED.BK=DB.BK=* 
 HB.BG ; therefore 4AC.AD : DB.BK : : AD 2 : ED 2 . Now AD : ED : : 
 R : sin. EAC=sin. £ BAC (1. Trig.) and AD 2 : ED 2 : : R 2 : (sin. £ BAC) 2 : 
 therefore, (11.5.) 4AC.AD : HB.BG : : R 2 : (sin. £ BAC) 2 , or since AB 
 =.AD, 4AC.AB : HB.BG : : R 2 : (sin. £ BAC) 2 . Now 4AC.AB is four 
 times the rectangle contained by the sides of the triangle ; HB.BG is that 
 contained by BC+(AB— AC) and BC— (AB— AC). 
 
 Cor. Hence 2 ^AC.AD : ^HB.BG • • R : sin } BAC.
 
 PLANE TRIGONOMETRY. 
 
 227 
 
 PROP. VIII. THEOR. 
 
 F(A» imes the rectangle contained by any two sides of a triangle, is to the 
 rectangle contained by two straight lines, of which one is the sum of those 
 sides increased by the base of the triangle, and the other the sum of the same 
 sides diminished by the base, as the square of the radius to the square Oj 
 the cosine of half the angle included between the two sides of the triangle. 
 
 Let ABC be a triangle, of which BC is the base, and AB the greater o' 
 the other two sides, 4ABAC : (AB + AC + BC) (AB + AC-BC) : : R 2 
 fcos. \ RAG) 2 . 
 
 From the centre C, with the radius CB, describe the circle BLM, meet- 
 ing AC, produced, in L and M. Produce ALto N, so that AN = AB ; let 
 AD = AB ; draw AE perpendicular to BD ; join BN, and let it meet the 
 circle again in P ; let CO be perpendicular to BN ; and let it meet AE in R. 
 
 It is evident that MN = AB-f AC + BC ; and that LN=AB + AC — 
 BC. Now, because BD is bisected in E, and DN in A, BN is parallel to 
 AE ; and is therefore perpendicular to BD, and the triangles DAE, DNB 
 are equiangular ; wherefore, since DN=2AD.BN=2AE, and BP=2BO 
 =2RE ; also PN=2AR. 
 
 But because the triangles ARC and AED are equiangular, AC : AD : : 
 AR : AE, and because rectangles of the same altitude are as their base* 
 
 (1. 6.), ACAD : AD 2 : : AR.AE : AE 2 ,and alternately ACAD : AR.AE 
 : : AD 2 : AE 2 , and 4AC.AD : 4AR.AE : : AD 2 : AE 2 . But 4AR AE = 
 2ARx2AE = NP.NB = MN.NL; therefore 4AC.A I) : MN.NL :: AD 2 : 
 AE 2 . But AD : AE : : R : cos. DAE (1) =scos. I (BAC): Wherefor* 
 4AC.AD : MN.NL : : R 2 : (cos. \ BAC) 2
 
 228 PLANE TRIGONOMETRY. 
 
 Now 4 AC AD is four times the rectangle under the sides AC and AB, 
 (for AD=AB), and MN.NL is the rectangle under the sum of the sides 
 increased by the base, and the sum of the sides diminished by the base 
 
 Cor. I . Hence 2 ^AC.AB : y/ MN.NL : : R : cos. ± BAC. 
 
 Cor. 2. Since by Prop. 7. 4AC.AB : (BC+(AB— AC)) (BC— (AB 
 — BC)) : : R 2 : (sin. ^ BAC) 2 ; and as has been now proved 4AC.AB : 
 (AB + AC+BC) (AB + AC-BC) : : R 2 : (cos. $ BAC) 2 ; therefore, ex 
 jequo, (AB + AC + BC) (AB + AC-BC) : (BC + (AB-AC)) (BC- 
 (AB— AC)) : : (cos. A BAC) 2 : (sin. £ BAC) 2 . But the cosine of any arc 
 is to the sine, as the radius to the tangent of the same arc ; therefore, (AB 
 + AC + BC) (AB + AC-BC) : (BC+(AB-AC)) BC-(AB-AC)) : : 
 R 2 : (tan. % BAC) 2 ; and 
 
 V (AB+AC+BC) (AB + AC-BC : 
 
 V(BC + AB-AC) (BC— (AB— AC)) : : R : tan. \ BAC. 
 
 LEMMA II. 
 
 If there be two unequal magnitudes, half their difference added to half then 
 sum is equal to the greater ; and half their difference taken from half their 
 sum is equal to the less. 
 
 Let AB and BC be two unequal magnitudes, of which AB is the great- 
 er ; suppose AC bisected in D, and AE 
 
 equal to BC. It is manifest that AC is A e J) g q 
 
 the sum, and EB the difference of the 
 
 magnitudes. And because AC is bisected in D, AD is equal to DC : but 
 AE is also equal to BC, therefore DE is equal to DB, and DE or DB is 
 half the difference of the magnitudes. But AB is equal to BD and DA, 
 that is, to half the difference added to half the sum ; and BC is equal to 
 the excess of DC, half the sum above DB, half the difference. 
 
 Cor. Hence, if the sum and the difference of two magnitudes be given, 
 the magnitudes themselves may be found ; for to half the sum add half the 
 difference, and it will give the greater : from half the sum subtract half 
 the difference, and it will give the less. 
 
 SCHOLIUM. 
 
 This property is evident from the algebraical sum and difference of the 
 two quantities a and b, of which a is the greater ; let their sum be denoted 
 by s, and their difference by d : then, 
 a-\-b=s > 
 a-b=d $ 
 /. by addition, 2a=s+d\ 
 
 and a=±+± 
 
 By subtraction, 2b=zs — d ; 
 
 — — ~.
 
 PLANE TRIGONOMETRY 
 
 SECTION II. 
 
 OF THE RULES OF TRIGONOMETRICAL 
 CALCULATION. 
 
 The General Problem which Trigonometry proposes to resolve is • 
 /•i any plane triangle, of the three sides and the three angles, any three being 
 given, and one of these three being a side, to find any of the other three. 
 
 The things here said to be given are understood to be expressed by their 
 numerical values: the angles, in degrees, minutes, &c; and the sides in 
 feet, or any other known measure. 
 
 The reason of the restriction in this problem to those cases in which at 
 least one side is given, is evident from this, that by the angles alone being 
 given, the magnitudes of the sides are not determined. Innumerable tri- 
 angles, equiangular to one another, may exist, without the sides of any 
 one of them being equal to those of any other ; though the ratios of their 
 sides to one another will be the same in them all (4. 6.). If therefore, only 
 the three angles are given, nothing can be determined of the triangle but 
 the ratios of the sides, which may be found by trigonometry, as being the 
 same with the ratios of the sines of the opposite angles. 
 
 For the conveniency of calculation, it is usual to divide the general pro- 
 blem into two ; according as the triangle has, or has not, one of the angles 
 a right anglo. 
 
 PROBLEM I. 
 
 In a right angled triangle, of the three sides, and three angles, any two being 
 given, besides the right angle, and one of those two being a side, it is required 
 to find the other three. 
 
 It is evident, that when one of the acute angles of aright angled trianglo 
 is given, the other is given, being the complement of the former to a right 
 angle ; it is also evident that the sine of any of the acute angles is th« 
 cosine of the other. 
 
 This problem admits of several cases, and the solutions, or rules for cal- 
 culation, which all depend on the first Proposition, may be conveniently 
 exhibited in the form of a table ; where the first column contains the things 
 given ; the second, the things required ; and the third, the rules or propo- 
 sitions by which they are found.
 
 230 
 
 PLANE TRIGONOMETRY. 
 
 r 
 
 QITSN. 
 
 SOUGHT. 
 
 SOLUTION. 
 
 
 CB and B, the 
 hypotenuse and 
 angle. 
 
 AC. 
 AB. 
 
 R : sin B : : CB : AC. 
 R : cos B : : CB : AB. 
 
 1 
 2 
 
 3 
 
 4 
 
 AC and C, a 
 side and one of 
 the acute angles. 
 
 BC. 
 AB. 
 
 Cos C : R : : AC : BC. 
 R : tan C : : AC : AB. 
 
 CB and BA, 
 the hypotenuse 
 land a side. 
 
 C 
 AC. 
 
 CB : BA : : R : sin C. 
 R : cos C : : CB : AC. 
 
 5 
 6 
 
 AC and AB ; 
 the two sides. 
 
 C. 
 CB. 
 
 AC : AB : : R : tan C. 
 Cos C : R : : AC : CB. 
 
 7 
 8 
 
 Remarks on the Solutions in the table. 
 
 In the second case, when AC and C are given to find the hypotenuse 
 BC, a solution may also be obtained by help of the secant, for CA : CB : : 
 R : sec. C. ; if, therefore, this proportion be made R : sec. G : : AC : CB, 
 CB will be found. 
 
 In the third case, when the hypotenuse BC and the side AB are given 
 to find AC, this may be done either as directed in the Table, or by the 
 47th of the first ; for since AC 2 = BC 2 — BA 2 , AC = /BC 2 — BA 2 . 
 This value of AC will be easy to calculate by logarithms, if the quantity 
 BC 2 — B A 2 be separated into two multipliers, which may be done ; because 
 (C or. 5. 2. ), BC 2 -BA 2 =(BC + BA) . (BC-BA). Therefore AC a 
 V(BC + BA) (BC-BA). 
 
 When AC and AB are given, BC may be found from the 47th, as in the 
 preceding instance, for BC= -v/BA 2 4-AC 2 . But BA 2 + AC 2 cannot be 
 separated into two multipliers ; and therefore, when BA and AC are large 
 numbers, this rule is inconvenient for computation by logarithms. It is 
 best in such cases to seek first for the tangent of C, by the analogy in the 
 Table, AC : AB : : R : tan. C ; but if C itself is not required, it is sufficient, 
 having found tan. C by this proportion, to take from the Trigonometric
 
 PLANE TRIGONOMETRY. *31 
 
 Tables tne cosine that corresponds to tan. C, and then to compute CB from 
 the proportion cos. C : R : : AC : CB. 
 
 PROBLEM II. 
 
 In an oblique angled triangle, of the three sides and three angles, any thre* 
 being given, and one of these three being a side, it is required to find tk$ 
 other three. 
 
 This problem has four cases, in each of which the solution depends oa 
 tome of the foregoing propositions. 
 
 CASE I. 
 
 Two angles A and B, and one side AB, of a triangle ABC, being giren, 
 to find the other sides. 
 
 SOLUTION. 
 
 Because the angles A and B are given, C is also given, being the sup 
 plement of A+B ; and, (2.) 
 
 Sin. C : sin. A : : AB : BC ; also, 
 Sin. C : sin. B : : AB : AC. 
 
 CASE II. 
 
 Two sides AB and AC, and the angle B opposite to one of them, being 
 given, to find the other angles A and C, and also the other side BC. 
 
 SOLUTION. 
 
 The angle C is found from this proportion, AC : AB : : sin. B : sin. C. 
 Also, A=180°— B— C ; and then, sin. B : sin. A ; : AC : CB, by Case 1. 
 
 In this case, the angle C may have two values ; for its sine being found 
 by the proportion above, the angle belonging to that sine may either be that 
 which is found in the tables, or it may be the supplement of it (Cor. def. 4.). 
 This ambiguity, however, does not arise from any defect in the solution, 
 but from a circumstance essential to the problem, viz. that whenever AC 
 is less than AB, there are two triangles which have the sides AB, AC, and 
 the angle at B of the same magnitude in each, but which are nevertheless 
 unequal, the angle opposite to A B in the one, being the supplement o( tnat 
 which is opposite to it in the other. The truth of this appears by describ- 
 ing from the centre A with the radius AC, an arc intersecting BC in C
 
 »32 PLANE TRIGONOMETRY. 
 
 A. 
 
 and C ; then, if AC and AC be drawn, it is evident that the triangles 
 ABC, ABC have the side AB and the angle at B common, and the sides 
 AC and AC equal, but have not the remaining side of the one equal to the 
 remaining side of the other, that is, BC to BC, nor their other angles equal, 
 viz. BCA to BCA, nor BAC to BAC. But in these triangles the angles 
 ACB, ACB are the supplements of one another. For the triangle CAC 
 is isosceles, and the angle ACC=AC'C, and therefore, ACB, which is 
 the supplement of ACC, is also the supplement of ACC or ACB ; and 
 these two angles, ACB, ACB are the angles found by the computation 
 above. 
 
 From these two angles, the two angles BAC, BAC will be found : the 
 angle BAC is the supplement of the two angles ACB, ABC (32. 1.), and 
 therefore its sine is the same with the sine of the sum of ABC and ACB 
 But BAC is the difference of the angles ACB, ABC : for it is the diffe- 
 rence of the angles ACC and ABC, because ACC, that is, ACC is equal 
 jo the sum of the angles ABC, BAC (32. 1.). Therefore, to find BC, 
 having found C, make sin. C : sin. (C+B) : : AB : BC ; and again, sin. 
 C : sin. (C-B) : : AB : BC. 
 
 Thus, when AB is greater than AC, and C consequently greater than 
 B, there are two triangles which satisfy the conditions of the question. 
 But when AC is greater than AB, the intersections C and C fall on oppo- 
 site sides of B, so that the two triangles have not the same angle at B com- 
 mon to them, and the solution ceases to be ambiguous, the angle required 
 being necessarily less than B, and therefore an acute angle. 
 
 CASE III. 
 
 Two sides AB and AC, and the angle A, between them, being given to 
 find the other angles B and C, and also the side BC. 
 
 SOLUTION. 
 
 First, make AB+AC : AB— AC : : tan. J (C + B) : tan. £ (C-B). 
 Then, since ^ (C+B) and ^ (C — B) are both given, B and C may be f )und. 
 For B=£ (C+B)+£ (C-B), and C=£ (C+B)-£ (C-B). (Lem. 2.) 
 
 To find BC. 
 
 Having found B, make sin. B : sin. A : : AC : BC. 
 But BC may also be found without seeking for the angle B and C , fot 
 BC= VAB 2 — 2cos. AxAB.AC+AC 2 , Prop 6
 
 PLANE TRIGONOMETRY. 
 
 233 
 
 This method of finding BC is extremely useful in many geometrical in 
 restigations, but it is not very well adapted for computation by logarithms 
 because the quantity under the radical sign cannot be separated into sim 
 pie multipliers. Therefore, when AB and AC are expressed by large 
 numbtrs, the other solution, by finding the angles, and then computing BC, 
 is preferable. 
 
 CASE IV. 
 The three sides AB, BC, AC, being given, to find the angles A, B, C. 
 
 solution I. 
 
 Take F such that BC : BA+AC : : BA— AC : F, then F is either the 
 sum or the difference of BD, DC, the segments of the base (5.). If F be 
 greater than BC, F is the sum, and BC the difference of BD, DC ; but, if 
 F be less than BC, BC is the sum, and F the difference of BD and DC. 
 In either case, the sum of BD and DC, and their difference being given. 
 BD and DC are found. (Lem. 2.) 
 
 Then, (1.) CA : CD : : R : cos. C ; and BA : BD : : R : cos. B ; where 
 fore C and B are given, and consequently A. 
 
 D C B 
 
 SOLUTION II. 
 
 C D 
 
 Let D be the differen ce of the sides AB, AC. Then (Cor. 7.) 2 y/ AB.AC 
 V(BC+D) (BC-D) : : R : sin. } BAC. 
 
 SOLUTION III. 
 
 Let S be the sum of the sides BA and AC. Then (1. Cor. 8.) 2 ^/AB.AC 
 V(S+BC)(S-BC) : : R : cos. $ BAC. 
 
 80LUTION IV. 
 
 S and D retaining the significations above, (2.Cor.8.) - v / (S- r -BC)(S -BCJ) 
 : y/{UC + D) (BC-D) : : R : tan. $ BAC. 
 
 It may be observed of these four solutions, that the first has the advan- 
 tage of being easily remembered, but that the others are rather more expe- 
 ditious in calculation. The second solution is preferable to the third, when 
 the angle sought is less than a right angle ; on the other hand, the third 
 js preferable to the second, when the angle sought is greater than a right 
 
 30
 
 834 PLANE TRIGONOMETRY. 
 
 angle , and in extreme cases, that is, when the angle sought is very acute 
 or very obtuse, this distinction is very material to be considered. The 
 reason is, that the sines of angles, which are nearly = 90°, or the cosines 
 of angles, which are nearly = 0, vary very little for a considerable varia- 
 tion in the corresponding angles, as may be seen from looking into the ta- 
 bles of sines and cosines. The consequence of this is, that when the sine 
 or cosine of such an angle is given (that is, a sine or cosine nearly equal to 
 the radius,) the angle itself cannot be very accurately found. If, for in 
 stance, the natural sine .9998500 is given, it will be immediately per- 
 ceived from the tables, that the arc corresponding is between 89°, and 89° 
 1' ; but it cannot be found true to seconds, because the sines of 89° and ol 
 89° 1', differ only by 50 (in the two last places,) whereas the arcs them- 
 selves differ by 60 seconds. Two arcs, therefore, that differ by 1", or even 
 by more than 1", have the same sine in the tables, if they fall in the last 
 degree of the quadrant. 
 
 The fourth solution, which finds the angle from its tangent, is not liable 
 to this objection ; nevertheless, when an arc approaches very near to 90°, 
 the variations of the tangents become excessive, and are too irregular to 
 allow the proportional parts to be found with exactness, so that when the 
 angle sought is extremely obtuse, and its half of consequence very near to 
 90, the third solution is the best. 
 
 It may always be known, whether the angle sought is greater or less 
 than a right angle by the square of the side opposite to it being greater oi 
 less than the squares of the other two sides. 
 
 SECTION III, 
 
 CONSTRUCTION OF TRIGONOMETRICAL TABLES. 
 
 In all the calculations performed by the preceding rules, tables of sines 
 and tangents are necessarily employed, the construction of which remains 
 to be explained. 
 
 The tables usually contain the sines, &c. to every minute of the quad- 
 rant from 1' to 90°, and the first thing required to be done, is to compute 
 the sine of 1', or of the least arc in the tables. 
 
 1. If ADB be a circle, of which the centre is C, DB, any arc of that cir- 
 cle, and the arc DBE double of DB ; and if the chords DE, DB be drawn, 
 also the perpendiculars to them from C, viz. CF, CG, it has been demon- 
 strated (8. 1. Sup.), that CGis a mean proportional between AH, half the 
 radius, and AF, the line made up of the radius and the perpendicular CF. 
 Now CF is the cosine of the arc BD, and CG the cosine of the half of BD ; 
 whence the cosine of the half of any arc BD, of a circle of which the ra- 
 dius = 1, is a mean proportional between £ and 1 + cos. BD. Or, for the 
 greater generality, supposing A = any arc, cos. ^ A is a mean proportional
 
 PLANE TRIGONOMETRY. 
 
 235 
 
 between £ and 1+cos. A, and therefore (cos. £ A) 2 =| (1+cos. A) or cot 
 I A = V£ (1 + cos. A). 
 
 2. From this theorem, (which is the same that is demonstrated (8. 1. 
 Sup.), only that it is here expressed trigonometrically,) it is evident, that if 
 the cosine of any arc be given, the cosine of half that arc may be found. 
 Let BD, therefore, be equal to 60°, so that the chord BD=radius, then the 
 cosine or perpendicular CF was shewn (9. 1. Sup.) to be =^, and there- 
 
 fore cos. £ BD, or cos. 30°= -y/jU+i) 3 Vi="^- ^ n tne same man " 
 
 ner, cos. 15°= v4 (1+cos. 30°), and cos. 7°, 30'= ■/£ (1+cos. 15°), &c. 
 In this way the cosine of 3°, 45 , of 1°, 52', 30", and so on, will be com- 
 puted, till after twelve bisections of the arc of 60°, the cosine of 52". 44'". 
 93"". 45 v . is found. But from the cosine of an arc its sine may be 
 found, for if from the square of the radius, that is, from l,the square of 
 the cosine be taken away, the remainder is the square of the sine, and its 
 square root is the sine itself. Thus the sine of 52". 44'". 03"". 45 v . is 
 found. 
 
 3. But it is manifest, that the sines of very small arcs are to one another 
 nearly as the arcs themselves. For it has been shewn that the number of 
 the sides of an equilateral polygon inscribed in a circle may be so great, 
 that the perimeter of the polygon and the circumference of the circle may 
 differ by a line less than any given line, or, which is the same, may be 
 nearly to one another in the ratio of equality. Therefore their like parts 
 will also be nearly in the ratio of equality, so that the side of the polygon 
 will be to the arc which it subtends nearly in the ratio of equality ; and 
 therefore, half the side of the polygon to half the arc subtended by it, that 
 is to say, the sine of any very small arc will be to the arc itself, nearly in 
 the ratio of equality. Therefore, if two arcs are both very small, the first 
 will be to the second as the sine of the first to the sine of the second 
 Hence, from the sine of 52". 54'". 03"". 45 v . being found, the sine of 1
 
 436 
 
 PLANE TRIGONOMETRY. 
 
 becomes known , for, as 52". 44'". 03"". 45 y . to l,so is the sine of the 
 former arc to the sine of the latter. Thus the sine of 1' is found as 
 0.0002908882. 
 
 4. The sine 1' being thus found, the sines of 2', of 3', or of any number 
 of minutes, may be found by the following proposition. 
 
 THEOREM. 
 
 Let AB, AC, AD be three such arcs, that BC the difference of the first 
 and second is equal to CD the difference of the second and third ; the ra- 
 dius is to the cosine of the common difference BC as the sine of AC, the 
 middle arc, to half the sum of the sines of AB and AD, the extreme arcs. 
 
 Draw CE to the centre : let BF, CG, and DH perpendicular to AE, be 
 the sines of the arcs AB, AC, AD. Join BD, and let it meet CE in I ; 
 draw IK perpendicular to AE, also BL and 
 IM perpendicular to DH. Then, because 
 the arc BD is bisected in C, EC is at right 
 angles to BD, and bisects it in I ; also BI is 
 the sine, and EI the cosine of BC or CD. 
 And, since BD is bisected in I, and IM is 
 parallel to BL (2. 6.), LD is also bisected in 
 M. Now BF is equal to HL, therefore BF 
 +DH=DH + HL = DL+2LH = 2LM+ 
 2LH=2MH or 2KI ; and therefore IK is 
 half the sum of BF and DH. But because 
 the triangles CGE, IKE are equiangular, 
 CE : EI : : CG : IK, and it has been shewn that EI=cos. BC, and IK:a 
 I (BF+DH) ; therefore R : cos. BC : : sin. AC : I (sin. AB-f sin. AD). 
 
 Cor. Hence, if the point B coincide with A, 
 R : cos. BC : : sin. BC : ^ sin. BD, that is, the radius is to the cosine oi 
 any arc as the sine of the arc is to half the sine of twice the arc ; or if any 
 arc=A, I sin. 2A=sin. Axcos. A, or sin. 2A=2 sin. Ax cos A. 
 
 Therefore also, sin. 2'=2' sin. 1' x cos. 1' : so that from the sine at 1 
 cosine of one minute the sine of 2' is found. 
 
 Again, 1', 2', 3', being three such arcs that the difference between the 
 first and second is the same as between the second and third, R : cos. 1' : : 
 sin. 2 : 1 (sin. l'+sin. 3'), or sin. l'-f-sin. 3'=2 cos. l'+sin. 2', and taking 
 sin. 1' from both, sin. 3'=2 cos. l'xsin. 2' — sin. 1. 
 
 In like manner, sin. 4'=2' cos. l'xsin. 3'— sin. 2, 
 sin. 5'=2' cos. l'xsin. 4'— sin. 3, 
 sin. G'=2' cos. l'xsin. 5'— sin. 4, &c. 
 
 Thus a table containing the sines for every minute 6f the quadrant may 
 be computed ; and as the multiplier, cos. 1' remains always the same, the 
 calculation is easy. 
 
 For computing the sines of arcs that differ by more than 1', the method 
 is the same. Let A, A-j-B, A+2B be three such arcs, then, by this the- 
 orem, R : cos.B : : sin. (A+B) : \ (sin A+sin. (A-J-2B)) ; and therefor© 
 making the radius 1,
 
 PLANE TRIGONOMETRY. 23? 
 
 «in. A+sin. (A+2B)=2 cos. Bxsin. (A+B), 
 
 or sin. (A+2B)=2 cos. Bxsin. (A+B)— sin. A. 
 By means of these theorems, a table of the sines, and consequently ah*« 
 af the cosines, of arcs of any number of degrees and minutes, from to 90, 
 
 maybe constructed. Then, because tan. A= — '—, the table of tangent* 
 
 is computed by dividing the sine of any arc by the cosine of the same arc. 
 When the tangents have been found in this manner as far as 45°, the tan- 
 gents for the other half of the quadrant may be found more easily by an- 
 other rule. For the tangent of an arc above 45° being the co-tangent of 
 an arc as much under 45° ; and the radius being a mean proportional be- 
 tween the tangent and co-tangent of any arc (1. Cor. def. 9), it follows, il 
 the difference between any arc and 45° be called D, that tan. (45° — D) : 
 
 1 : : 1 : tan. (45°+D), so that tan. (450+D) ^ (4 \ Q _ D y 
 
 Lastly, the secants are calculated from (Cor. 2. def. 9) where it i« 
 shewn that the radius is a mean proportional between the cosine and the 
 
 secant of any arc, so that if Abe any arc, sec. A= r-. 
 
 COS. A. 
 
 The versed sines are found by subtracting the cosines from the radius. 
 
 5. The preceding Theorem is one of four, which, when arithmetically 
 expressed, are frequently used in the application of trigonometry to the so- 
 lution of problems. 
 
 I/no, If in the last Theorem, the arc AC=A, the arc BC=B, and the 
 radius EC=I, then AD = A + B, and AB = A— B ; and by what has just 
 been demonstrated, 
 
 1 : cos. B : : sin. A : £ sin. (A+B)-f£ sin. (A— B), 
 
 and therefore, 
 sin. Ax cos. B=i sin. (A + B)+£ (A— B). 
 2do, Because BF, IK, DH are parallel, the straight lines BD and FH 
 are cut proportionally, and therefore FH, the difference of the straight lines 
 FE and HE, is bisected in K ; and therefore, as was shewn in the last 
 Theorem, KE is half the sum of FE and HE, that is, of the cosines of the 
 arcs AB and AD. But because of the similar triangles EGC, EKI, EC 
 • EI : : GE : EK ; now, GE is the cosine of AC, therefore, 
 R : cos. BC : : cos. AC : £ cos. AD-J-^ cos. AB, 
 or 1 : cos. B : : cos. A : $ cos. (A-f-B)+^ cos. (A — B) ; 
 and therefore, 
 cos. Ax cos. B=£ cos. (A + B)+£ cos. (A— B); 
 3tio, Again, the triangles I DM, CEG are equiangular, for the angles 
 KIM, Ell) are equal, being each of them right angles, and therefore, tak- 
 ing away the angle EIM, the angle DIM is equal to the angle EIK, that 
 is, to the angle ECG; and the angles DMI, CGE aro also equal, being 
 both rijjlit angle*, and therefore the triangles 1 DM, CGE have the sides 
 about their equal angles proportionals, and consequently, EC : CG : : Dl 
 : IM ; now, IM is half the difference of the cosines FE and EH, therefore, 
 R : sin. AC : : sin. BC : } cos. AB— 1 cos. AD, 
 or 1 : sin. A : : sin. B : } cos. (A— B)— \ cos. (A-J-B) ■
 
 338 PLANE TRIGONOMETRY 
 
 and also, 
 
 sin. Ax sin. B=% cos. 'A— B)— ^ cos. (A+B). 
 4*6, Lastly, in the same triangles ECG, DIM, EC : EG : : ID : DM; 
 now, DM is half the difference of the sines DH and BE, therefore, 
 R : cos. AC : : sin. BC : } sin. AD— i sin. AB, 
 or 1 : cos. A : : sin. B : i sin. (A+B)— \ sin. (A+B) ; 
 and therefore, 
 cos. Ax sin. B=£ sin. (A+B)— \ sin. (A— B). 
 
 6. If therefore A and B be any two arcs whatsoever, the radius being 
 supposed 1 ; 
 
 I. sin. Axcos B=i sin. (A + B)+^ sin. (A— B). 
 
 II. cos. Axcos. B=|cos. (A— B)+j cos. (A + B) 
 
 III sin. Axsin. B=|cos.(A- B) — \ cos. (A+B). 
 
 IV. cos. Axsin. B==| sin. (A+B)— |sin. (A B). 
 
 From these four Theorems are also deduced other four. 
 
 For adding the first and fourth together, 
 sin. Axcos. B + cos. Axsin. B=sin. (A+B). 
 
 Also, by taking the fourth from the first, 
 sin. Axcos. B — cos. Axsin. B=sin. (A — B). 
 Again, adding the second and third, 
 cos. Axcos. B+sin. Axsin. B=cos. (A— B) ; 
 And, lastly, subtracting the third from the second, 
 cos. Axcos. B— sin. Ax sin. B=cos. (A+B). 
 
 7. Again, since by the first of the above theorems, 
 
 sin. AX cos. B=4 sin. (A+B)+£ sin. (A— B), if A+B=S, and A— B=D. 
 
 V /T ON a S + D jr> S ~ D u t ■ S + D 
 
 then (Lem. 2.) A-= — - — , and B= — - — ; wherefore sm. — — X cos. 
 
 49 . 40 2t 
 
 § J) 
 
 — - — =^ sin. S+£ D. But as S and D may be any arcs whatever, to 
 
 preserve the former notation, they may be called A and B, which also ex- 
 press any arcs whatever : thus, 
 
 A+B A-B . . ■ . 
 
 sm. — - — x cos. — - — =^ sin. A+-J- sm. B, or 
 
 _ . A+B A-B ..-'.'„ 
 
 2 sin. — - — Xcos. — - — =sin. A+sin. B. 
 -o 2 
 
 In the same manner, from Theor. 2 is derived, 
 
 A+B A— B _ . „ 
 
 2 cos. — - — x cos. — - — =cos. B+cos. A. From the 3d, 
 
 & z 
 
 _ . A+B . A— B - 
 
 2 sin. — - — xsin. — - — =cos. B— cos. A; and from the 4tn, 
 
 A+B A-B . ; ' ■ 
 
 2 cos. — - — Xsin. — -— =sin. A— sm. B. 
 
 In all these Theorems, the arc B is supposed less than A. 
 
 8. Theorems of tne same Kind with respect to the tangents of arcs miy 
 be deduced from the preceding. Because the tangent of any arc is equa) 
 to the sine of the arc divided by its cosine,
 
 PLANE TRIGONOMETRY. 239 
 
 tan. (A + B)= — — — „\ . But it has just been shewn, that 
 v ' cos. (A+B) 
 
 sin. (A+B)=sin. Axcos. B+cos. Axsin. B, and that 
 
 cos. (A+B)=cos. Axcos. B— sin. Axsin.B; therefore, tan. (A+B) =a 
 
 sin. Axcos. B+cos. Ax sin. B , .. . .. , , , , , 
 
 . — - — ■ : — — , and dividing both the numerator and ueno- 
 
 cos. Axcos. B— sin. Ax sin. B 
 
 — tin \ i tin I • 
 
 minator of this fraction by cos. Axcds. B, tan. (A + B)=- — k rr 
 
 J v ' 1 tan. Ax tan. B 
 
 r ii / « r.* tan - A tan. B 
 
 In like manner, tan. (A — B) = - ■ =-,. 
 
 v ' 1+tan. AX tan. B 
 
 9. If the Theorem demonstrated in Prop. 3, be expressed in the same 
 manner with those above, it gives 
 
 sin. A+sin. B _ tan. \ (A + B) 
 
 sin. A — sin. B ™ tan. ^ (A— B)' 
 Also by Cor. 1, to the 3d, 
 
 cos. A-f-cos. B _ cot. \ (A+B) 
 
 cos. A— cos. B tan. \ (A — B)* 
 
 And by Cor. 2, to the same proposition, 
 
 sin. A+sin. B tan. 1 (A+B) _ . , 
 
 T-r- u = -jz ', or since R is here supposed =k 1, 
 
 COSt l\ ™y~ COS * D t\ 
 
 sin. A+sin. B . . . , „. 
 
 c-3rA+c^TB = ,an -^ A+B »- 
 
 10. In all the preceding Theorems, R, the radius, is supposed = 1 , be- 
 cause in this way the propositions are most concisely expressed, and aro 
 also most readily applied to trigonometrical circulation. But if it be re- 
 quired to enunciate any of them geometrically, the multiplier R, which 
 has disappeared, by being made = 1, must be restored, and it will always 
 be evident from inspection in what terms this multiplier is wanting. Thus, 
 Theor. 1,2 sin. A Xcos. B=sin. (A+B)+sin. (A — B), is atrue proposition, 
 taken arithmetically ; but taken geometrically, is absurd, unless we sup- 
 ply the radius as a multiplier of the terms on the right hand of the sine of 
 equality. It then becomes 2 sin. Axcos. B = R(sin. (A+B) + sin. (A — B)); 
 or twice the rectangle under the sine of A, and the cosine of B equal to the 
 rectangle under the radius, and the sum of the sines of A+B and A — B. 
 
 In general, the number of linear multipliers, that is, of lines whose nume- 
 rical values are multiplied together, must be the same in every term, other- 
 wise we will compare unlike magnitudes with one another. 
 
 The propositions in this section are useful in many of the higher branches 
 of the Mathematics, and are the foundation of what is called the Arithmetit 
 $f Sines.
 
 ELEMENTS 
 
 OF 
 
 SPHERICAL 
 
 TRIGONOMETRY. 
 
 PROP. I. 
 
 If a sphere be cut by a plane through the centre, the section is a circle, having the 
 same centre with the sphere, and equal to the circle by the revolution of which 
 the sphere was described- 
 
 For all the straight lines drawn from the centre to the superficies of the 
 sphere are equal to the radius of the generating semicircle, (Def. 7. 3. 
 Sup.). Therefore the common section of the spherical superficies, and of 
 a plane passing through its centre, is a line, lying in one plane, and hav- 
 ing all its points equally distant from the centre of the sphere ; therefore it 
 is the circumference of a circle (Def. 11. 1.), having for its centre the cen- 
 tre of the sphere, and for its radius the radius of the sphere, that is, of the 
 semicircle by which the sphere has been described. It is equal, therefore, 
 to the circle of which that semicircle was a part. 
 
 DEFINITIONS. 
 
 1 . Any circle, which is a section of a sphere by a plane through ts centre, 
 is called a great circle of the sphere. 
 
 Cor. All great circles of a sphere are equal ; and any two of them bisect 
 
 one another. 
 They are all equal, having all the same radii, as has just been shewn ; and 
 ■ any two of them bisect one another, for as they have the same centre, 
 
 their common section is a diameter of both, and therefore bisects both. 
 
 2. The pole of a great circle of a sphere is a point in the superficies of the 
 sphere, from which all strai T ht lines drawn to the circumference of the 
 circle are equal. 
 
 3. A spherical angle is an angle on the superficies of a sphere, contained 
 by the arcs of two great circles which intersect one another ; and is the 
 same with the inclination of the planes of these great circles
 
 SPHERICAL TRIGONOMETRY. 
 
 241 
 
 4. A spherical triangle is a figure, upon the superficies of a sphere, com- 
 prehended by three arcs of three great circles, each of which is less than 
 a semicircle. 
 
 PROP. II 
 
 The arc of a great circle, between the pole and the circumference of another 
 great circle, is a quadrant. 
 
 Let } BC be a great circle, and D its pole ; if DC, an arc of a great 
 circle, pass through D, and meet ABC in C, the arc DC is a quadrant. 
 
 Let the circle, of which CD is an arc, meet ABC again in A, and let 
 AC be the common section of the planes 
 of these great circles, which will pass 
 through E, the centre of the sphere : Join 
 DA, DC. Because AD=DC, (Def. 2.), 
 and equal straight lines, in the same cir- 
 cle, cut off equal arcs (28. 3.), the arc AD 
 = the arc DC ; but ADC is a semicircle, 
 theiefore the arcs AD, DC are each of 
 them quadrants. 
 
 Cor. 1. If DE be drawn, the angle AED is a right angle ; and DE 
 being therefore at right angles to every line it meets with in the plane of 
 the circle ABC, is at right angles to that plane (4. 2. Sup.). Therefore 
 the straight line drawn from the pole of any great circle to the centre of the 
 sphere is at right angles to the plane of that circle ; and, conversely, a 
 straight line drawn from the centre of the sphere perpendicular to the plane 
 of any greater circle, meets the superficies of the sphere in the pole of that 
 circle. 
 
 Cor. 2. The circle ABC has two poles, one on each side of its plane, 
 which are the extremities of a diameter of the sphere perpendicular to the 
 plane ABC ; and no other points but these two can be poles of the circle 
 ABC. 
 
 PROP. III. 
 
 If the pole of a great circle be the same with the intersection of other two gteai 
 circles : the arc of the first mentioned circle intercepted between the other 
 two, is the measure of the spherical angle which the same two circles make 
 with one another. 
 
 Let the great circles BA, CA on the superficies 
 of a sphere, of which the centre is D, intersect one 
 another in A, and let BC be an arc of another great 
 circle, of which the pole is A ; BC is the measure 
 of the spherical angle BAC. 
 
 Join AD, DB, DC ; since A is the pole of BC, 
 AB, AC are quadrants (2.), and the angles ADB, 
 \DC are right angles : therefore (4. def. 2. Sup.), 
 he angle CDB is the inclination of the planes of 
 
 31
 
 242 
 
 SPHERICAL TRIGONOMETRY. 
 
 the circles VB, AC, and is (def. 3.) equal to the spherical angle BAG 
 but the arc BC measures the angle BDC, therefore it also measures the 
 spherical angle BAC* 
 
 Cor. If two arcs of great circles, AB and AC, which intersect one an- 
 other in A, be each of them quadrants, A will be the pole of the great cir- 
 cle which passes through E and C the extremities of those arcs. For 
 since the arcs AB and AC are quadrants, the angles ADB, ADC are right 
 angles, and AD is therefore perpendicular to the plane BDC, that is, to the 
 plane of the great circle which passes through B and C. The point A is 
 therefore (1. Cor. 2.) the pole of the great circle which passes through B 
 and C. 
 
 PROP. IV. 
 
 If the planes of two great circles of a sphere be at right angles to one another 
 the circumference of each of the circles passes through the poles of the 
 other ; and if the circumference of one great circle pass through the poles 
 of another, the planes of these circles are at right angles. 
 
 Let ACBD, AEBF be two great circles, the planes of which are right 
 angles toone another, the poles of the circle AEBF are in the circumference 
 ACBD, and the poles of the circle ACBD in the circumference AEBF. 
 
 From G the centre of the sphere, draw GC in the plane ACBD perpen- 
 dicular to AB. Then because GC in the plane ACBD, at right angles 
 to the plane AEBF, is at right angles 
 to the common section of the two 
 planes, it is (Def. 2- 2. Sup.) also at 
 right angles to the plane AEBF, and 
 therefore (1. Cor. 2.) C is the pole of 
 the circle AEBF ; and if CG be pro- 
 duced in D, D is the other pole of the 
 circle AEBF. 
 
 In the same manner, by drawing 
 GE in the plane AEBF, perpendicu- 
 lar to AB, and producing it to F, it has 
 shewn that E and F are the poles of 
 the circle ACBD. Therefore, the 
 poles of each of these circles are in 
 the circumference of the other. 
 • Again, If C be one of the poles of the circle AEBF, the great circie 
 ACBD which passes through C, is at right angles to the circle AEBF. 
 For, CG being drawn from the pole to the centre of the circle AEBF, is 
 at right angles (1. Cor. 2.) to the plane of that circle ; and therefore, every 
 plane passing through CG (17. 2. Sup.) is at right angles to the plane 
 AEBF ; now, the plane ACBD passes through CG. 
 
 Cor. 1 . If of two great circles, the first passes through the poks of the 
 
 • When in any reference no mention is made of a Book, or of the Plane 1 :igonomeiry, UM 
 Soher'ca' Trigonometry is meant.
 
 SPHERICAL TRIGONOMETRY. 243 
 
 second, the second also passes through the poles of the first. For, if the 
 first passes through the poles of the second, the plane of the first must b« 
 at right angles to the plane of the second, by the second part of this propo- 
 sition ; and therefore, by the first part of it, the circumference of each 
 passes through the poles of the other. 
 
 Cor. 2. All greater circles that have a common diameter have theii 
 poles in the circumference of a circle, the plane of which is perpendiculai 
 to that diameter. 
 
 PROP. V. 
 In isosceles spherical triangles the angles at the base are equal. 
 
 Let ABC be a spherical triangle, having the side AB equal to the side 
 AC ; the spherical angles ABC and ACB are equal. 
 
 Let C be the centre of the sphere ; join 
 DB, DC, DA, and from A on the straight 
 lines DB, DC, draw the perpendiculars AE, 
 AF ; and from the points E and F draw in 
 the plane DBC the straight lines EG, FG 
 perpendicular to DB and DC, meeting one 
 another in G : Join AG. 
 
 Because DE is at right angles to each of 
 the straight lines AE, EG, it is at right angles 
 to the plane AEG, which passes through 
 AE, EG (4. 2. Sup.) ; and therefore, every 
 plane that passes through DE is at right angles to the plane AEG (17. 2. 
 Sup.) ; wherefore, the plane DBC is at right angles to the plane AEG. 
 For the same reason, the plane DBC is at right angles to the plane AFG, 
 and therefore AG, the common section of the planes AFG, AEG is at 
 right angles (18. 2. Sup.) to the plane DBC, and the angles AGE, AGF 
 are consequently right angles. 
 
 But since the arc AB is equal to the arc AC, the angle ADB is equal 
 to the angle ADC. Therefore the triangles ADE, ADF, have the angles 
 EDA, FDA, equal, as also the angles AED, AFD, which are right an- 
 gles; and they have the side AD common, therefore the other sides are 
 equah, viz. AE to AF(26. 1.), and DE to DF. Again, because the angles 
 AGE, AGF are right angles, the squares on AG and GE are equal to the 
 square of AE ; and the squares of AG and GF to the square of AF. But 
 the squares of AE and AF are equal, therefore die squares of AG and GE 
 are equal to the squares of AG and GF, and taking away the common 
 square of AG, the remaining squares of GE and GF are equal, and GE is 
 therefore equal to GF. Wherefore, in the triangles AFG, AEG, the side 
 GF is equal to the side GE, and AF has been proved to be equal to AE, 
 and the base AG is common ; therefore, the angle AFG is equal to the 
 angle AEG (8. 1.). But the angle AFG is the angle which the plane 
 ADC makes with the plane DBC (I. <lef. 2. Sup.), because FA and FG, 
 which are drawn in these planes, are at right angles to DF, the common 
 •action of the planes. The angle AFG (3. def.) is therefore equal to the
 
 244 SPHERICAL TRIGONOMETRY. 
 
 spheneal angle ACB ; and, for the same reason, the angle AEG ib vqxiai 
 to the spherical angle ABC. But the angles AFG, AEG are equal 
 Therefore the spherical angles ACB, ABC are also equal. 
 
 PROP. VI. 
 
 If the angles at the base of a spherical triangle be equal, the triangle is isosceles. 
 
 Let ABC be a spherical triangle having the angles ABC, ACB equal 
 to one another ; the sides AC and AB are also equal. 
 
 Let D be the centre of the sphere ; join DB, DC, DA, and from A on 
 the straight lines DB, DC, draw the perpendiculars AE, AF ; and from 
 the points E and F, draw in the plane DBC 
 the straight lines EG, FG perpendicular to 
 DB and DC, meeting one another in G ; 
 join AG. 
 
 Then, it may be proved, as was done in 
 the last proposition, that AG is at right an- 
 gles to the plane BCD, and that therefore 
 the angles AGF, AGE are right angles, and 
 also that the angles AFG, AEG are equal 
 to the angles which the planes DAC, DAB 
 make with the plane DBC. But because 
 the spherical angles ACB, ABC are equal, the angles which the planes 
 DAC, DAB make with the plane DBC are equal (3. def.), and therefore 
 the angles AFG, AEG are also equal. The triangles AGE, AGF have 
 therefore two angles of the one equal to two angles of the other, and they 
 have also the side AG common, wherefore they are equal, and the side AF 
 is equal to the side AE. 
 
 Again, because the triangles ADF, ADE are right angled at F and E, 
 the squares of DF and FA are equal to the square of DA, that is, to the 
 squares of DE and DA ; now, the square of AF is equal to the square of 
 AE, therefore the square of DF is equal to the square of DE, and the side 
 DF to the side DE. Therefore, in the triangles DAF, DAE, because DF 
 is equal to DE and DA common, and also AF equal to AE, the angle 
 ADF is equal to the angle ADE ; therefore also the arcs AC and AB, 
 which are the measures of the angles ADF, and ADE, are equal to one 
 another ; and the triangle ABC is isosceles. 
 
 PROP. VII. 
 
 Any two sides of a spherical triangle are greater than the third. 
 
 Let ABC be a spherical triangle, any two sides AB, BC are greater than 
 ine third side A.C.
 
 SPHERICAL TRIGONOMETRY. 245 
 
 Let D be the centre of the sphere ; 
 join DA, DM, DC. 
 
 The solid angle at D is contained by 
 three plane angles ADB, ADC, BDC ; 
 any two of which, ADB, BDC are 
 greater (20. 2. Sup.) than the third 
 ADC ; and therefore any two of the 
 »rcs AB, AC, BC, which measure 
 these angles, as AB and BC must also 
 be greater than the third AC. 
 
 PROP. VIII. 
 
 The three sides of a spherical triangle are less than the circumference of a 
 
 great circle. 
 
 Let ABC be a spherical triangle as before, the three sides AB, BC, AC 
 are less than the circumference of a great circle. 
 
 Let D be the centre of the sphere : The solid angle at D is contained 
 by three plane angles BDA, BDC, ADC, which together are less than 
 four right angles (21.2. Sup.) therefore the sides AB, BC, AC, which are 
 the measures of these angles, are together less than four quadrants describ- 
 ed with the radius AD, that is, than the circumference of a great circle. 
 
 PROP. IX. 
 
 In a spherical triangle the greater angle is opposite to the greater side ; and 
 
 conversely. 
 
 Let ABC be a spherical triangle, the greater angle A is opposed to the 
 greater side BC. 
 
 Let the angle BAD be made equal 
 to the angle B, and then BD, DA will 
 be equal (6.), and therefore AD, DC 
 are equal to BC ; but AD, DC are 
 greater than AC (7.), therefore BC is 
 greater than AC, that is, the greater 
 angle A is opposite to the greater side 
 BC. The converse is demonstrated as 
 Prop. 19. 1. Elem. 
 
 PROP. X. 
 
 According as the sum of two of the sides of a spherical triangle, is greater than 
 a semicircle, equal to it, or less, each of the interior angles at the base is greater 
 than the exterior and opposite angle at the base, equal to it, or less ; and also 
 the sum o*~ the two interior angles at the base greater than two right angles, 
 equal to two right angles, or less than two right angles. 
 
 5 et ABC be a spherical triangle, of which the sides are AB and BC ,
 
 216 
 
 SPHERICAL TRIGONOMETRY. 
 
 produce any of the two sides as AB, and the base AC, till they meet again 
 in D; then, the arc ABD is a semicircle, and the spherical angles at A 
 and D are equal, because each of them is the inclination of the circle AR D 
 to the circle ACD. mtu 
 
 1. If AB, BC be equal to a 
 semicircle, that is, to A D, BC will 
 be equal to BD, and therefore (5.) 
 the angle D, or the angle A, will 
 be equal to the angle BCD, that 
 is, the interior angle at the base 
 equal to the exterior and oppo- 
 site. 
 
 2. If AB, BC together be greater than a semicircle, that is, greater than 
 ABD, BC will be greater than BD ; and therefore (9.), the angle D, that 
 Js, the angle A, is greater than the an^le BCD. 
 
 3. In the same manner it is shewn, if AB, BC together be less than a 
 semicircle, that the angle A is less than the angle BCD. 
 
 Now, since the angles BCD, BCA are equal to two right angles, if th*s 
 angle A be greater than BCD, A and ACB together will be greater than 
 two right angles. If A be equal to BCD, A and ACB together, will be 
 equal to two right angles ; and if A be less than BCD, A and ACB will 
 be less than two right angles. 
 
 PROP. XI. 
 
 If the angular points of a spherical triangle be made the poles of three great 
 circles, these three circles by their intersections will form a triangle, which 
 is said to be supplemental to the former ; and the two triangles are such, 
 that the sides of the one are the supplements of the arcs which measure the 
 angles of the other. 
 
 Let ABC be a spherical triangle ; and from the points A, B, and C as 
 poles, let the great circles FE, ED, DF be described, intersecting one an- 
 other in F, D and E ; the sides of the triangle FED are the supplement of 
 the measures of the angles A, B, C, viz. FE of the angle BAC, DE of the 
 angle ABC, and DF of the angle ACB : And again, AC is the supplement 
 of the angle DFE, AB of the angle FED, and BC of the angle EDF 
 
 Let AB produced meet DE, EF in G, M ; 
 let AC meet FD, FE in K, L ; and let BC 
 meet FD, DE in N, H. 
 
 Since A is the pole of FE, and the circle 
 AC passes through A, EF will pass through 
 the pole of AC (1. Cor. 4.) and since AC 
 passes through C, the pole of FD, FD will 
 pass through the pole of AC ; therefore the 
 pole of AC is in the point F, in which the 
 arcs DF, EF intersect each other. In the 
 same manner, D is the pole of BC, and E 
 :he pole of AB. 
 
 And since F, E are the poles of AL, AM, the arcs FL and EM (2.) ?r«
 
 SPHERICAL TRIGONOMETRY. 247 
 
 quadrants, and FL, EM together, that is, FE and ML togethei, are equai 
 to a semicircle. But since A is the pole of ML, ML is the measuro o r the 
 angle BAC (3.), consequently FE is the supplement of the measure of the 
 angle BAC. In the same manner, ED, DF are the supplements of th» 
 measures of the angles ABC, BCA. 
 
 Since likewise CN, BH are quadrants, CN and BH together, that is, 
 NH and BC together, are equal to a semicircle ; and since D is the polo of 
 NH, NH is the measure of the angle FDE, therefore the measure of the 
 angle FDE is the supplement of the side BC. In the same manner, it is 
 shewn that the measures of the angles DEF, EFD are the supplements 
 of the sides AB, AC in the triangle ABC. 
 
 PROP. XII. 
 
 The three angles of a spherical triangle are greater than two, and less than sir 
 
 right angles. 
 
 The measure of the angles A, B, C, in the triangle ABC, together with 
 the three sides of the supplemental triangle DEF, are (11.) equal to three 
 semicircles ; but the three sides of the triangle FDE, are (8.) less than two 
 semicircles ; therefore the measures of the angles A, B, C, are greater than 
 a semicircle ; and hence the angles A, B, C are greater than two righ* 
 angles. 
 
 And because the interior angles of any triangle, together with the exte- 
 rior, are equal to six right angles, the interior alone are less than six right 
 angles. 
 
 PROP. XIII. 
 
 If to the circumference of a great circle, from a point in the surface of the sphere, 
 which is not the pole of that circle, arcs of great circles be drawn ; the greatest 
 >f these arcs is that which passes through the pole of the first-mentioned cir- 
 cle, and the supplement of it is the least ; and of the other arcs, that which is 
 nearer to the greatest is greater than that which is more remote. 
 
 Let ADB be the circumference of a great circle, of which the pole is H, 
 *nd let C be any other point ; through C and H let the semicircle ACB be 
 drawn meeting the circle ADB in A and B ; and let the arcs CD, CE, CF 
 also be described. From C draw CG perpendicular to AB, and then, be- 
 cause the circle AHCB which passes 
 through H, the pole of the circle ADB, 
 is at right angles to ADB, CG is per- 
 pendicular to the plane ADB. Join 
 GD, GE, GF, CA, CD, CE, CF, CB. 
 
 Because AB is the diameter of the 
 circle ADB, and G a point in it, which 
 is not the centre, (for the centre is ir 
 thft point where the perpendicular from 
 H meets A 15), therefore AG, the part 
 of the diameter in which the centre is
 
 2*8 
 
 SPHERICAL TRIGONOMETRY. 
 
 is the greatest (7. 3.), and GB the least of all the straight lines that can be 
 drawn from G to the circumference ; and GD, which is nearer to AB, is 
 greater than GE, which is more remote. But the triangles CGA, CGD 
 are right angled at G, and therefore AC 2 =AG 2 +GC 2 , and DC 2 =DG 2 + 
 GC 2 ; but AG 2 +GC 2 7DG 2 +GC 2 ; because AG7DG; therefore AC 2 
 7 DC 2 , and AC 7 DC. And because the chord AC is greater than the 
 chord DC, the arc AC is greater than the arc DC. In the same manner, 
 since GD is greater than GE, and GE than GF, it is shewn that CD is 
 greater than CE, and CE than CF. Wherefore also the arc CD is greater 
 than the arc CE, and the arc GE greater than the arc CF, and CF than 
 CB, that is, of all the arcs of greater circles drawn from C to the circum- 
 ference of the circle ADB, AC which passes through the pole H, is the 
 greatest, and CB its supplsment is the least ; and of the others, that which 
 is nearer to AC the greatest, is greater than that which is more remote. 
 
 PROP. XIV. 
 
 In a right angled spherical triangle, the sides containing the right angle are q> 
 the same affection with the angles opposite to them, that is, if the sides be 
 greater or less than quadrants, the opposite angles will be greater or less than 
 right angles, and conversely- 
 
 L>et ABC be a spherical triangle, right angled at A, any side AB will 
 be of the same affection with the opposite angle ACB. 
 
 Produce the arcs AC, AB, till they meet again in D, and bisect AD in 
 E. Then ACD, ABD are semicircles, and AE an arc of 90°. Also, be- 
 cause CAB is by hypothesis a right angle, the plane of the circle ABD is 
 perpendicular to the plane of the 
 circle ACD, 'so that the pole of. 
 ACD is in ABD, (1. Cor. 4.), 
 and is therefore the point E. Let 
 EC be an arc of a great circle 
 passing through E and C. 
 
 Then because E is the pole of 
 the circle ACD, EC is a (2.) 
 quadrant, and the plane of the 
 circle EC (4.) is at right angles 
 to the plane of the circle ACD, 
 that is, the spherical angle ACE 
 is a right angle ; and therefore, 
 when AB is less than AE, the 
 angle ACB, being less than 
 ACE, is less than a right angle. 
 But when AB is greater than 
 AE, the angle ACB is greater 
 than ACE, or than a right an- 
 gle. In the same way may the 
 converse be demonstrated.
 
 SPHERICAL TRIGONOMETRY 249 
 
 PROP. XV. 
 
 If the two sides of a right angled spherical triangle about the right angle be oj 
 the same affection, the hypotenuse will be less than a quadrant ; and if they be 
 of different affection, the hypotenuse will be greater than a quadrant. 
 
 Let ABC be a right angled spherical triangle ; according as the two 
 Bid n .s AB, AC are of the same or of different affection, the hypotenuse BC 
 * ill be less, or greater than a quadrant. 
 
 The construction of the last proposition remaining, bisect the semicircle 
 ACD in G, then AG will be an arc of 90°, and G will be the pole of the 
 circle ABD. 
 
 1. Let AB, AC be each less than 90°. Then, because C is a point on 
 the surface of the sphere, which i3 not the pole of the circle ABD, the arc 
 CGD, which passes through Gthe pole of ABD is greater than CE (13.), 
 and CE greater than CB. But CE is a quadrant, as was before shewn, 
 therefore CB is less than a quadrant. Thus also it is proved of the right 
 angled triangle CDB, (right angled at D), in which each of the sides CD, 
 DB is greater than a quadrant, that the hypotenuse BC is less than a 
 quadrant. 
 
 2. Let AC be less, and AB greater than 90°. Then because CB falls 
 between CGD and CE, it is greater (12.) than CE, that is, than a quad- 
 rant. 
 
 Cor. 1. Hence conversely, if the hypotenuse of a right angled triangle 
 be greater or less than a quadrant, the sides will be of different or the same 
 affection. 
 
 Cor. 2. Since (14.) the oblique angles of a right angled spherical trian- 
 gle have the same affection with the opposite sides, therefore, according as 
 the hypotenuse is greater or less than a quadrant, the oblique angles will 
 be different, or of the same affection. 
 
 Cor. 3. Because the sides are of the same affection with the opposite 
 angles, therefore when an angle and the side adjacent are of the same affec 
 ion, the hypotenuse is less than a quadrant : and conversely. 
 
 PROP. XVL 
 
 In any spherical triangle, if the perpendicular upon the base from the opposite 
 angle fall within the triangle, the angles at the base are of the same affection; 
 and if the perpendicular fall without the triangle, the angles at the base are of 
 different affection. 
 
 I et ABC be a spherical triangle, and let the arc CD be drawn from C 
 oerpendicular 'o the base AB. 
 
 \. Let CD fall within the triangle ; then, since ADC, BDC are right 
 angled spherical triangles, the angles A, B must each be of the same affec- 
 tion with CD (14.). 
 
 32
 
 250 
 
 SPHERICAL TRIGONOMETRY. 
 
 c 
 
 2 Let CD fall without the triangle ; then (14.) the angle B is of the 
 lame affection with CD ; and the angle CAD is of the same affection with 
 CD ; therefore the angle CAD and B are of the same affection, and the 
 angle CAB and B are therefore of different affections. 
 
 Cor. Hence, if the angles A and B be of the same affection, the per- 
 pendicular will fall within the base ; for if it did not, A and B would be of 
 different affection. And if the angles A and B be of different affection, 
 the perpendicular will fall without the triangle ; for, if it did not, the angles 
 A. and B would be of the same affection, contrary to the supposition. 
 
 PROP. XVII. 
 
 If to the base of a spherical triangle a perpendicular be drawn from the opposite 
 angle, tvhich either falls within the triangle, or is the nearest of the two that 
 fall without; the least of the segments of the base is adjacent to the least of 
 the sides of the triangle, or to the greatest, according as the sum of the sides 
 is less or greater than a semicircle. 
 
 Let ABEF be a great circle of a sphere, H its pole, and GHD any cir- 
 cle passing through H, which therefore is perpendicular to the circle 
 ABEF. Let A and B be two points in the circle ABEF, on opposite 
 sides of the point D, and let D be nearer 
 to A than to B, and let C be any point 
 in the circle GHD between H and D. 
 Through the points A and C, B and C, 
 let the arcs AC and BC be drawn, and 
 let them be produced till they meet the 
 circle ABEF in the points E and F, 
 then the arcs ACE, BCF are semicir- 
 cles. Also ACB, ACF, CFE, ECB, 
 are four spherical triangles continued 
 by arcs of the same circles, and having 
 the same perpendiculars CD and CG. 
 
 I . Now because CE is nearer to the arc CHG tr an CB is, OE is greater 
 than CA, and therefore CE and CA are greater than CB and CA, where 
 fore CB and CA are less than a semicircle ; but because AD is by sup 
 Dosition less than DB, AC is also less than CB (13.), and therefore in this 
 case, viz. when the perpendicular falls within the triangle, and whf n he
 
 SPHERICAL TRIGONOMETRY. 251 
 
 sum of the sides is less than a semicircle, the least segment isadjaceni to the 
 least side. 
 
 2. Again, in the triangle FCA the two sides FC and CA are less than 
 a semicircle ; for since AC is less than CB, AC and CF are less than BC 
 and CF. Also, AC is less than CF, because it is more remote from CHG 
 than CF is ; therefore in this case also, viz. when the perpendicular falls 
 without the triangle, and when the sum of the sides is less than a semicir- 
 cle, the least segment of the base AD is adjacent to the least side. 
 
 3. But in the triangle FCE the two sides FC and CE are greater than 
 a semicircle ; for, since FC is greater than CA, FC and CE are greater 
 than AC and CE. And because AC is less than CB, EC is greater than 
 CF, and EC is therefore nearer to the perpendicular CHG man CF is, 
 wherefore EG is the least segment of the base, and is adjacent to the 
 greater side. 
 
 4. In the triangle ECB the two sides EC, CB are greater than a semi 
 circle ; for, since by supposition CB is greater than CA, EC and CB are 
 greater than EC and CA. Also, EC is greater than CB, wherefore in 
 this case, also, the least segment of the base EG is adjacent to the greatest 
 side of the triangle. Therefore, when the sum of the sides is greater than 
 a semicircle, the least segment of the base is adjacent to the greatest side, 
 whether the perpendicular fall within or without the triangle : and it has 
 been shewn, that when the sum of the sides is less than a semicircle, the 
 least segment of the base is adjacent to the least of the sides, whether the 
 perpendicular fall within or without the triangle. 
 
 PROP. XVIII. 
 
 in right angled spherical triangles, the sine of either of the sides about the rxgltt 
 angle, is to the radius of the sphere, as the tangent of the remaining side is 
 to the tangent of tlie angle opposite to that side. 
 
 Let ABC be a triangle, having the right angle at A ; and let AB be 
 either of the sides, the sine of the side AB will be to the radius, as the tan- 
 gent of the other side AC to the tangent of the angle ABC, opposite to AC. 
 Let D be the centre of the sphere ; join AD, BD, CD, and let AF be drawn 
 perpendicular to BD, which therefore will be the sine of the arc AB, and 
 from the point F, let there be drawn in the plane BDC the straight line 
 FE at right angles to BD, meeting DC in 
 E, and let AE be joined. Since therefore 
 the straight line DB is at right angles to 
 both FA and FE, it will also be at right 
 angles to the plane AEF (4. 2. Sup.) ; 
 wherefore the plane ABD, which passes 
 through DF, is perpendicular to the plane 
 AEF (17. 2. Sup.), and the plane AEF 
 perpendicular to ABD : But the plane 
 ACD or AED, is also perpendicular to 
 the same ABD, because the spherical an- 
 gle BAC is a right angle . Therefore AE, 
 the common section of the planes AED,
 
 252 SPHERICAL TRIGONOMETRY 
 
 AEF, is at right angles to the plane ABD (18. 2. Sup.), and EAF, EAD 
 are right angles. Therefore AE is the tangent of the arc AC ; and in the 
 rectilineal triangle AEF, having a right angle at A, AF is to the radius as 
 AE to the tangent of the angle AFE (1. PI. Tr. ) ; but AF is the sine ol 
 the arc AB, and AE the tangent of the arc AC ; and the angle AFE is 
 the inclination of the planes CBD, ABD (4. def. 2. Sup.), or is equal to the 
 spherical angle ABC : Therefore the sine of the arc AB is to the radius as 
 the tangent of the arc AC to the tangent of the opposite angle ABC. 
 
 Cor. Since by this proposition, sin. AB : R : : tan. AC : tan. ABC ; 
 and because R : cot. ABC : : tan. ABC : R (1 Cor. def. 9. PI. Tr.) by 
 equality, sin. AB : cot. ABC : : tan. AC .: R. 
 
 PROP. XIX. 
 
 In right angled spherical triangles the sine of the hypotenuse is to the radius as 
 the sine of cither side is to the sine of the angle opposite to that side. 
 
 Let the triangle ABC be right angled at A, and let AC be either of the 
 sides ; the sine of the hypotenuse BC will be to the radius as the sine of 
 the arc AC is to the sine of the angle ABC. 
 
 Let D be the centre of the sphere, and let CE be drawn perpendicular 
 to DB, which will therefore be the sine of the hypotenuse BC ; and from 
 the point E let there be drawn in the 
 plane ABD the straight line EF per- 
 pendicular to DB, and let CF be joined ; 
 then CF will be at right angles to the 
 plane ABD, because as was shewn of 
 EA in the preceding proposition, it is 
 the common section of two planes DCF, 
 ECF, each perpendicular to the plane 
 ADB. Wherefore CFD, CFE are right 
 angles, and CF is the sine of the arc 
 AC ; and in the triangle CFE having 
 the right angle CFE, CE is to the radius, as CF to the sine of the angle 
 CEF (1. PI. Tr.). But, since CE, FE are at right angles to DEB, which 
 is the common section of the planes CBD, ABD, the angle CEF is equal 
 to the inclination of these planes (4. def. 2. Sup.), that is, to the spherical 
 angle ABC. Therefore the sine of the hypotenuse CB, is to the radius, as 
 the sine of the side AC to the sine of the opposite angle ABC 
 
 PROP. XX. 
 
 In right angled spherical triangles, the cosine of the hypotenuse is to the radius 
 as the cotangent of either of the angles is to the tangent of the remaining 
 angle. 
 
 Let ABC be a spherical triangle, having a right angle at A, the cosine 
 of the hypotenuse BC is to the radius as the cotangent of the angle ABC 
 10 the tangent of the angle ACB
 
 SPHERICAL TRIGONOMETRY. 253 
 
 Describe the circle DE, of which B is the pole, and let it meei AC in 
 f and the circle BC in E ; and since the circle BD pases through the 
 
 pole B, of the circle DF, DF must pass through the pole of BD (4.). An 
 since AC is perpendicular to BD, the plane of the circle AC is perpend) 
 cular to the 'plane of the circle BAD, and therefore AC must also (4.) pass 
 through the pole of BAD ; wherefore, the pole of the circle BAD is in the 
 point F, where the circles AC, DE, intersect. The arcs FA, FD are 
 therefore quadrants, and likewise the arcs BD, BE. Therefore, in the tri- 
 angle CEF, right angled at the point E, CE is the complement of BC, the 
 hypotenuse of the triangle ABC ; EF is the complement of the arc ED, 
 the measure of the angle ABC, and FC, the hypotenuse of the triangle 
 CEF, is the complement of AC, and the arc AD, which is the measure ol 
 the angle CFE, is the complement of AB. 
 
 But (18.) in the triangle CEF, sin. CE : R : : tan. EF : tan. ECF, that 
 is, in the triangle ACB, cos. BC : R : : cot. ABC : tan. ACB. 
 
 Cor. Because cos. BC : R : : cot. ABC : tan. ACB, and (Cor. 1. def. 9. 
 PI. Tr.) cot. ABC : R : : R : tan. ABC, ex aequo, cot. ACB : cos. BC : : R 
 : cot. ABC. 
 
 PROP. XXI. 
 
 In right angled spherical triangles, the cosine of an angle is to the radius as the 
 tangent of the side adjacent to that angle is to the tangent of the hypotenuse 
 
 The same construction remaining ; In the triangle CEF, sin. FE : R : . 
 tan. CE : tan. CFE (18.): but sin. EF=cos. ABC ; tan. CE=cot. BC,and 
 tan. CFE=cot. AB, therefore cos. ABC : R : : cot. BC : cot. AB. Now 
 because (Cor. 1. def. 9. PI. Tr.) cot. BC : R : : R : tan. BC, and cot. AB : 
 R : : R : tan. AB, by equality inversely, cot. BC : cot. AB : tan. AB : 
 BC ; therefore (11. 5.) cos. ABC : R : : tan. AB : tan. BC. 
 
 Cor. '. From the demonstration it is manifest, that the tangents of any 
 two arcs \B. BC are reciprocally proportional to their cotangents.
 
 *54 SPHERICAL TRIGONOMETRY 
 
 Cor. 2 Because cos. ABC : R : : tan. AB : tan. BC, and R : cos. BC : : 
 tnn. BC : R, by equality, cos. ABC : cot. BC : : tan. AB : R. That is, the 
 cosine of any of the oblique angles is to the cotangent of the hypotenuse, 
 as the tangent of the side adjacent to the angle is to the radius.- 
 
 PROP. XXII. 
 
 In right angled spherical triangles, the cosine of either of the sides is to the ra- 
 dius, as the cosine of the hypotenuse is to the cosine of the other side. 
 
 The same construction remaining : In the triangle CEF, sin. CF : R : : 
 gin. CE : sin. CFE (19.) ; but sin. CF=cos. CA, sin. CE=cos. BC, and 
 sin. CFE=cos. AB ; therefore cos. CA : R : : cos. BC : cos. AB. 
 
 PROP. XXIII. 
 
 In right angled spherical triangles, the cosine of either of the sides is to the ra- 
 dius, as the cosine of the angle opposite to that side is to the sine of the other 
 angle. 
 
 The same construction remaining : In the triangle CEF, sin. CF : R : : 
 sin. EF : sin. ECF (19.); but sin. CF = cos. CA,sin. EF=cos. ABC, and 
 sin. ECF=sin. BCA : therefore, cos. CA : R : : cos. ABC : sin. BCA. 
 
 PROP. XXIV. 
 
 In spherical triangles, whether right angled or oblique angled, the sines of the 
 sides are proportional to the sines of the angles opposite to them. 
 
 First, let ABC be a right angled triangle, having a right angle at A , 
 herefore (19.; tne sine of the hypotenuse BC is to the radius, (or the sjmp
 
 SPHERICAL TRIGONOMETRY. 
 
 255 
 
 of the right angle at A), as the sine of 
 the side AC to the sine of the angle B. 
 And, in like manner, the sine of BC is 
 to the sine of the angle A, as the sine 
 of AB to the sine of the angle C ; 
 wherefore (11. 5.) the sine of the side 
 AC is to the sine of the angle B, as the 
 sine of A B to the sine of the angle C. 
 
 Secondly, Let ABC be an oblique angled triangle, the sine of any of the 
 sides BC will be to the sine of any of the other two AC, as the sine of the 
 angle A opposite to BC, is to the sine of the angle B opposite to AC. 
 Through the point C, let there be drawn an arc of a great circle CD per- 
 pendicular to AB ; and in the right angled triangle BCD, sin. BC : R : . 
 
 sin. CD : sin. B (19.) ; and in the triangle ADC, sin. AC : R : : sin. CD : 
 Rin. A ; wherefore, by equality inversely, sin. BC : sin. AC : : sin. A : sin. 
 B. In the same manner, it maybe proved that sin. BC : sin. AB : : sin. 
 A : sin. C, &c. 
 
 PROP. XX\. 
 
 In oblique angled spherical triangles, a perpendicular arc being drawn Jrom 
 any of the angles upon the opposite side, the cosines of the angles at the base 
 are proportional to the sines of the segments of the vertical angle. 
 
 Let ABC be a triangle, and the arc CD perpendicular to the base BA , 
 the cosine of the angle B will be to the cosine of the angle A, as the sine 
 of the angle BCD to the sine of the angle ACD. 
 
 For having drawn CD perpendicular to AB, in the right angled triangle 
 BCD (23.), cos. CD : R : : cos. B : sin. DCB ; and in the right angled 
 triangle ACD, cos. CD : R : : cos. A : sin. ACD ; therefore (11. 5.) cos. 
 B : sin. DCB : : cos. A : sin. ACD, and alternately, cos. B ■ cos. A : : sin. 
 BCD : sin. ACD. 
 
 PROP. XXVI. 
 
 The same things remaining, the cosines of the sides BC, CA, are proportwna 
 to the cosines q/"BD, DA, the segments of the base. 
 
 For in the triangle BCD (22.) cos. BC : cos. BD : : cos. DC : R, and in
 
 256 SPHERICAL TRIGONOMETRY. 
 
 the triangle ACD, cos. AC : cos. AD : : cos. DC : R ; therefore (11. 5.) 
 cos. BC : cos. BD : : cos. AC : cos. AD, and alternately, cos. BC : cos 
 AC : : cos. BD : cos. AD. 
 
 i 
 
 PROP. XXVII. 
 
 The same construction remaining, the sines of BD, DA, the segments of the 
 base are reciprocally proportional to the tangents ofB and A, the angles 
 at the base. 
 
 In the triangle BCD (18.), sin. BD : R : : tan. DC : tan. B ; and in the 
 triangle ACD, sin. AD : R : : tan. DC : tan. A ; therefore, by equality in- 
 versely sin. BD : sin. AD : : tan. A : tan. B. 
 
 PROP. XXVIII. 
 
 The same construction remaining, the cosines of the segments of the vertical 
 angle are reciprocally proportional to the tangents of the sides. 
 
 Because (21.), cos. BCD : R : : tan. CD : tan. BC, and also cos. ACD 
 R : : tan. CD : tan. AC, by equality inversely, cos. BCD : cos ACD • • 
 tan. AC : tan. BC. 
 
 PROP. XXIX. 
 
 If from an angle of a spherical triangle there be drawn a perpendicular to the 
 opposite side, or base, the rectangle contained by the tangents of half the 
 sum, and of half the difference of the segments of the base is equal to the 
 rectangles contained by the tangents of half the sum, and of half the diffe- 
 rence of the two sides of the triangle. 
 
 Let ABC be a spherical triangle, and let the arc CD be drawn from the 
 angle C at right angles to the base AB, tan. ^ (m-\-n) Xtan. £ (m — n)=% 
 tan. (a-\-b)X^ tan. (a — b). 
 
 Let BC=a, AC=b ; BD=m, AD=n. Because (26.) cos. a : cos. b : . 
 cos. m : cos. n(E. 5.), cos. a-\-b : cos. a — cos. b : : cos. tti-J-cos. n ; cos. m — 
 cos. n. But (1. Cor. 3. PI. Trig.), cos. a-|- cos - ^ : cos - ° — cos - ° '• '• cot - \ 
 (a-\-b) : tan. % (a — b), and also, cos. wi+cos. n : cos. m — cos. n : : cot. \ 
 (m-\-n) : tan. ^ (m—n). Therefore, (11. 5.) cot. ^ (a-\-b) : tan. \ (a — b) 
 : . cot. ^ (m-\-n) : tan. ^ (m — n). And because rectangles of the same al-
 
 SPHERICAL TRIGONOMETRY. 
 
 257 
 
 titude are as their bases, tan. £ (a+6)X cot. £(a+i) : tan. \ (a+5)Xtan 
 I (a— b) :: tan. £ (m+n)x cot. \ (m-\-n) : tan. ^ (mX«)+tan. £(m — n) 
 Now the first and third terms of this proportion are equal, being each equa 
 to the square of the radius (1. Cor. PL Trig.), therefore the remaining twc 
 are equal (9. 5.), or tan. £(m+n)x tan. ^ (»» — n)=tan. % (a+6)xtan. 1 
 (a—b) ; that is, tan. \ (BD+AD)xtan. I (BD-AD)=tan. $ (BC+AC) 
 Xtan. ^(BC-AC). 
 
 Cor. 1. Because the sides of equal rectangles are reciprocally propor- 
 tional, tan. £ (BD-f AD) : tan. J (BC+AC) : : tan. \ (BC — AC) : tan. } 
 fBD— AD). 
 
 Cor. 2. Since, when the perpendicular CD falls within the triangle, 
 BD+AD=AB, the base ; and when CD falls without the triangle BD — 
 AD=AB, therefore, in the first case, the proportion in the last corollary 
 becomes tan. ^(AB) : tan.} (BC+AC) •- tan.$(BC-AC) : tan. $ (BO- 
 AD) ; and in the second case, it becomes by inversion and alternation, tan. 
 \ (AB) : tan. $ (BC+AC) : : tan. \ (BC-AC) : tan. \ (BD+AD). 
 
 SCHOLIUM. 
 
 The preceding proposition, which is very useful in spherical trigonome 
 try, may be easily remembered from its analogy to the proposition in plane 
 trigonometry, that the rectangle under half the sum, and half the difference 
 of the sides of a plane triangle, is equal to the rectangle under half the 
 sum, and half the difference of the segments of the base. See (K. 6.), also 
 4th Case PL Tr. We are indebted to Napier for this and the two follow- 
 ing theorems, which are so well adapted to calculation by Logarithm*, 
 that they must be considered as three of the most valuable proposition* in 
 Trigonometry. 
 
 33
 
 258 
 
 SPHERICAL TRIGONOMETRY. 
 
 PROP. XXX. 
 
 If a perpendicular be drawn from an angle of a spherical triangle to the oppo- 
 site side or base, the sine of the sum of the angles at the base is to the sine 
 of their difference as the tangent of half the base to the tangent of half the 
 difference of its segments, when the perpendicular falls within; but as the 
 co-tangent of half the base to the co-tangent of half the sum of the segments, 
 when the perpendicular falls without the triangle : And the sine of the sum 
 of the two sides is to the sine of their difference as the co-tangent of half 
 the angle contained by the sides, to the tangent of half the difference of 
 the angles which the perpendicular makes with the same sides when it falls 
 within, or to the tangent of half the sum of these angles, when it falls with- 
 out the triangle. 
 
 If ABC be a spherical triangle, and AD a perpendicular to the base BC, 
 sin. (C+B) : sin.(C-B) : : tan. ^BC : tan. £ (BD -DC), when AD falls 
 within the triangle; but sin. (C+B) : sin. (C— B) : : cot. £ BC : cot. \ 
 (BD+DC), when AD falls without. And again, 
 
 A 
 
 B 
 
 T> 
 
 sin. (AB+AC) : sin. (AB— AC) : : cot. i BAC : tan. \ (BAD— CAD), 
 
 when AD falls within ; but when AD falls without the triangle, 
 sin. (AB + AC) : sin. (AB— AC) : : cot. £ BAC : tan. £(BAD+CAD). 
 For in the triangle BAC (27.), tan. B : tan. C : : sin. CD : sin. BD,and 
 therefore (E. 5.), tan. C+tan. B : tan. C— tan. B : : sin. BD+sin. CD : 
 sin. BD— sin. CD. Now (by the annexed Lemma), tan. C + tan. B : tan. 
 C— tan. B : : sin. (C+B) : sin. (C— B), and sin. BD+sin. CD : sin. BD 
 -sin. CD : : tan. £ (BD+CD) : tan. \ (BD— CD), (3. PI. Trig.), there- 
 fore because ratios which are equal to the same ratio are equal to one 
 another (11. 5.), sin. (C+B) : sin. (C-B) : : tan. \ (BD+CD) : tan \ 
 (BD— CD).
 
 SPHERICAL TRIGONOMETRY. 259 
 
 Now when AD is within the triangle, BD+CD=BC, and therefore sin 
 (C+B) : sin. (C-B) : : tan. £ BC : tan. \ (BD— CD). And again, when 
 AD is without the triangle, BD— CD=BC, and therefore sin. (C-f-B) : sin 
 (C — B) : : tan. £ (BD + CD) : tan. \ BC, or because the tangents of ain 
 two arcs are reciprocally as their co-tangents, in (C + B) : sin. (C— B) : : 
 cot. \ BC : cot. £ (BD + CD). 
 
 The second part of the proposition is next to be demonstrated. Because 
 (28.) tan. AB : tan. AC : : cos. CAD : cos. BAD, tan. AB + tan. AC : tan. 
 AH— tan AC :: cos. CAD+cos. BAD : cos. CAD— cos. BAD. But 
 (Lemma) tan. AB + tan. AC : tan. AB— tan. AC : : sin. (AB + AC) : sin. 
 (AB — AC), and (1. cor. 3. PI. Trig.) cos. CAD+cos. BAD : cos. CAD — 
 cos. BAD : : cot. \ (BAD + CAD) : tan. \ (BAD — CAD). Therefore (11. 
 5.) sin. (AB + AC) : sin. (AB — AC) :: cot. £ (BAD + CAD) : tan. \ (BAD 
 —CAD). Now, when AD is within the triangle, BAD+CAD=BAC, 
 and therefore sin. (AB + AC) : sin. (AB-AC) : : cot. \ BAG : tan. \ (BAD 
 -CAD.) 
 
 But if AD be without the triangle, BAD— CAD=BAC, and therefore 
 sin. (AB + AC) : sin. (AB-AC) : : 
 cot. i (BAD + CAD) : tan. \ BAC ; or because 
 cot. | (BAD + CAD) : tan. \ BAC : : cot. \ BAC : 
 tan. I (BAD + CAD), sin. (AB + AC) : sin. (AB— AC) : : cot. \ BAC : 
 <an. J (BAD + CAD). 
 
 LEMMA. 
 
 The sum of the tangents of any two arcs, is to the difference of their tangents, 
 as the sine of the sum of the arcs, to the sine of their difference. 
 
 Let A and B be two arcs, tan. A+tan. B : tan. A— tan. B : : sin. (A+B) 
 : (A-B). 
 
 For, by § 6. page 232, sin. A X cos. B + cos. A x sin. B =sin. ( A + B), and 
 
 i. r r -r ii u a r> sin - A i sm - B - sin - ( A + B ) .v . 
 therefore dividing all by cos. A cos. B, -i — = - i=:,that 
 
 cos. A cos. B cos. A X cos. B 
 
 . sin. A _ sin. (A + B) T . 
 
 is, because r =tan. A, tan. A+tan. B= ^ -=;. In the same 
 
 cos. A cos. A X cos. B 
 
 manner it is proved that tan. A —tan. B = -^ —r. Therefore tan. A 
 
 cos. A X cos. B 
 
 +tan. B : tan. A— tan. B : : sin. (A + B) : sin. (A— B). 
 
 PROP. XXXI. 
 
 The sine of half the sum of any two angles of a spherical triangle is to i/u 
 sine of half their difference, as the tangent of half the side adjacent to these 
 angles is to the tangent of half the difference of the sides opposite to them ; 
 and the cosine of half the sum of the same angles is to the cosine of half 
 their difference, as the tangent of half the side adjacent to them, to the tan- 
 gent of half the sum of the sides opposite. 
 
 Let C+B=2S, C— B=2D, the base BC=2B, and the difference oi
 
 SPHERICAL TRIGONOMETRY. 
 
 the segments of the base, or BD— CD=2X. Then, because (30.) sin 
 (C+ 6) : sin. (C— B) : : tan. £ BC : tan. £ (BD— CD), sin. 2S : sin. 2D 
 : : tan. B : tan. X. Now, sin. 2S=sin. (S-j S)==2 sin. Sx cos. S, (Sect 
 III. cor. PI. Tr.). In the same manner, nin. 2D=2 sin. Dxcos. D 
 Theiefore sin. Sxcos. S : sin. Dxcos. D : : tan. B : tan. X 
 
 Again, in the spherical triangle ABC it has been proved, that sin. C+ 
 sin. B : sin. C — sin. B : : sin. AB+sin. AC : sin. AB — sin. AC, and since 
 sin. C+sin. B=2 sin. A (C+B)+cos. A (C— B), (Sect. III. 7. PI. Tr.)= 
 2 sin. S X cos. D ; and sin. C— sin. B=2 cos. 1 (C + B) X sin. A (C— B)=r 
 2 cos. S X sin. D. Therefore 2 sin. S X cos. D : 2 cos. S X sin. D : : sin. 
 AB+sin. AC : sin. AB— sin. AC. But (3. PI. Tr.) sin. AB+sin. AC : 
 sin. AB— sin. AC : : tan. A (AB + AC) : tan. A (AB— AC) : : tan. 2 : tan. 
 d, 2 being equal to A (AB+AC) and d to A (AB— AC). Therefore sin. 
 
 S X cos. D : cos. S X sin. D : : tan. 2 : tan. d. Since then — —rz = 
 
 tan. B 
 
 sin. Dxcos. D . tan. 4 cos. Sxsin. D . ... . , , 
 
 sin. Sxcos. S ; and t -a^ = sin.Sxcos.D ' ** "^P 1 ^ e 1 uals »* 
 
 . tan. X tan. ^_(sin. D) 2 Xcos. Sxcos. D_(sin. D) 2 
 
 6qUa S ' tan. B X tan. 2~ (sin. S) 2 xcos. Sxcos. D — (sin. S) 2 ' 
 
 But m \ tan.^(BD-DC) _ tan.^(AB+AC) tan. X_tan. 2 
 
 " U K ' tan. A(AB-AC) - tan. A BC ' that 1S ' tanT^-tan^B' 
 
 , . . tan. X tan. 2x tan. 4 tan. X tan. 4 tan. J 2 
 
 and therefore, == - =—. — , as also — = -=-. =rrr. 
 
 tan. B (tan. B) 2 tan. B tan. 2 (tan. B) 2 
 
 „ tan. X tan. J (sin. D) 2 , (tan. J) 2 (sin. D) 2 , tan. J 
 
 But -X ^=7-^ ^ 5 whence^ 7^=7-. 7^ 5 an d ™ 
 
 tan. B tan. 2 (sin. S) 2 (tan. B) 2 (sin. S) 2 tan. B 
 
 rr~ . ' or sin. S : sin. D : : tan. B : tan. J, that is, sin. (C+B) : sin. 
 sin. o 
 
 C— B) : : tan. \ BC : tan. A (AB— AC) ; which is the first part of the 
 
 . tan. 4 cos. S X sin. D . . tan. 2 
 
 proposition. Again, since =— — ~ =r, or inversely - = 
 
 r r 5 ' tan. 2 sin. S X cos. D J tan. -4 
 
 sin. Sxcos. D , . tan. X sin. Dxcos. D . . .... 
 
 r; : — =r ; and since ?r=- — f; « > therefore by multiph- 
 
 cos. Sxsin. D tan. B sin. Dxcos. S ' ■ 
 
 tan. X tan. 2 (cos. D) 2 
 
 cation, 5 x -—) ^ 
 
 tan. B tan 4 (cos. S) 3
 
 SPHERICAL TRIGONOMETRY. 201 
 
 _ , , tan. X tan. 2 x tan. J , . . 
 
 But it was already shewn that =t=j — ; =7^ — , wherefore also 
 
 3 tan. B (tan. B) 2 
 
 tan. X tan. ^_(tan. 2f 
 tanTB X tan. ^~(tan. B) 2 ' 
 
 tan. X tan. 2 (cos. D) 2 , . . , 
 
 Now, S X r=7 =ac, as has just been shewn. 
 
 tan. B tan. d (cos. S)^ 
 
 __ , (cos D) 2 (tan. 2V . , cos. D tan. 2 
 
 Therefore ) ^0=7: t7T2. and consequently -=- — , or cos. 
 
 (cos. S) 2 (tan. B) 2 cos. S tan. B 
 
 S : cos. D : : tan. B : tan. 2, that is, cos. (C + B) : cos. (C — B) : : tan. $ 
 
 BC : tan. J (C+B) ; which is the second part of the proposition. 
 
 Cor. 1. By applying this proposition to the triangle supplemental to 
 ABC (11.) and by considering, that the sine of half the sum or half the 
 diflerence of the supplements of two arcs, is the same with the sine of half 
 the sum or half the difference of the arcs themselves : and that the same 
 is true of the cosines, and of the tangents of half the sum or half the dif- 
 ference of the supplements of two arcs : but that the tangent of half the 
 supplement of an arc is the same with the cotangent of half the arc itself: 
 it will follow, that the sine of half the sum of any two sides of a spherical 
 triangle, is to the sine of half their difference as the cotangent of half the 
 angle contained between them, to the tangent of half the difference of the 
 angles opposite to them : and also that the cosine of half the sum of these 
 sides, is to the cosine of half their difference, as the cotangent of half the 
 angle contained between them, to the tangent of half the sum of the angles 
 opposite to them. 
 
 Cor. 2. If therefore A, B, C, be the three angles of a spherical trian- 
 gle, a, b, c the sides opposite to them, 
 
 I. sin. I (A + B) : sin. I (A— B) : : tan. I c : tan. I (a— b). 
 II. cos. if (A -j- B) : cos. \ (A— B) : : tan.|c : tan. \ (a+4). 
 
 III. sin. I {a+b) : sin. I (a—b) : : tan. I C : tan. I (A— B). 
 
 IV. cos. I (a+b) : cos. \ {a—b) : : tan. \ C : tan. \ (A+B).
 
 V62 
 
 SPHERICAL TRIGONOMETRY. 
 
 PROBLEM I. 
 
 In a right angled spherical triangle, of the three sides and three angles, any 
 two being given, besides the right angle, to find the other three. 
 
 This problem has sixteen cases, the solutions of which are contained 
 in the following table, where ABC is any spherical triangle right angled 
 at A. 
 
 GIVEN. 
 
 SOUGHT. 
 
 SOLUTION. 
 
 
 BC and B. 
 
 AC. 
 
 AB. 
 
 C. 
 
 R : sin BC : : sin B : sin AC, (19). 
 R : cos B : : tan BC : tan AB, (21). 
 R : cos BC : : tan B : cot C, (20). 
 
 1 
 2 
 3 
 
 AC and C. 
 
 AB. 
 
 BC. 
 
 B. 
 
 R : sin AC : : tan C : tan AB, (18). 
 cos C : R : : tan AC : tan BC, (21). 
 R : cos AC : : sin C : cos B, (23). 
 
 4 
 5 
 6 
 
 AC and B. 
 
 AB. 
 
 BC. 
 
 C. 
 
 tan B : tan AC : : R : sin AB, (18). 
 sin B : sin AC : : R : sin BC, (19). 
 cos AC : cos B : : R : sin C, (23). 
 
 7 
 8 
 9 
 
 AC and BC. 
 
 AB. 
 B. 
 C. 
 
 cos AC : cos BC : : R : cos AB, (22). 
 sin BC : sin AC: : R: sin B, (19). 
 tan BC : tan AC : : R : cos C, (21). 
 
 10 
 11 
 12 
 
 AB and AC. 
 
 BC. 
 B. 
 C. 
 
 R : cos AB : : cos AC : cos BC, (22). 
 sin AB : R:: tan AC : tan B, (18). 
 sin AC : R : : tan AB : tan C, (18). 
 
 13 
 14 
 14 
 
 B and C. 
 
 - 
 
 AB. 
 AC. 
 BC. 
 
 sin B : cos C : : R : cos AB, (23). 
 sin C : cos B : : R : cos AC, (23). 
 tan B : cot C : : R : cos BC, (20). 
 
 15 
 15 
 16
 
 SPHERICAL TRIGONOMETRY. 
 
 263 
 
 T vBLE for determining the affections of the Sides and Angles found by 
 the preceding rules. 
 
 AC and B of the same affection. 
 
 1 
 
 If BC/ 90°, AB and B of the same affection, otherwise dif- 
 
 
 ferent, (Cor. 15.) 
 
 2 
 
 If BC^/90 , C and B of the same affection, otherwise diffe- 
 
 
 rent, (15.) 
 
 3 
 
 J 
 
 AB and C are of the same affection, (14.) 
 
 4 
 
 If AC and C are of the same affection, BC / 90° , otherwise 
 
 
 BC/90°, (Cor. 15.) 
 
 5 ! 
 
 B and AC are of the same affection, (14.) 
 
 6 
 
 Ambiguous. 
 
 7 
 
 Ambiguous. 
 
 8 
 
 Ambiguous. 
 
 9 
 
 When BC /_ 90°, AB and AC of the same ; otherwise of dif- 
 
 
 ferent affection, (15-) 
 
 10 
 
 AC and B of the same affection, (14.) 
 
 11 
 
 When BC/90°, AC and C of the same ; otherwise of dif- 
 
 
 ferent affection, (Cor. 15.) 
 
 12 
 
 BC/90°, when AB and AC are of the same affection, 
 
 
 (1. Cor. 15.) 
 
 13 
 
 B and AC of the same affection, (14.) 
 
 14 
 
 C and AB of the same affection, (14.) 
 
 14 
 
 AB and C of the same affection, (14.) 
 
 15 
 
 AC and B of the same affection, (14.) 
 
 15 
 
 When B and C are of the same affection, BC/90°, other- 
 
 
 wise, BC 790°, (15.) 
 
 16 
 
 The cases marked ambiguous are those in which the thing sought has 
 two values, and may either be equal to a certain angle, or to the supple- 
 ment of that angle. Of these there are three, in all of which the things 
 given are a side, and the angle opposite to it ; and accordingly, it is easy to 
 shew that two right angled spherical triangles may always be found that 
 have a side and the angle opposite to it the same in both, but of which the 
 remaining sides, and the remaining angle of the one, are the supplements 
 of the remaining sides and the remaining angle of the other, each of each. 
 
 Though the affection of the arc or angle found may in all the other cases 
 be determined by the rules in the second of the preceding tables, it is of 
 use to remark, that all these rules except two, may be reduced to one, vix. 
 that when the thing found by the rules in the first table is either a tangent or 
 a cosine ; and when, of the tangents or cosines employed in '.he computation m 
 it, one only belongs to an obtuse angle, the angle required is also obtuse
 
 264 SPHERICAL TRIGONOMETRY. 
 
 Thus, in the 15th case, when cos AB is found, if C be an obtuse angle, 
 because of cos C, AB must be obtuse ; and in case 16, if either B or C be 
 obtuse, BC is greater than 90°, but if B and C are either both acute, or 
 Woth obtuse, BC is less than 90°. 
 
 It is evident, that this rule does not apply when that which is found is 
 the sine of an arc ; and this, besides the three ambiguous cases, happens 
 also in other two, viz. the 1st and 11th. The ambiguity is obviated, in 
 these two cases, by this rule, that the sides of a spherical right angled tri 
 angle are of the same affection with the opposite angles. 
 
 Two rules are therefore sufficient to remove the ambiguity in all the 
 eases of *he right angled triangle in which it can possibly be removed.
 
 SPHERICAL TRIGONOMETRY. 
 
 265 
 
 It may be useful to express the same solutions as in the annexed table. 
 Let A be at the right angle as in the figure, and let the side opposite to it 
 be a; let b be the side opposite to B, and c the side opposite to C 
 
 OIVKN. 1 SOUGHT. 
 
 SOLUTION. 
 
 
 a and B. 
 
 b. 
 c. 
 C. 
 
 sin b = sin a x sin B. 
 tan c — tan a X cos B 
 cot C = cos a X tan B. 
 
 1 
 2 
 3 
 
 4 
 5 
 6 
 
 b and C. 
 
 c. 
 a. 
 B. 
 
 tan c = sin b X tan C. 
 
 tan b 
 
 tan a = *• 
 
 cos u 
 
 cos B = cos b X sin C. 
 
 b and B. 
 
 c. 
 a. 
 C. 
 
 tan b 
 
 sin c = =. 
 
 tanB 
 
 sin b 
 
 sin a = -: — =r. 
 
 sin B 
 
 . _, cosB 
 
 sinu= r. 
 
 cos b 
 
 7 
 8 
 9 
 
 a and b. 
 
 c. 
 
 B. 
 
 C. 
 
 cos a 
 
 10 
 11 
 12 
 
 cos b 
 
 sin b 
 
 sm B = —. . 
 
 sin a 
 
 _ tan b 
 
 cos C = . 
 
 tan a 
 
 b and c. 
 
 a. 
 B. 
 
 C. 
 
 cos a = cos 6 X cos c. 
 
 tan B = -^ . 
 
 sm c 
 
 ,_ tan c 
 
 tan C = -: — r. 
 
 sm b 
 
 13 
 14 
 
 14 
 
 B and C. 
 
 c. 
 b 
 
 m. 
 
 cosC 
 
 cos c = —. — =,. 
 
 sin B 
 
 , cos B 
 
 COS b = -. — t=. 
 
 sm C 
 
 cotC 
 
 cos a = - — =. 
 
 tanB 
 
 15 
 15 
 16 
 
 34
 
 266 
 
 SPHERICAL TRIGONOMETRY. 
 
 PROBLEM II. 
 
 In any oblique angled spherical triangle, of the three sides and three angles^ 
 any three being given, it is required to find the other three. 
 
 In this Table the references (c. 4.), (c. 5.), &c. are to the cases In the 
 preceding Table, (16.), (27.), &c. to the propositions in Spherical Trigo- 
 nometry. 
 
 
 GIVEN. 
 
 SOUGHT. 
 
 SOLUTION. 
 
 1 
 
 2 
 
 _ 
 
 Two sides 
 AB, AC, 
 and the in- 
 cluded angle 
 A. 
 
 One of the 
 
 other angles 
 
 B. 
 
 Let fall the perpendicular CD from 
 the unknown angle, not requir- 
 ed, on AB. 
 
 R : cos A : : tan AC : tan AD, 
 (c. 2.) ; therefore BD is known, 
 and sin BD : sin AD : : tan A : 
 tan B, (27.) ; B and A are of 
 the same or different affection, 
 according as AB is greater or 
 less than BD, (16.). 
 
 The third 
 side 
 BC. 
 
 Let fall the perpendicular CD from 
 one of the unknown angles on 
 the side AB. 
 
 R : cos A : : tan AC : tan AD, 
 (c. 2) ; therefore BD is known, 
 and cos AD : cos BD : : cos AC 
 : cos BC, (26.) ; according as 
 the segments AD and DB are of 
 the same or different affection, 
 AC and CB will be of the same 
 or different affection.
 
 SPERICAL TRIGONOMETRY. 
 
 267 
 
 TABLE continued. 
 
 OITKN. 
 
 SOUGHT. 
 
 SOLUTION. 
 
 3 
 
 Two angles, 
 A and ACB, 
 
 and 
 
 AC, 
 the side be- 
 tween them. 
 
 4 
 
 The side 
 BC. 
 
 From C the extremity of AC near 
 the side sought, let fall the per- 
 pendicular CD on AB. 
 
 R : cos AC : : tan A : cot ACD, 
 (c. 3.) ; therefore BCD is known, 
 and cos BCD : cos ACD : : tan 
 AC : tan BC, (28.). BC is less 
 or greater than 99°, according 
 as the angles A and BCD are 
 of the same, or different affec- 
 tion. 
 
 The third 
 
 angle 
 
 B. 
 
 Let fall the perpendicular CD from 
 one of the given angles on the 
 opposite side AB. 
 
 R : cos AC : : tan A : cot ACD, 
 (c. 3.) ; therefore the angle BCD 
 is given, and sin ACD : sin BCD 
 : : cos A : cos B, (25.) ; B and 
 A are of the same or differ- 
 ent affection, according as CD 
 falls within or without the tri- 
 angle, that is, according as ACB 
 is greater or less than BCD, 
 (16.).
 
 268 
 
 SPHERICAL TRIGONOMETRY. 
 
 TABLE continued. 
 
 GIVKN. 
 
 SOUGHT. 
 
 _ _ — 
 
 SOLUTION. 
 
 5 
 
 Two sides 
 
 AC and BC, 
 
 and an angle 
 
 A 
 
 opposite to 
 6 
 one of them, 
 
 BC. 
 
 7 
 
 The angle 
 
 B 
 opposite to 
 the other gi- 
 ven side 
 AC. 
 
 Sin BC : sin AC : : sin A : sin B, 
 (24.) The affection of B is am- 
 biguous, unless it can be deter- 
 mined by this rule, that accord- 
 ing as AC + BC is greater or 
 less than 180°, A+B is greater 
 or less than 180°, (10.) 
 
 The angle 
 
 ACB 
 
 contained by 
 
 the given 
 
 sides 
 
 AC and BC. 
 
 From ACB the angle sought draw 
 CD perpendicular to AB ; then 
 R : cos AC : : tan A : cot ACD, 
 (c 3.) ; and tan BC : tan AC : : 
 cos ACD : cos BCD, (28.) ACD 
 ± BCD = ACB, and ACB is 
 ambiguous, because of the am- 
 biguous sign + or — . 
 
 The third 
 side 
 AB. 
 
 Let fall the perpendicular CD from 
 the angle C, contained by the 
 given sides, upon the side AB. 
 R : cos A : : tan AC : tan AD, 
 (c. 2.) ; cos AC : cos BC : : cos 
 AD : cos BD, (26.) 
 AB=AD±BD, wherefore AB 
 is ambiguous.
 
 SPHERICAL TRIGONOMETRY. 
 
 269 
 
 TABLE continued. 
 
 
 • IVKN. 
 
 SOUGHT. 
 
 80LDTI0N. 
 
 
 The side 
 
 Sin B : sin A : : sin AC : sin BC, 
 
 
 
 BC 
 
 (24) ; the affection of BC is un- 
 
 
 
 opposito 
 
 certain, except when it can be de- 
 
 8 
 
 
 to the 
 
 termined by this rule, that accord- 
 
 
 
 other 
 
 ing as A+B is greater or less than 
 
 
 
 given an- 
 
 180°, AC+BC is also greater or 
 
 
 
 gle A. 
 
 less than 180°, (10.). 
 
 
 Two angles 
 
 
 
 
 
 From the unknown angle C, draw 
 
 
 A,B, 
 
 The side 
 
 CD perpendicular to AB ; then 
 
 
 
 AB 
 
 R : cos A : : tan AC : tan AD, 
 
 
 and a side 
 
 adjacent 
 
 (c. 2.) ; tan B : tan A : : sin AD : 
 
 
 
 to the 
 
 sin BD. BD is ambiguous ; and| 
 
 9 
 
 AC 
 
 given 
 
 therefore AB = AD ± BD may- 
 
 
 
 angles 
 
 have four values, some of which 
 
 
 opposite to 
 
 A, B. 
 
 will be excluded by this condition, 
 that AB must be less than 180°. 
 
 
 one of them, 
 
 
 
 
 
 From the angle required, C, draw CD 
 
 
 B. 
 
 
 perpendicular to AB. 
 
 
 
 The third 
 
 R : cos AC : : tan A : cot ACD, 
 (c. «?.), cos A : cos B : ; sin ACD :j 
 
 
 
 angle 
 
 sin BCD, (25.). The affection of 
 
 10 
 
 
 
 BCD is uncertain, and therefore 
 
 
 
 ACB. 
 
 ACB = ACD ± BCD, has four 
 values, some of which may be ex- 
 cluded by the condition, that ACB 
 is less than 180°. 
 
 
 
 From C one of the angles not requir- 
 
 
 The three 
 
 
 ed, draw CD perpendicular to AB. 
 Find an arc E such that tan £ AB 
 
 
 sides, 
 
 
 : tan } (AC+BC) : : tan 1 (AC— 
 BC) : tan £ E ; then, if AB be 
 
 11 
 
 
 One of the 
 
 
 AB, AC, 
 
 
 greater than E, AB is the sum, and 
 
 
 
 angles 
 
 E the difference of AD and DB ; 
 
 
 and 
 
 
 but if AB be less than E, E is the 
 
 
 
 A. 
 
 sum and AB the difference of AD, 
 
 
 BC. 
 
 
 DB, (29.). In either case, AD and 
 BD are known, and tan AC : tan 
 AD : : R : cos A.
 
 270 
 
 SPHERICAL TRIGONOMETRY. 
 
 TABLE continued. 
 
 
 GIVEN. 
 
 SOUGHT. 
 
 1 
 
 SOLUTION. 
 
 
 
 
 Suppose the supplements of the 
 
 
 
 
 three given angles, A, B, C, to 
 
 
 
 
 be a, b, c, and to be the sides of 
 
 
 The three 
 
 One of the 
 
 a spherical triangle. Find, by 
 the last case, the angle of this 
 
 12 
 
 angles 
 
 sides 
 
 triangle, opposite to the side a, 
 and it will be the supplement of 
 
 
 A, B, C. 
 
 BC. 
 
 _ 
 
 the side of the given triangle op- 
 posite to the angle A, that is, of 
 BC, (11.) ; and therefore BC is 
 found. 
 
 In the foregoing table, the rules are given for ascertaining the affection 
 of the arc or angle found, whenever it can be done : Most of these rules 
 are contained in this one rule, which is of general application, viz. that 
 when the thing found is either a tangent or a cosine, and, of the tangents or 
 cosines employed in the computation of it, either one or three belong to obtuse ' 
 angles, the angle found is also obtuse. This rule is particularly to be attend- 
 ed to in cases 5 and 7, where it removes part of the ambiguity. 
 
 It may be necessary to remark wivh respect to the 11th case, that the 
 segments of the base computed there are those cut off by the nearest per- 
 pendicular; and also, that when the sum of the sides is less than 180°, 
 the least segment is adjacent to the least side of the triangle ; otherwise 
 to the greatest, (17.).
 
 SPHERICAL TRIGONOMETRY. 
 
 271 
 
 The last table may also be conveniently expressed in the following 
 manner, denoting the side opposite to the angle A, by o, to B by b, and to 
 C by c ; and also the segments of the base, or of opposite angle, by « 
 and y. 
 
 Two sides 
 
 b and c, and 
 
 the angle 
 
 between 
 
 them A. 
 
 Angles 
 
 A and C 
 
 and 
 
 side b 
 
 Sides 
 a and b 
 
 and 
 angle A. 
 
 B 
 
 B 
 
 B 
 
 Find x, so that 
 
 tan x=tan bx cos A ; then 
 
 _ sin x x tan A 
 
 tan B=^— r . 
 
 sin (c — x) 
 
 Find x, as above, 
 
 , cos b x cos (c — x) 
 then cos c= - . 
 
 Find x, so that 
 cot x=cos ixtan A ; then 
 tan b X cos x 
 
 tan a=- 
 
 cos (c — x) 
 
 Find x, as above, 
 
 then cos B = 
 
 cos A X sin (c — x) 
 
 sin B = 
 
 sin Ax sin A 
 
 Find x, so that 
 
 cot x=cos ixtan A ; then 
 
 _ cos x X tan b 
 
 cos C= . 
 
 tan a 
 
 Find x, so that 
 
 tan x=tan b X cos A ; and find 
 
 y, so that 
 
 cos aXcos x 
 cos y= 
 
 cos b 
 
 :=*±y-
 
 %TZ 
 
 SPHERICAL TRIGONOMETRY. 
 
 TABLE continued. 
 
 The angles 
 
 A and B 
 
 and the 
 
 side b. 
 
 10 
 
 sin ixsin A 
 sin B 
 
 Find x, so that 
 
 tan x=tan bxcos A ; and y, so 
 
 that 
 
 sin arXtan A 
 sin y= 
 
 tan B 
 
 c=a:±y. 
 
 Find a;, so that 
 
 cot a:=cos b X tan A ; and also y, 
 
 so that 
 
 sin ffXcos B 
 sin y= 
 
 cos A 
 
 c=xJty. 
 
 11 
 
 a, b, e. 
 
 sin 
 
 Let a+b-\~c=s. 
 I . _ -y/sin (Is— &)xsin (§s— c) 
 ■/sin b x sin c 
 
 or cos | A 
 
 _ -v/sin %s X sin (j*— a) 
 
 •v/sin ixsin c 
 
 12 
 
 A, B, C. 
 
 Let A+B + C=S. 
 
 . . ./ cos ^ S X cos (I S— A) 
 
 in $a=— * V3 1 
 
 -/sin B x sin C 
 
 sin 
 
 or cos 
 
 _ - v /cos(£S--B)^cos(S -C) 
 ^sin B X sin C
 
 APPENDIX 
 
 TO 
 
 SPHERICAL 
 
 TRIGONOMETRY, 
 
 CONTAINING 
 
 NAPIER'S RULES OF THE CIRCULAR PARTS. 
 
 The rule of the Circular Parts, invented by Napier, is of great use in 
 Spherical Trigonometry, by reducing all the theorems employed in the 
 solution of right angled triangles to two. These two are not new proposi- 
 tions, but are merely enunciations, which, by help of a particular arrange- 
 ment and classification of the parts of a triangle, include all the six propo- 
 sitions, with their corollaries, which have been demonstrated above fror* 
 the 1 8th to the 23d inclusive. They are perhaps the happiest example w 
 artificial memory that is known. 
 
 DEFINITIONS. 
 
 1. If in a spherical triangle, we set aside the right angle, and consider only 
 the five remaining parts of the triangle, viz. the three sides and the two 
 oblique angles, then the two sides which contain the right angle, and 
 the complements of the other three, namely, of the two angles and the 
 hypotenuse, are called the Circular Parts. 
 
 Thus, in the triangle ABC right angled at A, the circular parts are AC, 
 AB with the complements of B, BC, and C. These parts are called 
 circular ; because, when they are named in the natural order of theii 
 succession, they go round the triangle. 
 
 2. When of the five circular parts any one is taken, for the middle part, 
 then of the remaining four, the two which are immediately adjacent to 
 it, on the right and left, are called the adjacent parts ; and the other two, 
 each of which is separated from the middle by an adjacent part, are call- 
 ed opposite parts. 
 
 Thus in the right annled triangle ABC, A, being the right angle, AC, AB, 
 90° — B. 90°- BC. 9(P— C, are the circular parts, by Def. ) ; and il 
 
 35
 
 274 APPENDIX TO 
 
 anyone, as AC, be reckoned the middle part, then AB and 90°— C, which 
 
 are contiguous to it on different sides, are called adjacent parts ; and 90° 
 
 -B. 90° — BC are the opposite parts. In like manner if AB is taken f <w 
 
 the middle part, AC and 90° — B are the adjacent parts : 90°— BC, and 
 90°— C are the opposite. Or if 90°— BC be the middle part, 90— B, 
 90° — C are adjacent ; AC and AB opposite, &c. 
 
 This arrangement being made, the rule of the circular part is contained 
 in the following 
 
 PROPOSITION. 
 
 In a right angled spherical triangle, the rectangle under the radius and the sine 
 of the middle part, is equal to the rectangle under the tangents of the adjacent 
 parts ; or, to the rectangle under the cosines of the opposite parts 
 
 The truth of the two theorems included in this enunciation may be 
 easily proved, by taking each of the five circular parts in succession for 
 the middle part, when the general proposition will be found to coincide 
 with some one of the analogies in the table already given for the resolution 
 of the cases of right angled spherical triangles. Thus, in the triangle ABC, 
 if the complement of the hypotenuse BC be taken as the middle part, 90° 
 — B, and 90° — C, are the adjacent parts, AB and AC the opposite. Then 
 the general rule gives these two theorems, Rxcos BCi=cot Bxcot C, 
 and Rxcos BC=cos AB X cos AC. The former of these coincides with 
 the cor. to the 20th ; and the latter with the 22d. 
 
 To apply the foregoing general proposition to resolve any case of a right 
 angled spherical triangle, consider which of the three qualities named 
 (the two things given and the one required) must be made the middle term, 
 in order that the other two may be equi-distant from it, that is, may be 
 both adjacent, or both opposite ; then one or other of the two theorems 
 contained in the above enunciation will give the value of the thing re- 
 quired. 
 
 Suppose, for example, that AB and BC are given, to find C ; it is evi- 
 dent that if AB be made the middle part, BC and C are the opposite parts, 
 and therefore Rxsin AB=sin Cxsin BC, for sin C=cos (90° — C), and 
 
 cos (90° — BC)=sin BC, and consequently sin C=- — =-tl. 
 
 v J sm BC 
 
 Again, suppose that BC and C are given to find AC ; it is obvious thai 
 C is in the middle between the adjacent parts AC and (90° — BC), there-
 
 SPHERICAL TRIGONOMETRY. 375 
 
 fow R X cos C=tan AC X cot BC, or tan AC=-^ 57; r=c08 C + tan BC •, 
 
 cot dL 
 
 because, as has been shewn above, =r-=r=tan BC. 
 
 cot BC 
 
 In the same way may all the other cases be resolved. One or two trials 
 will always lead to the knowledge of the part which in any given case is 
 to be assumed as the middle part ; and a little practice will make it easy, 
 eTen without such trials, to judge at once which of them is to be so as- 
 sumed. It may be useful for the learner to range the names of the five 
 circular parts of the triangle round the circumference of a circle, at equal 
 distances from one another, by which means the middle part will be imme 
 diately determined. 
 
 Besides the rule of the circular parts, Napier derived from the last of the 
 three theorems ascribed to him above, (schol. 29.) the solutions of all the 
 cases of oblique angled triangles. These solutions are as follows : A, B, 
 C, denoting the three triangles of a spherical triangle, and a, b, c, the sides 
 opposite to them. 
 
 I. 
 
 Given two sides b, c, and the angle A between them. 
 To find the angles B and C. 
 
 tan i (B— C)=cot |Ax T? {*7 C l ( 31 cor - *• 
 
 a * ' ■ sin £ (o+c) 
 
 tan i (B4-C)=cot $ Ax °° S \ ^.~7 e \ . (31.) cor. 1. 
 " x a cos $ {b+c) v " 
 
 To find the third side a. 
 sin B : sin A : : sin b : sin a. 
 
 II. 
 
 Given the two sides b, c, and the angle B opposite to one of them. 
 
 To find C, and the angle opposite to the other side. 
 
 sin b : sin c : : sin B : sin C. 
 
 To find the contained angle A. 
 
 cot i A=Un i (B-C) x 8i " \ ( * +c | . (31 .) cor. 1. 
 
 * a sin £ (b— c) ' ' 
 
 To find the third side a. 
 sin B : sin A : : sin b : sin a. 
 
 III. 
 
 Given two angles A and B, and the side c between them. 
 To find the other two sides a, b.
 
 876 APPENDIX TO 
 
 -J (»+.,_*, l,x£}£=§. (31.) 
 
 To find the third angle C. 
 sin a : sin c : : sin A : sin C. 
 
 IV. 
 
 Given two angles A and B, and the side a, opposite to one of them 
 
 To find l, the side opposite to the other. 
 
 sin A : sin B : : sin a : sin b. 
 
 To find e, the side between the given angles. 
 
 uni«=*„l(«->)x£i£±!> (31.) 
 
 To find the third angle C. 
 sin a : sin c : : sin A : sin C. 
 
 The other two cases, when the three sides are given to find the angles, 
 or when the three angles are given to find the sides, are resolved by the 
 29th, (the first of Napier's Propositions,) in the same way as in the table 
 already given for the case of the oblique angled triangle. 
 
 There is a solution of the case of the three sides being given, which it 
 is often very convenient to use, and which is set down here, though the 
 proposition on which it depends has not been demonstrated. 
 
 Let a, b, c, be the three given sides, to find the angle A, contained be- 
 tween b and c. 
 
 If Rad = 1, and a + b -f- e = s, 
 
 8in * A r J™g*=!VX^n Stt . or> 
 
 •/sin b X sin c 
 
 cos 4 A = y sin -(i* xsin £(*- a )) , 
 ^/sin b X sin c 
 
 In like manner, if the three angles, A, B, C are given to find o the side 
 between A and B.
 
 SPHERICAL TRIGONOMETRY. 877 
 
 Le< A + B + C = S, 
 
 ■in A c=— 3 v g = '- ; or, 
 
 v^sin B X sin C 
 
 cos \ c= Vc ° S ( * S - B Tx~cos g S-C) 
 yf sin B X sin C. 
 
 These theorems, on account of the facility with which Logarithms are 
 applied to them, are the most convenient of any for resolving the two cases 
 to which they refer. When A is a very obtuse angle, the second theorem, 
 which gives the value of the cosine of its half, is to be used ; otherwise 
 the first theorem, giving the value of the sine of its half its preferable. 
 The same is to be observed with respect to the side c, the reason of which 
 ■ • explained, Plane Trig. Schol. 
 
 EKD OF SPHERICAL TRIGONOMETRY
 
 NOTES 
 
 ON THE 
 
 FIRST BOOK OF THE ELEMENTS. 
 
 DEFINITIONS. 
 I. 
 
 In the definitions a few changes have been made, of which it is necea- 
 »ary to give some account. One of these changes respects the firs', defini- 
 tion, that of a point, which Euclid has said to be, ' That which has no 
 parts, or which has no magnitude.' Now, it has been objected to this defi- 
 nition, that it contains only a negative, and that it is not convertible, aa 
 every good definition ought certainly to be. That it is not convertible is 
 evident, for though every point is unextended, or without magnitude, yet 
 every thing unextended or without magnitude, is not a point. To this it 
 is impossible to reply, and therefore it becomes necessary to change the 
 definition altogether, which is accordingly done here, a point being defined 
 to be, that which has position but not magnitude. Here the Affirmative part 
 includes all that is essential to a point, and the negative part includes 
 every thing that is not essential to it. I am indebted for this definition to 
 a friend, by whose judicious and learned remarks I have often profited. 
 
 II. 
 
 After the second definition Euclid has introduced the following, " the 
 " extremities of aline are points." 
 
 Now, this is certainly not a definition, but an inference from the defini- 
 tions of a point and of a line. That which terminates a line can have no 
 breadth, as the line in which it is has none ; and it can have no length, as it 
 would not then be a termination, but a part of that which is supposed to 
 terminate. The termination of a line can therefore have no magnitude, and 
 having necessarily position, it is a point. But as it is plain, that in all this 
 we are drawing a consequence from two definitions already laid down, and 
 not giving a new definition, I have taken the liberty of putting it down as 
 a corollary to the second definition, and have added, that the intersections oj 
 one line with another are points, as this affords a good illustration of the nature 
 of a point, and is an inference exactly of the same kind with the preceding. 
 The same thing nearly has been done with the fourth definition, where 
 that which Euclid gave as a separate definition is made a corollary to the
 
 280 
 
 NOTES. 
 
 fourth, Lecause it is in fact an inference deduced from comparing the deft 
 unions of a superficies and a line. 
 
 As it is impossible to explain the relation of a superficies, a line, and a 
 point to one another, and to the solid in which they all originate, bettei 
 than Dr. Simson has done, I shall here add, with very little change, the 
 illustration given by that excellent Geometer. 
 
 " It is necessary to consider a solid, that is, a magnitude which has 
 ength, breadth, and thickness, in order to understand aright the definitions 
 of a point, line and superficies ; for these all arise from a solid, and exist in 
 it ; The boundary, or boundaries which contain a solid, are called superfi- 
 cies, or the boundaiy which is common to two solids which are cuiitiguous, 
 or which divides one solid into two contiguous parts, is eallea a supcrfi ■ 
 cies ; Thus, it BCGF be one of the boundaries which contain the solid 
 ABCDEFGH, or which is the common boundary of this solid, and the solid 
 BKLCFNMG, and is therefore in the one as well as the other solid, it is 
 called a superficies, and has no thickness ; For if it have any, thi B thick- 
 ness must either be a part of the thickness of the solid AG, or the sobd BM, 
 or a part of the thickness of each of them. It cannot be a part of the '.hick- 
 ness of the solid BM ; because, if this solid be removed from the solid AG, 
 the superficies BCGF, the boundary of the solid AG, remains sti'l the 
 same as it was. Nor can it be a part of the thickness of the solid AG : 
 because if this be removed from the solid BM, the superficies BCG1 , the 
 boundary of the solid BM, docs nevertheless remain; therefore the super- 
 ficies BCGF has no thickness, but only length and breadth. 
 
 " The boundary of a superficies is called a line ; or a line is the common 
 boundary of two superficies that are contiguous, or it is that which divides 
 one superficies into two contiguous parts : Thus, if BC be one of the boun- 
 daries which contain the superficies ABCD, or which is the common born 
 dary of this superficies, and of the superficies KBCL, which is contiguous 
 to it, this boundary BC is called a line, and has no breadth ; For, if it hav» 
 any, this must be part either of the breadth of the superficies ABCD o» 
 of the superficies KBCL, or part of 
 each of them. It is not part of the 
 breadth of the superficies KBCL ; 
 for if this superficies be removed from 
 the superficies ABCD, the line BC 
 which is the boundary of the super- 
 ficies ABCD remains the same as it 
 was. Nor can the breadth that BC 
 is supposed to have, be a part of the 
 breadth of the superficies ABCD ; be- 
 cause, if this be removed from the su- 
 perficies KBCL, the line BC, which 
 is the boundary of the superficies 
 KBCL, does nevertheless remain : Therefore the line BC has no breadth 
 And because the line BC is in a superficies, and that a superficies has nc 
 thickness, as was shown ; therefore a line has neither breadth nor thick- 
 ness, out only length. 
 
 " The boundary of a line is called a point, or a point is t common boun 
 dary or extremity of two lines that are contiguous : Thus if B be the ex-
 
 NOTES 
 
 28. 
 
 P 
 
 Iremity of the line AB, or the common extremity of the two lines AB, KB, 
 
 this extremity is called a point, and has no length : For if it have any, thin 
 
 length must either be part of the JI £J- 
 
 length of *.l»; line AB, or of the line 
 
 KB. It is not part of the length of 
 
 KB ; for if the line KB be removed 
 
 from AB, the point B, which is the E 
 
 extremity of the line AB, remains the 
 
 tame as it was ; Nor is it part of the 
 
 length of the line AB ; for if A B be 
 
 .emoved from the line KB, the point 
 
 13, which is the extremity of the lino 
 
 KB, does nevertheless remain : 
 
 Therefore the point B has no length ; v~ 
 
 And because a point is in a line, and 
 
 a line has neither breadth nor thickness, therefore a point has no length, 
 
 breadth, nor thickness. And in this manner the definition of a point, line, 
 
 uid superficies are to be understood." 
 
 1} 
 
 III. 
 
 Euclid has defined a straight line to be a line which (as we translate it) 
 u lies evenly between its extreme points." This definition is obviously 
 faulty, the word evenly standing as much in need of an explanation as the 
 word straight, which it is intended to define. In the original, however, it 
 must be confessed, that this inaccuracy is at least less striking than in oui 
 translation ; for the word which we render evenly is slian, equally, and is ac- 
 cordingly translated ex cequo, and cqualiter by Commandine and Gregory. 
 The definition, therefore, is, that a straight linp is one which lies equally 
 between its extreme points : and if by this we understand a line that lies 
 between its extreme points so as to be related exactly alike to the space 
 on the one side of it, and to the space on the other, we have a definition 
 that is perhaps a little too metaphysical, but which certainly contains in it 
 the essential character of a straight line. That Euclid took the definition 
 in this sense, however, is not certain, because he has not attempted to 
 deduce from it any property whatsoever of a straight line ; and indeed, it 
 should seem not easy to do so, without employing some reasonings of a 
 more metaphysical kind than he has any where admitted into his Elements. 
 To supply the defects of his definition, he has therefore introduced the 
 Axiom, that two straight lines cannot inclose a space; on which Axiom it is, 
 and not on his definition of a straight line, that his demonstrations are 
 founded. As this manner of proceeding is certainly not so regular and 
 scientific as that of laying down a definition, from which the properties of 
 the thing defined maybe logically deduced, I have substituted another defi- 
 nition of a straight line in the room of Euclid's. This definition of a straight 
 line was suggested by a remark of Boscovich, who, in his Notes on the 
 philosophical Poem of Professor Stay, says, " Rectam lineam rectse con- 
 *' gruere totam toti in infinitum productum si bina puncta unius binis al- 
 • jerius congruani, patet ex ipsa admodum clara rectitudinis idea quaro 
 
 36
 
 S82 NOTE?. 
 
 "habemus." (Supplementum in lib. 3. § 550.) Now, that which Mr. 
 Boscovich would consider as an inference from our idea of straightness, 
 eeems itself to be the essence of that idea, and to afford the best criterion 
 for judging whether any given line be straight or not. On this principle 
 we have given the definition above, If there be two lines which cannot coin~ 
 cide in two points, without coinciding altogether, each of them is called a straight 
 line. 
 
 This definition was otherwise expressed in the two former editions ; it 
 was said, that lines are straight lines which cannot coincide in part, with 
 out coinciding altogether. This was liable to an objection, viz. that it de 
 fined straight lines, but not a straight line ; and though this in truth is but 
 a mere cavil, it is better to leave no room for it. The definition in the form 
 now given is also more simple. 
 
 From the same definition, the proposition which Euclid gives as an 
 Axiom, that two straight lines cannot inclose a space, follows as a neces- 
 sary consequence. For, if two lines inclose a space, they must intersect 
 one another in two points, and yet, in the intermediate part, must not coin- 
 cide ; and therefore by the definition they are not straight lines. It follows 
 in the same way, that two straight lines cannot have a common segment, 
 or cannot coincide in part, without coinciding altogether. 
 
 After laying down the definition of a straight line, as in the first Edition, 
 I was favoured by Dr. Reid of Glasgow with the perusal of a MS. contain- 
 ing many excellent observations on the first Book of Euclid, such as might 
 be expected from a philosopher distinguished for the accuracy as well as 
 the extent of his knowledge. He there defined a straight line nearly as 
 has been done here, viz. " A straight line is that which cannot meet ano- 
 " ther straight line in more points than one, otherwise they perfectly coincide, 
 " and are one and the same." Dr. Reid also contends, that this must have 
 been Euclid's own definition ; because, in the first proposition of the 
 eleventh Book, that author argues, " that two straight lines cannot have a 
 " common segment, for this reason, that a straight line does not meet a 
 " straight line in more points than one, otherwise they coincide." Whether 
 this amounts to a proof of the definition above having been actually 
 Euclid's, I will not take upon me to decide ; but it is certainly a proof 
 that the writings of that Geometer ought long since to have suggested this 
 definition to his commentators ; and it reminds me, that I might have learn- 
 ed from these writings what I have acknowledged above to be derived from 
 a remoter source. 
 
 There is another characteristic, and obvious property of straight lines, 
 by which 1 have often thought that they might be very conveniently defin* 
 ed, viz. that the position of the whole of a straight line is determined by the 
 position of two of its points, in so much that, when two points of a straight 
 line continue fixed, the line itself cannot change its position. It might 
 therefore be said, that a straight line is one in which, if the position of tioo 
 points be determined, the position of the whole line is determined. But this de- 
 finition, though it amount in fact to the same thing with that already gn en, 
 is rather more abstract, and not so easily made the foundation of reason- 
 ing. I therefore thought it best to lay it aside, and to adopt the definition 
 given in the text.
 
 NOTES. 283 
 
 V. 
 
 The definition of a plane is given from Dr. Simson, Euclid's being liable 
 to the same objections with his definition of a straight line ; for, he says, 
 that a plane superficies is one which " lies evenly between its extreme 
 " lines." The defects of this definition are completely removed in that which 
 Dr. Simson has given. Another definition different from both might have 
 been adopted, viz. That those superficies are called plane, which are inch, 
 .hat if three points of the one coincide with three points of the other, the 
 whole of the one must coincide with the whole of the other. This defini- 
 tion, as it resembles that of a straight line, already given, might, perhaps 
 have been introduced with some advantage ; but as the purposes of demon- 
 stration cannot be better answered than by that in the text, it has been 
 thought best to make no farther alteration. 
 
 VI. 
 
 In Euclid, the general definition of a plane angle is placed before thatol 
 a rectilineal angle, and is meant to comprehend those angles which are 
 formed by the meeting of the other lines than straight lines. A plane 
 angle is said to be " the inclination of two lines to one another which 
 M meet together, but are not in the same direction." This definition is 
 omitted here, because that the angles formed by the meeting of curve lines, 
 though they may become the subject of geometrical investigation, certainly 
 do not belong to the Elements ; for the angles that must first be considered 
 are those made by the intersection of straight lines with one another. 
 The angles formed by the contact or intersection of a straight line and a 
 tircle, or of two circles, or two curves of any kind with one another, 
 could produce nothing but perplexity to beginners, and cannot possibly be 
 understood till the properties of rectilineal angles have been fully explained. 
 On this ground, I am of opinion, that in an elementary treatise it may 
 fairly be omitted Whatever is not useful, should, in explaining the ele- 
 ments of a science, be kept out of sight altogether •• for, if it does not assist 
 the progress of the understanding, it will certainly retard it 
 
 AXIOMS 
 
 Among the Axioms there have been made only two alterations. The 
 I Oth Axiom in Euclid is, that " two straight lines cannot inclose a space ;" 
 which, having become a corollary to our definition of a straight line, ceases 
 of course to be ranked with self-evident propositions. It is therefore re- 
 moved from among the Axioms. 
 
 The 12th Axiom of Euclid is, that " if a straight line meets two straight 
 
 lines, so as to make the two interior angles on the same side of it taken 
 
 44 together less than two right angles, these straight lines being continually 
 
 • produced, shall at length meet upon that side on which are the angles
 
 284 NOTES. 
 
 " which ire less than two right angles." Instead of this proposition, 
 which, though true, is by no means self-evident ; another that appeared 
 more obvious, and better entitled to be accounted an Axiom, has been in 
 troduced, viz. " that two straight lines, which intersect one another, can- 
 u not be both parallel to the same straight line." On this subject, ho w« 
 ever, a fuller explanation is necessary, for which see the note on -he 29th 
 Prop 
 
 PROP. IV. and VIII. B. I. 
 
 The IV. and VIII. propositions of the first book are the foundation of all 
 that follows with respect to the comparison of triangles. They are de- 
 monstrated by what is called the method of superaposition, that is, by lay 
 ing the one triangle upon the other, and proving that they must coincide 
 To this some objections have been made, as if it were ungeometrical to 
 suppose one figure to be removed from its place and applied to another 
 figure. " The laying," says Mr. Thomas Simson in his Elements, " of 
 " one figure upon another, whatever evidence it may afford, is a mechanical 
 " consideration, and depends on no postulate." It is not clear what Mr. 
 Simson meant here by the word mechanical : but he probably intended only 
 to say, that the method of superaposition involves the idea of motion, which 
 belongs rather to mechanics than geometry ; for I think it is impossible 
 that such a Geometer as he was could mean to assert, that the evidence 
 derived from this method is like that which arises from the use of instru- 
 ments, and of the same kind with what is furnished by experience and ob- 
 servation. The demonstrations of the fourth and eighth, as they are given 
 by Euclid, are as certainly a process of pure reasoning, depending solely 
 on the idea of equality, as established in the 8th Axiom, as any thing in 
 geometry. But, if still the removal of the triangle from its place be consi- 
 dered as creating a difficulty, and as inelegant, because it involves an idea, 
 that of motion, not essential to geometry, this defect may be entirely re- 
 medied, provided that, to Euclid's three postulates, we be allowed to add 
 the following, viz. That if there be two equal straight lines, and if any figure 
 whatsoever be constituted on the one, a figure every way equal to it may be con- 
 stituted on the other. Thus if AB ana DE be two equal straight lines, and 
 ABC a triangle on the base AB, a triangle DEF every way equal to ABC 
 may be supposed to be constituted on DE as a base. By this it is not 
 meant to assert that the method of describing the triangle DEF is actually 
 known, but merely that the triangle DEF may be conceived to exist in 
 all respects equal to the triangle ABC. Now, there is no truth whatso- 
 ever that is better entitled than this to be ranked among tr e Postulates or 
 Axioms of geometry ; for the straight lines AB and DE being even' way 
 equal, there can be nothing belonging to the one that may not also belong 
 to the other. 
 
 On the strength of this Postulate the IV. proposition is thus demonstrated 
 If ABC, DEF be two triangles, such that the two sides AB and AC oi 
 the one are equal to the two ED, DF of the other, and the angle BAC, 
 contained by the sides AB, AC of the one, equal to the angle EDF, con 
 lained by the sides ED, DF of the other ; the triangles ABC and EDF are 
 every wav equal.
 
 NOTES. 
 
 .285 
 
 Chi AB let a triangle be constituted every way equal to the triangle DEF ; 
 then if this triangle coincide with the triangle ABC, it is evident that the 
 proposition is true, for it is equal to DEF by hypothesis, and to ABC, be- 
 cause it coincides with it ; wherefore ABC, DEF are equal to one another 
 But if it does not coincide with ABC, let it have the position ABG ; and first 
 suppose G not to fall on AC ; then the angle BAG is not equal to the angle 
 BAC. But the angle BAG is equal to the angle EDF, therefore EDF 
 and ABC are not equal, and they are also equal by hypothesis, which is 
 impossible. Therefore the point G must fall upon AC ; now, if it fall upon 
 AC but not at C, then AG is not equal to AC ; but AG is equal to DF, 
 therefore DF and AC are not equal, and they are also equal by supposition, 
 which is impossible. Therefore G must coincide with C, and the triangle 
 AGB with the triangle ACB. But AGB is every way equal to DEF, 
 therefore ACB and DEF are also every way equal. 
 
 By help of the same postulate, the fifth may also be very easily de- 
 monstrated. 
 
 Let ABC be an isosceles triangle, in which AB, AC are the equal sides , 
 the angle ABC, ACB opposite to these sides are also equal. 
 
 Draw the straight line EF equal to BC, and suppose that on EF the tri 
 angle DEF is constituted every way equal to the triangle ABC, that is. 
 having DE equal to AB, DF to AC, the angle EDF to the angle BAC. the 
 angle ACB to the angle DFE, &c. 
 
 Then because DE is equal to AB, and AB is equal to AC, DE is equa 
 to AC ; and for the same reason, DF is equal to AB. And because DF is 
 equal to AB, DE to AC, and the angle FDE to the angle BAC, the angle 
 ABC is equal to the angle DFE But the angle ACB is also, by hy- 
 pothesis, equal to the angle DFE • there r ore the angles ABC \CU art 
 equal to one another.
 
 286 NOTES. 
 
 Such demonstrations, it must, however, be acknowledged, trespass 
 against a rule which Euclid has uniformly adhered to throughout the Ele- 
 ments, except where he was forced by necessity to depart from it ; This 
 rule is, that nothing is ever supposed to be done, the manner of doing which 
 has not been already taught, so that the construction is derived either di- 
 rectly from the three postulates laid down in the beginning, or from pro- 
 blems already reduced to those postulates. Now, this rule is not essential 
 to geometrical demonstration, where, for the purpose of discovering the 
 properties of figures, we are certainly at liberty to suppose any figure to be 
 constructed, or any line to be drawn, the existence of which does not in- 
 volve an impossibility. The only use, therefore, of Euclid's rule is tc 
 guard against the introduction of impossible hypotheses, or the taking for 
 granted that a thing may exist which in fact implies contradiction ; from 
 such suppositions, false conclusions might, no doubt, be deduced, and the 
 rule is therefore useful in as much as it answers the purpose of excluding 
 them. But the foregoing postulatum could never lead to suppose the 
 actual existence of any thing that is impossible ; for it only assumes the 
 existence of a figure equal and similar to one already existing, but in a dif- 
 ferent part of space from it, or having one of its sides in an assigned posi- 
 tion. As there is no impossibility in the existence oi one of these figures 
 it is evident that there can be none in the existence of the other. 
 
 PROP. XXI. THEOR. 
 
 It is essential to the truth of this proposition, that the straight hn^s 
 drawn to the point within the triangle be drawn from the two extremities 
 of the base ; for, if they be drawn from other points of the base, their sum 
 may exceed the sum of the sides of the triangle in any ratio that is less 
 than that of two to one. This is demonstrated by Pappus Alexandiinus 
 in the 3d Book of his Mathematical Collections, but the demonstration is of a 
 kind that does not belong to this place. If it be required simply to show, 
 that in certain cases the sum of the two lines drawn to the point withm the 
 triangle may exceed the sum of the sides of the triangle, the demonstra- 
 tion is easy, and is given nearly as follows by Pappus, and also by Proclus, 
 in the 4th Book of his Commentary on Euclid. 
 
 Let ABC be a triangle, having the angle at A a right angle : let D be 
 any point in AB ; join CD, then CD will be greater than AC, because in 
 the triangle ACD the angle CAD is greater than the angle ADC From 
 DC cut off DE equal to AC ; bisect CE 
 in F, and join BF ; BF and FD are greater 
 than BC and CA. 
 
 Because CF is equal to FE, CF and FB 
 are equal to EF and FB, but CF and FB 
 are greater than BC, therefore EF and FB 
 are greater than BC. To EF and FB add 
 ED, and to BC add AC, which is equal to 
 BD by construction, and BF and FD will 
 be greater than BC and CA.
 
 NOTES. 287 
 
 It is evident, that if the angle BAC be obtuse, the same reasoning may 
 be applied. 
 
 This proposition is a sufficient vindication of Euclid for having demon- 
 strated the 21st. proposition, which some affect to consider as self-evident ; 
 for it proves that the circumstance on which the truth of that proposition 
 depends is not obvious, nor that which at first sight it is supposed to be, viz. 
 that of the one triangle being included within the other. For this reason I 
 cannot agree with M. Clairaut, that Euclid demonstrated this proposition 
 only to avoid the cavils of the Sophists. But I must, at the same time, ob- 
 serve, that what the French Geometer has said on the subject has certain 
 ly been misunderstood, and in one respect, unjustly censured by Dr. Simson. 
 The exact translation of his words is as follows : "If Euclid has taken the 
 "trouble to demonstrate, that a triangle included within another has the 
 " sum of its sides less than the sum of the sides of the triangle in which it 
 " is included, we are not to be surprised. That Geometer had to do with 
 " those obstinate Sophists, who made a point of refusing their assent to the 
 " most evident truths," &c. (Elements de Geometrie par M. Clairaut. 
 Pref.) 
 
 Dr. Simson supposes M. Clairaut to mean, by the proposition which he 
 enunciates here, that when one triangle is included in another, the sum of 
 the two sides of the included triangle is necessarily less than the sum of the 
 two sides of the triangle in which it is included, whether they be on the 
 same base or not. Now this is not only not Euclid's proposition, as Dr 
 Simson remarks, but it is not true, and is directly contrary to what has 
 just been demonstrated from Proclus. But the fact seems to be, that M. 
 Clairaut's meaning is entirely different, and that he intends to speak not of 
 two of the sides of a triangle, but of all the three ; so that his proposition 
 is, " that when one triangle is included within another, the sum of all the 
 " three sides of the included triangle is less than the sum of all the three 
 sides of the other," and this is without doubt true, though I think by no 
 means self-evident. It must be acknowledged also, that it is not exactly 
 Euclid's proposition, which, however, it comprehends under it, and is the 
 general theorem, of which the other is only a particular case. Therefore, 
 though M. Clairaut may be blamed for maintaining that to be an Axiom 
 which requires demonstration, yet he is not to be accused of mistaking a 
 false proposition for a true one. 
 
 PROP. XXII. PROB. 
 
 Thomas Simson in his Elements has objected to Euclid's demonstratioi 
 of this proposition, because it contains no proof, that the two circles made 
 use of in the construction of the Problem must cut one another ; and Dr. 
 Simson on the other hand, always unwilling to acknowledge the smallest 
 blemish in the works of Euclid, contends that the demonstration is perfect. 
 The truth, however, certainly is, that the demonstration admits of some 
 improvement ; for the limitation that is made in the enunciation of any 
 Problem ought always to be shewn to be necessarily connected with the 
 construction of it, and this is what Euclid has neglected to do in the pro- 
 sent instance. The defect may easily be supplied, and Dr. Simson him-
 
 288 NOTES. 
 
 aelf has done it in effect in his note on this proposition, though he denies it 
 to be necessary. 
 
 Because that of the three straight lines DF, FG, GH, any two are great- 
 er than the third, by hypothesis, FD is less than FG and GH, that is, 
 than FH, and therefore the circle described from the centre F, with the 
 distance FD must meet the hne FE between F and H ; and, for the like 
 
 reason, the circle described from the centre G at the distance GH, must 
 meet DG between D and G, and therefore the one of these circles can- 
 not be wholly within the other. Neither can the one be wholly without 
 the other, because* DF and GH are greater than FG ; the two circles 
 must therefore intersect one another. 
 
 PROP. XXVII. and XXVIII. 
 
 Euclid has been guilty of a slight inaccuracy in the enunciations of 
 these propositions, by omitting the condition, that the two straight lines on 
 which the third line falls, making the alternate angles, &c. equal, must 
 be in the same plane, without which they cannot be parallel, as is evident 
 from the definition of parallel lines. The only editor, I believe, who has re- 
 marked this omission, is M. de Foix Due de Candalle, in his transla- 
 tion of the Elements published in 1566. How it has escaped the notice of 
 subsequent commentators is not easily explained, unless because they 
 thought it of little importance to correct an error by which nobody was 
 likely to be misled. 
 
 PROP. XXIX. 
 
 The subject of parallel lines is one of the most difficult in the Elements 
 of Geometry. It has accordingly been treated of in a great variety of differ- 
 ent ways, of which, perhaps, there is none that can be said to have given 
 entire satisfaction. The difficulty consists in converting the 27th and 28th of 
 Euclid, or in demonstrating, that parallel straight lines, or such as do not 
 meet one another, when they meet a third line, make the alternate angles 
 with it equal, or, which comes to the same, are equally inclined to it, and 
 make the exterior angle equal to the interior and opposite. In order to de-
 
 NOTES. 28& 
 
 aionstrate this proposition, Euclid assumed it as an Axiom, that " if a 
 • straight line meet two straight lines, so as to make the interior angles on 
 " the same side of it less than two right angles, these straight lines bein# 
 •* continually produced, will at length meet on the side on which the angles 
 ' are that are less than two right angles." This proposition, however, is 
 not self-evident, and ought the less to be received without proof, that, as 
 Proclus has observed, the converse of it is a proposition that confessedly 
 requires to be demonstrated. For the converse of it is, that two straight 
 lines which meet one another make the interior angles, with any third line, 
 less than two right angles ; or, in other words, that the two interior angles 
 of any triangle are less than two right angles, which is the 17th of the 
 First Book of the Elements : and it should seem, that a proposition can 
 never rightly be taken for an Axiom, of which the converse requires a de- 
 monstration. 
 
 The methods by which Geometers have attempted to remove this 
 blemish from the Elements are of three kinds. 1. By a new definition of 
 parallel lines. 2. By introducing a new Axiom concerning parallel lines, 
 more obvious than Euclid's. 3. By reasoning merely from the definition 
 of parallels, and the properties of lines already demonstrated without the 
 ssumption of any new Axiom. 
 
 1. One of the definitions that has been substituted for Euclid's is, that 
 straight lines are parallel, which preserve always the same distance from 
 one another, by the word distance being understood, a perpendicular drawn 
 to oneof the lines from anypoint whatever in the other. If these perpendicu- 
 lars be every where of the same length, the straight lines are called parallel 
 This is the definition given by Wolfius, by Boscovich, and by Thomas 
 Simson, in the first edition of his Elements. It is however a faulty defi- 
 nition, for it conceals an Axiom in it, and takes for granted a property of 
 straight lines, that ought either to be laid down as self-evident, or demonstrat- 
 ed, if possible, as a Theorem. Thus, if from the three points, A, B, and C 
 of the straight line AC, perpendiculars AD, BE, CF be drawn all equal 
 to one another, it is implied in the definition 
 that the points D, E and F are in the same 
 straight line, which, though it be true, it was 
 not the business of the definition to inform us 
 of. Two perpendiculars, as AD and CF, are 
 alone sufficient to determine the position of the 
 •traight line D*F, and therefore the definition ought to be, "that two straighi 
 " lines are parallel, when there are two points in the one, from which the 
 " perpendiculars drawn to the other are equal, and on the same side of it." 
 
 This is the definition of parallels which M. D'Alembert seems to prefei 
 to all others ; but he acknowledges, and very justly, that it still remains a 
 matter of difficulty to demonstrate, that all the perpendiculars drawn from 
 the one of these lines to the other are equal. (Encyclopedic, Art. Parallclc.) 
 
 Another definition that has been given of parallels is, that tiny are lines 
 which make equal angles with a third line, toward the same parts, or such 
 as make the exterior angle equal to the interior and opposite. Varignon 
 Bezout, and several other mathematicians, have adopted this definition 
 vhich, it must be acknowledged, is a perfectly good one, if it be undersn>od 
 
 37
 
 290 NOTES. 
 
 by it, that the two lines called parallel, are such as make equal angle* w*tt 
 
 a certain third line, but not with any line that falls upon them. It remains 
 therefore, to be demonstrated, That if AB and CD make equal angles with 
 GH, they will do so also with any other line whatsoever. The definition, 
 therefore, must be thus understood, That parallel lines are such as make 
 equal angles, with a certain third line, or, more simply, lines which are per- 
 pendicular to a given line. It must then be proved, 1. That straight lines 
 which are equally inclined to a certain line or perpendicular to a certain line, 
 must be equally inclined to all the other lines that fall upon them ; and also, 
 2. That two straight lines which do not meet when produced, must make 
 equal angles with any third line that meets them. 
 
 The demonstration of the first of these propositions is not at all facilitated 
 by the new definition, unless it be previously shown that all the angles of a 
 triangle are equal to two right angles. 
 
 The second proposition would hardly be necessary if the new definition 
 were employed ; for when it is required to draw a line that shall not meet 
 a given line, this is done by drawing a line that shall have the same incli- 
 nation to a third line that the first or given line has. It is known that lines 
 so drawn cannot meet. It would no doubt be an advantage to have a defi- 
 nition that is not founded on a condition purely negative. 
 
 2. As to the Mathematicians who have rejected Euclid's Axiom, and in- 
 troduced another in its place, it is not necessary that much should be said. 
 Clavius is one of the first in this class ; the Axiom he assumes is, " That a 
 " line of which the points are all equidistant from a certain straight line in 
 " the same plane with it, is itself a straight line." This proposition he does 
 not, however, assume altogether, as he gives a kind of metaphysical proof 
 of it, by which he endeavours to connect it with Euclid's definition of a 
 straight line, with which proof at the same time he seems not very well 
 satisfied. His reasoning, after this proposition is granted (though it ought 
 not to be granted as an Axiom), is logical and conclusive, but is prolix and 
 operose, so as to leave a strong suspicion that the road pursued is by no 
 means the shortest possible. 
 
 The method pursued by Simson, in his Notes in the First Book of Euclid, 
 is not very different from that of Clavius. He assumes this Axiom, " That 
 ' a straight line cannot first come nearer to another straight line, and then 
 " go farther from it without meeting it." (Notes, &c. English Edition.) By 
 coming nearer is understood, conformably to a previous definition, the dimi-
 
 NOTES 
 
 291 
 
 nution of the perpendiculars drawn from the one line to the other. Thi» 
 Axiom is more readily assented to than that of Clavius, from which, how- 
 ever, it is not very different : but it is not very happily expressed, as the idea 
 not merely of motionj but of time, seems to be involved in the notion of first 
 coming nearer, and then going farther off. Even if this inaccuracy is pass 
 ed over, the reasoning of Simson, like that of Clavius, is prolix, and evi 
 den'ly a circuitous method of coming at the truth. 
 
 Thomas Simson, in the second edition of his Elements, has presented 
 ♦.his Axiom in a simpler form. " If two points in a straight line are positeo 
 " at unequal distances from another straight line in the same plane, 
 " those two lines being indefinitely produced on the side of the least dis- 
 " tance will meet one another." 
 
 By help of this Axiom it is easy to prove, that if two straight lines AB, 
 CU are parallel, the perpendiculars to the one, terminated by the other, 
 are all equal, and are also perpendicular to both the parallels. That they 
 are equal is evident, otherwise the lines would meet by the Axiom. That 
 they are perpendicular to both, is demonstrated thus : 
 
 If AC and BD, which are perpendicular to AB, and equal to one another, 
 be not also perpendicular to CD, from C let CE ^ 
 be drawn at right angles to BD. Then, be- 
 cause AB and CE are both perpendicular to 
 BD, they are parallel, and therefore the perpen- 
 diculars AC and BE are equal. But AC is 
 equal to BD, (by hypotheses,) therefore BE and 
 BD are equal, which is impossible ; BD is therefore at right angles to CD. 
 
 Hence the proposition, that " if a straight line fall on two parallel lines, it 
 " makes the alternate angles equal," is easily derived. Let FH and GE be 
 
 perpendicular to CD, then they will be parallel to one another, and also ai 
 right angles to AB, and therefore FG and HE are equal to one another, 
 by the last proposition. Wherefore in the triangles EFG, EFH, the side* 
 HE and EF are equal to the sides GF and FE, each to each, and also the 
 third side HF to the third side EG, therefore the angle HEF is equal to 
 the angle EFG, and they are alternate angles. 
 
 This method of treating the doctrine of parallel lines is extremely plain 
 and concise, and is perhaps as good as any that can be followed, when a 
 new Axiom is assumed. In the text above, I have, however, followed a 
 different method, employing as an Axiom, "That two straight lines, which 
 " cut one another, cannot be both parallel to the same straight line." Tins 
 Axiom has been assumed by others, particularly by Ludlam, in his very 
 useful little tract, entitled Rudiments of Mathematics.
 
 192 NOTES. 
 
 It is a pioposition readily enough admitted as self-evident, and leads 
 'o the demonstration of Euclid's 29th Proposition, even with more brevity 
 than Simson's. 
 
 3. All the methods above enumerated leave the mind somewhat dissatis- 
 fied, as we naturally expect to discover the properties of parallel lines, as 
 we do those of other geometric quantities, by comparing the definition of 
 those lines, with the properties of straight lines already known. The most 
 ancient writer who appears to have attempted to do this is Ptolemy the as- 
 tronomer, who wrote a treatise expressly on the subject of Parallel Lines. 
 Proclus has preserved some account of this work in the Fourth Book of his 
 commentaries : and it is curious to observe in it an argument founded on the 
 principle which is known to the moderns by the name of the sufficient reason. 
 
 To prove, that if two parallel straight lines, AB and CD, be cut by a 
 third line EF, in G and H, the two interior angles AGH, CHG will be 
 
 equal to two right angles, Ptolemy reasons thus: If the angles AGH, 
 CHG be not equal to two right angles, let them, if possible, be greater 
 than two right angles : then, because the lines AG and CH are not more 
 parallel than the lines BG and DH, the angles BGH, DHG are also 
 greater than two right angles. Therefore, the four angles AGH, CHG, 
 BGH, DHG are greater than four right angles ; and they are also equal 
 to four right angles, which is absurd. In the same manner it is shewn, 
 that the angles AGH, CHG cannot be less than two right angles. There- 
 fore they are equal to two right angles. 
 
 But this reasoning is certainly inconclusive. For why are we to sup- 
 pose that the interior angles which the parallels make with the line cutting 
 them, are either in every case greater than two right angles, or in every 
 case less than two right angles 1 For any thing that we are yet supposed 
 to know, they may be sometimes greater than two right angles, and some* 
 times less, and therefore we are not entitled to conclude, because the angles 
 AGH, CHG are greater than two right angles, that therefore the angles 
 BGH, DHG are also necessarily greater than two right angles. It 
 may safely be asserted, therefore, that Ptolemy has not succeeded in his 
 attempt to demonstrate the properties of parallel lines without the assist- 
 ance of a new Axiom. 
 
 Another attempt to demonstrate the same proposition without the assist- 
 ance of a npw Axiom has been made by a modern geometer, Franceschini
 
 NuTES. 
 
 293 
 
 F N 
 
 Professor of Mathematics in the University of Bologna, in an essay, wliicl. 
 he entitles, La Teorta delie parallele rigorosamente dimonstrata, printed in 
 his Opuscoli Mathematics at Bassano in 1787. 
 
 The difficulty is there reduced to a proposition nearly the same with this, 
 That if BE make an acute angle with BD, and if DE be perpendicular tc 
 BD at any point, BE and DE, 
 if produced, will meet. To de- 
 monstrate this, it is supposed, 
 that BO, BC are two parts taken 
 in BE, of which BC is greater 
 than BO, and that the perpendi- 
 culars ON, CL are drawn to BD ; 
 then shall BL be greater than 
 BN. For, if not, that is, if the 
 perpendicular CL falls either at 
 N, or between B and N, as at 
 F ; in the first of these cases the 
 angle CNB is equal to the angle ONB, because they are both right angles, 
 which is impossible ; and, in the second, the two angles CFN, CNF of the 
 triangle CNF, exceed two right angles. Therefore, adds our author, since, 
 as BC increases, BL also increases, and since BC may be increased with- 
 out limit, so BL may become greater than any given line, and therefore may 
 be greater than BD ; wherefore, since the perpendiculars to BD from points 
 beyond D meet BC, the perpendicular from D necessarily meets it. 
 
 Now it will be found, on examination, that this reasoning is no more 
 conclusive than the preceding. For, unless it be proved, that whatever 
 multiple BC is of BO, the same is BL of BN, the indefinite increase of 
 BC does not necessarily imply the indefinite increase of BL,or that BL may 
 be made to exceed BD. On the contrary, BL may always increase, and 
 yet may do so in such a manner as never to exceed BD : In order that the 
 demonstration should be conclusive, it would be necessary to shew, that 
 when BC increases by a part equal to BO, BL increases always by a part 
 equal to BN ; but to do this will be found to require the knowledge of those 
 very properties of parallel lines that we are seeking to demonstrate. 
 
 Legendre, in his Elements of Geometry, a work entitled to the highest 
 praise, for elegance and accuracy, has delivered the doctrine of parallel lines 
 without any new Axiom. He has done this in two different ways, one in 
 the text, and the other in the notes. In tho former he has endeavoured to 
 prove, independently of the doctrine of parallel lines, that all the angles of 
 a triangle are equal to two right angles ; from which proposition, when 
 it is once established, it is not difficult to deduce every thing with respect to 
 parallels. But, though his demonstration of the property of triangles just 
 mentioned is quite logical and conclusive, yet it has the fault of being long 
 and indirect, proving first, that the three angles of a triangle cannot be 
 greater than two right angles, next, that they cannot be less, and doing 
 both by reasoning abundantly subtle, and not of a kind readily apprehend- 
 ed by those who are only beginning to study the Mathematics. 
 
 The demonstration which he has given in the notes is extremely ingeni- 
 ous, and proceeds on this very simple and undeniable Axiom, that we can- 
 not compare an angle and a 'ine, as to magnitude, or cannot have an cqua-
 
 294 NOTES. 
 
 tion o* any sort between them. This truth is involved in the Jistinction 
 between homogeneous and heterogeneous quantities, (Euc. v. def. 4.), 
 which has long been received in Geometry, but led only to negative con- 
 sequetices, till it fell into the hands of Legendre. The proposition which 
 he deduces from it is, that if two angles of one triangle be equal to two an- 
 gles of another, the third angles of these triangles are also equal. For, it 
 is evident, that when two angles of a triangle are given, and also the side 
 between them, the third angle is thereby determined ; so that if A and B 
 be any two angles of a triangle, P the side interjacent, and C the third an- 
 gle, C is determined, as to its magnitude, by A, B and P ; and, besides 
 these, there is no other quantity whatever which can affect the magnitude 
 of C. This is plain, because if A, B and P are given, the triangle can be 
 constructed, all the triangles in which A, B and P are the same, being equal 
 to one another. 
 
 But of the quantities by which C is determined, P cannot be one ; for if 
 it were, then C must be a function of the quantities A, B, P ; that is to sav, 
 the value of C can be expressed by some combination of the quantities A, 
 B and P. An equation, therefore, may exist between the quantities A, B, 
 C and P ; and consequently the value of P is equal to some combination, 
 that is, to some function of the quantities A, B and C ; but this is impossi- 
 ble, P being a line, and A, B, C being angles ; so that no function of the 
 first of these quantities can be equal to any function of the other three. The 
 angle C must therefore be determined by tne angles A and B alone, without 
 any regard to the magnitude of P, the side interjacent. Hence in all trian- 
 gles that have two angles in one equal to two in another, each to each, the 
 third angles are also equal. 
 
 Now, this being demonstrated, it is easy to prove that the three angles of 
 any triangle are equal to two right angles. 
 
 Let ABC be a triangle right angled at A, draw AD perpendicular to 
 BC. The triangles ABD, ABC have the an- \ 
 
 gles BAC, BDA right angles, and the angle 
 B common to both; therefore by what has just 
 been proved, their third angles BAD, BCA are 
 also equal. In the same way it is shewn, that 
 CAD is equal to CBA ; therefore the two an- 
 gles, BAD, CAD are equal to the two BCA, 
 CBA ; but BAD+CAD is equal to a right B V ^ 
 
 angle, therefore the angles BCA, CBA are together equal to a right angle, 
 and consequently the three angles of the right angled triangle ABC are 
 equal to two right angles. 
 
 And since it is proved that the oblique angles of every right angled 
 triangle are equal to one right angle, and since every triangle may be 
 divided into two right angled triangles, the four oblique angles of which are 
 equal to the three angles of the triangle, therefore the three angles of every 
 triar.gle are equal to two right angles. 
 
 Though this method of treating the subject is strictly demonstrative, yet, 
 as the reasoning in the first of the two preceding demonstrations is not per- 
 haps sufficiently simple to be apprehended by those just entering on mathe- 
 matical studies, I shall submit to the reader another method, n'»t liable to 
 *^e same objection, which I know, from experience, to be 0/ u*f in explain
 
 NOTES. 295 
 
 ing the Elements. It proceeds, like that of the French Geometer by de 
 monstrating, in the first place, that the angles of any triangle are together 
 equal to two right angles, and deducing from thence, that two lines, whicL 
 make with a third line the interior angles, less than two right angles, must 
 meet if produced. The reasoning used to demonstrate the first of these 
 propositions may be objected to by some as involving the idea of motion, and 
 the transference of a line from one place to another. This, however, is no 
 more than Euclid has done himself on some occasions ; and when it furnish- 
 es so short a road to the truth as in the present instance, and does not im- 
 pair the evidence of the conclusion, it seems to be in no respect inconsistent 
 with the utmost rigour of demonstration. It is of importance in explaining 
 the Elements of Science, to connect truths by the shortest chain possible ; 
 and till that is done, we can never consider them as being placed in their 
 natural order. The reasoning in the first of the following propositions is so 
 simple, that it seems hardly susceptible of abbreviation, and it has the ad- 
 vantage of connecting immediately two truths so much alike, that one 
 might conclude, even from the bare enunciations, that they are but differeut 
 cases of the same general theorem, viz. That all the angles about a point, 
 and all the exterior angles of any rectilineal figure, are constantly of the 
 same magnitude, and equal to four right angles. 
 
 DEFINITION. 
 
 If, while one extremity of a straight line re- 
 mains fixed at A, the line itself turns about that 
 point from the position AB to the position AC, it 
 is said to describe the angle BAC contained by 
 the line AB and AC. 
 
 Cor. If a line turn about a point from the position AC till it come into 
 the position AC again, it describes angles which are together equal to four 
 right angles. This is evident from the second Cor. to the 15th. 1. 
 
 PROP, ft, 
 
 All the exterior angles of any rectilineal figure are together equal to four 
 right angles. 
 
 1. Let the rectilineal figure be the triangle ABC, of which the exterior 
 angles are DCA, FAB, GBC ; these angles are together equal to four 
 right angles. 
 
 Let the line CD, placed in the direction of BC produced, turn about the 
 point C till it coincide with CE, a part of the side CA, and have described 
 the exterior angle DCE or DCA. Let it then be carried along the line 
 CA, till it be in the position AF, that is, in the direction of CA produced, 
 and the point A remaining fixed, let it turn about A till it describe the 
 angle FAB, and coincido with a part of the line AH Let it next be car- 
 ried along AB till it come into the position BO, and by turning about B
 
 296 
 
 NOTES. 
 
 C D 
 
 ie* it describe the angle GBC, so 
 as to coincide with a part of BC. 
 Lastly, Let it be carried along BC 
 till it coincide with CD, its first 
 position. Then, because the line 
 CD has turned about one of its 
 extremities till it has come into 
 the position CD again, it has by 
 the corollary to the above defini- 
 tion described angles which are 
 together equal to four right an- 
 gles ; but the angles which it 
 has described are the three ex- 
 terior angles of the triangle ABC, 
 therefore the exterior angles of 
 the triangle ABC are equal to 
 four right angles. 
 
 2. If the rectilineal figure have any number of sides, the proposition is 
 demonstrated just as in the case of a triangle. Therefore all the exterior 
 angles of any rectilineal figure are together equal to four right angles. 
 
 Cor. 1. Hence, all the interior angles of any triangle are equal to two 
 right angles. For all the angles of the triangle, both exterior and interior, 
 are equal to six right angles, and the exterior being equal to four right 
 angles, the interior are equal to two right angles. 
 
 Cor. 2. An exterior angle of any triangle is equal to the two interior and 
 opposite, or the angle DCA is equal to the angles CAB, ABC. For the 
 angles CAB, ABC, BCA are equal to two right angles ; and the angles 
 ACD, ACB are also (13. 1.) equal to two right angles ; therefore the three 
 angles CAB, ABC, BCA are equal to the two ACD, ACB ; and taking 
 ACB from both, the angle ACD is equal to the two angles CAB, ABC. 
 
 Cor. 3. The interior angles of any rectilineal figure are equal to twice 
 as many right angles as the figure has sides, wanting four. For all the 
 angles exterior and interior are equal to twice as many right angles as the 
 figure has sides ; but the exterior are equal to four right angles ; therefore 
 the interior are equal to twice as many right angles as the figure has sides 
 wanting four. 
 
 PROP. II. 
 
 Two straight lines, which make with a third line the interior angles on 
 the same side of it less than two right angles, will meet on that side, if pro- 
 duced far enough. 
 
 Let the straight lines AB, CD, make with AC the two angles BAC, 
 DCA less than two right angles ; AB and CD will meet if producedtoward 
 B and D. 
 
 In AB take AF=AC ; join CF, produce BA to H, and through C draw 
 CE, making the angle ACE equal to the angle CAH. 
 
 Because 4C is equal to AF, the angles AFC, ACF aie also equal (5
 
 NOTES. 
 
 297 
 
 I.) ; but the exterior angle HAC is equal to the two interior and opposite 
 angles ACF, AFC, and therefore it is double of either of them, as of ACF 
 Now ACE is equal to HAC by construction, therefore ACE is double o» 
 ACF, and is bisected by the line CF. In the same manner, if FG be taker 
 equal to FC, and if CG be drawn, it may be shewn that CG bisects tho 
 angle FCE, and so on continually. But if from a magnitude, as the an- 
 gle ACE, there be taken its half, and from the remainder FCE its 
 half FCG, and from the remainder GCE its half, &c. a remainder will at 
 length be found less than the given angle DCE. # 
 
 TI A 
 
 Let GCE be the angle, whose half ECK is lnss than DCE, then a 
 straight line CK is found, which falls between CD and CE, but never- 
 theless meets the line AB in K. Therefore CD, if produced, must meet 
 AB in a point between G and K. 
 
 This demonstration is indirect ; but this proposition, if the definition of 
 parallels were changed, as suggested at p. 291, would not be necessary , 
 and the proof, that lines equally inclined to any one line must be so to 
 every line, would follow directly from the angles of a triangle being equal 
 to two right angles. The doctrine of parallel lines would in this manner 
 be freed from all difficulty. 
 
 PROP. III. or 29. I.Euclid. 
 
 If a straight line fall on two parallel straight lines, it makes the alternate 
 angles equal to one another ; the exterior equal to the interior and oppo- 
 site on the same side ; and likewise the two interior angles, on the same 
 side equal to two right angles. 
 
 Let the straight line EF fall on 
 the parallel straight lines AB, 
 CD ; the alternate angles AGH, 
 GHD are equal, the exterior angle 
 EGB is equal to the interior and 
 opposite GHD ; and the two inte- 
 rior angles BGH, GHD are equal 
 to two light angles. 
 
 For if AGH be not equal to 
 GHD, let it be greater, then add- 
 *ig BGH to both, the angles 
 AGH, HGB are greater than the 
 
 * Prop. 1. ] Sup. The reference of this proposition involves nothing incons stent witk 
 gocd reasoning, is the demonstration of it does not depend on any thing that has gone before. 
 «o that it m:') he inti Mr iced in any part of the Elements. 
 
 38
 
 198 NOTES. 
 
 angles DUG, HGB. But AGH, HGB are equal to two right angles (13. 
 1 .) ; therefore BGH, GHD are less than two right angles, and therefore the 
 lines AB, CD will meet, by the last proposition, if produced toward B and 
 D. But they do not meet, for they are parallel by hypotheses, and there- 
 fore the angles AGH, GHD are not unequal, that is, they are equal to one 
 another. 
 
 Now the angle AGH is equal to EGB, because these are vertical, and 
 it has also been shewn to be equal to GHD, therefore EGB and GHD ire 
 equal. Lastly, to each of the equal angles EGB, GHD add the angle 
 BGH, then the two EGB, BGH are equal to the two DHG, BGH. But 
 EGB, BGH are equal to two right angles (13. l.),therefore BGH, GHD 
 are also equal to two right angles. 
 
 The following proposition is placed here, because it is more connected 
 with the First Book than with any other. It is useful for explaining the 
 nature of Hadley's sextant; and, though involved in the explanations usual- 
 ly given of that instrument, it has not, I believe, been hitherto considered as 
 a distinct Geometrical Proposition, though very well entitled to be so on ac 
 count of its simplicity and elegance, as well as its utility. 
 
 THEOREM. 
 
 If an exterior angle of a triangle be bisected, and also one of the interior 
 and opposite, the angle contained by the bisecting lines is equal to half the 
 other interior and opposite angle of the triangle. 
 
 Let the exterior angle ACD of the triangle ABC be bisected by the 
 straight line CE, and the interior and opposite ABC by the straight line 
 BE, the angle BEC is equal to half the angle BAC. 
 
 The line CE, BE will meet ; for since the angle ACD is greater than 
 ABC, the half of ACD is greater than the half of ABC, that is, ECD 
 is greater than EBC ; add 
 
 EGB to both, and the two Jjg 
 
 angles ECD, ECB are A 
 
 greater than EBC, ECB. 
 But ECD, ECB are equal 
 to two right angles ; there- 
 fore ECB, EBC are less 
 than two right angles, and 
 therefore the lines CE, BE 
 must meet on the same side 
 of BC on which the trian 
 gle ABC is. Let them meet in E. 
 
 Because DCE is the exterior angle of the triangle BCE, it is equal to 
 the two angles CBE, BEC, and therefore twice the angle DCE, that is, the 
 angle DCA is equal to twice the angles CBE and BEC. But twice the 
 angle CBE is equal to the angle ABC, therefore the angle DCA is equal 
 to the angle ABC, together with twice the angle BEC ; an<l the same an
 
 NOTES. 299 
 
 glo DCA being the exterior angle of the triangle ABC, is equal to the two 
 angles ABC, CAB, wherefore the two angles ABC, CAB are coual to 
 ABC and twice BEC. Therefore, taking away ABC from both, there 
 remains the angle CAB equal to.twice the angle BEC, or BEC equal to 
 the half of B AC. 
 
 BOOK II. 
 
 The Demonstrations of this Book are no otherwise changed than by in- 
 troducing into them some characters similar to those of Algebra, which is 
 always of great use where the reasoning turns on the addition or subtrac- 
 tion of rectangles. To Euclid's demonstrations, others are sometimes add- 
 ed, as Scholiums, in which the properties of the sections of lines are easilv 
 demonstrated by Algebraical formulas. 
 
 BOOK III. 
 
 DEFINITIONS. 
 
 The definition which Euclid makes the first of this Book is that of equal 
 circles, which he defines to be " those of which the diameters are equal." 
 This is rejected from among the definitions, as being a Theorem, the truth 
 of which is proved by supposing the circles applied to one another, so that 
 their centres may coincide, for the whole of the one must then coincide with 
 the whole of the other. The converse, viz. That circles which are equal 
 have equal diameters, is proved in the same way. 
 
 The definition of the angle of a segment is also omitted, because it does 
 not relate to a rectilineal angle, but to one understood to be contained be- 
 tween a straight line and a portion of the circumference of a circle. In like 
 manner, no notice is taken in the 16th proposition of the angle comprehend- 
 ed between the semicircle and the diameter, which is said by Euclid to be 
 greater than an acute rectilineal angle. The reason for these omissions has 
 already been assigned in the notes on the fifth definition of the first Book 
 
 PROP. XX. 
 
 It has been remarked of this demonstration, that it takes for granted, tha 
 •f two magnitudes be double of two others, each of each, the sum or diffor 
 ence of the first two is double of the sum or difference of the other two, 
 trhich are two cases of the 1st and 5th of the 5th Book. The justness o'
 
 300 NOTES. 
 
 this remark cannot be denied ; and though the cases of the Propositions here 
 referred to are the simplest of any, yet the truth of them ought not in strict- 
 ness to be assumed without proof. The proof is easily given. Let A and 
 B, C and D be four magnitudes, such that A=2C, and B=2D ; then A 
 + B=2(C+D). For since A=C+C, and B=D+ D, adding equals to 
 equals, A + B=(C+D)+(C-|-D)=2(C+D). So also, if A be greater 
 than B, and therefore C greater than D, since A=C4-C, and B=D-|-D, 
 taking equals from equals, A — B=(C — D)+(C — D), that is, A — B=2 
 (C-D). 
 
 BOOK V. 
 
 The subject of proportion has been treated so differently by those who 
 have written on elementary geometry, and the method which Euclid has fol- 
 lowed has been so often, and so inconsiderately censured, that in these notes 
 it will not perhaps be more necessary to account for the changes that I have 
 made, than for those that I have not made. The changes are but few, and 
 relate to the language, not to the essence of the demonstrations ; they will 
 be explained after some of the definitions have been particularly considered 
 
 DEF. III. 
 
 The definition of ratio given here has been greatly extolled by some au- 
 thors ; but whatever value it may have in the eyes of a metaphysician, it 
 has but little in those of the geometer, because nothing concerning the pro- 
 perties of ratios, can be deduced from it. Dr. Barrow has very judiciously 
 remarked concerning it, " that Euclid had probably no other design in mak- 
 " ing this definition, than to give a general summary idea of ratio to begin- 
 " ners, by premising this metaphysical definition to the more accurate defi- 
 " nitions of ratios that are equal to one another, or one of which is greater 
 " or less than the other ; I call it a metaphysical, for it is not properly a ma- 
 " thematical definition, since nothing in mathematics depends on it, or is de- 
 " duced, nor, as I judge, can be deduced, from it." (Barrow's Lectures, 
 Lect. 3.) Dr. Simson thinks the definition has been added by some unskil- 
 ful editor ; but there is no ground for that supposition, other than what ari- 
 ses from the definition being of no use. We may, however, well enough 
 imagine, that a certain idea of order and method induced Euclid to give 
 some general definition of ratio before he used the term in the definition of 
 equal ratios. 
 
 DEF. IV. 
 
 This definition is a little altered in the expression ; Euclid has it, tha. 
 magnitudes are said to have a ratio to one another, when the »ess cap be 
 multiplied so as to exceed the greater "
 
 NOTES. 
 
 D T '/. V. 
 
 One of the chief obstacles to the ready understanding of the 5th Book of 
 Euclid, is the difficulty that most people find of reconciling the idea of pro- 
 portion which they have already acquired, with the account of it that is 
 given in this definition. Our first ideas of proportion, or of proportionality, 
 are got by trying to compare together the magnitude of external bodies ; 
 and though they be at first abundantly vague and incorrect, they are usually 
 rendered tolerably precise by the study of arithmetic ; from which we learn 
 to call four numbers proportionals, when they are such that the quotient 
 which arises from dividing the first by the second, (according to the com- 
 mon rule for division), is the same with the quotient that arises from divid- 
 ing the third by the fourth. 
 
 Now, as the operation of arithmetical division is applicable as readily to 
 any two magnitudes of the same kind, as to two numbers, the notion of pro- 
 portion thus obtained .nay be considered as perfectly general. For, in arith- 
 metic, after finding how often the divisor is contained in the dividend, we 
 multiply the remainder by 10, or 100, or 1000, or any power, as it is called, 
 of 10, and proceed to inquire how oft the divisor is contained in this new 
 dividend ; and, if there be any remainder, we go on to multiply it by 10, 
 100, &c. as before, and to divide the product by the original divisor, and so 
 on, the division sometimes terminating when no remainder is left, and some- 
 times going on ad infinitum, in consequence of a remainder being left at each 
 operation. Now, this process may easily be imitated with any two mag- 
 nitudes A and B, providing they be of the same kind, or such that the one 
 can be multiplied so as to exceed the other. For, suppose that B is the 
 least of the two ; take B out of A as oft as it can be found, and let the quo- 
 tient be noted, and also the remainder, if there be any ; multiply this remain- 
 der by 10, or 100, &c. so as to exceed B, and let B be taken out of the quan- 
 tity produced by this multiplication as oft as it can be found ; let the quotient 
 be noted, and also the remainder, if there be any. Proceed with this remain- 
 der as before, and so on continually ; and it is evident, that we have an opera- 
 tion that is applicable to all magnitudes whatsoever, and that may be perform- 
 ed with respect to any two lines, any two plane figures, or any two solids, &c. 
 
 Now, when we have two magnitudes and two others, and find that the 
 first divided by the second, according to this method, gives the very same 
 series of quotients that the third does when divided by the fourth, we say of 
 these magnitudes, as we did of the numbers above described, that the first 
 is to the second as the third to the fourth. There are only two more cir- 
 cumstances necessary to be considered, in order to bring us precisely to 
 Euclid's definition. 
 
 First, It is known from arithmetic, that the multiplication of the succes- 
 sive remainders each of them by 10, is equivalent to multiplying the quantity 
 to be divided by the product of all those tens ; so that multiplying, for in- 
 stance, the first remainder by 10, the second by 10, and the third by 10, is. 
 the same thing, with respect to the quotient, as if the quantity to be divided 
 had beer at first multiplied by 1000 ; and therefore, our standard of the pro- 
 portionality of numbers may be expressed thus : If the first multiplied any 
 ^u"nber of times by 10, and then divided by the second, gives the same quo-
 
 302 NOTES. 
 
 tienf as when the third is muliplied as often by 10, and then divided by the 
 fourth, the four magnitudes are proportionals. 
 
 Again, it is evident, that there is no necessity in these multiplications for 
 confining ourselves to 10, or the powers of 10, and that we do so, in arith- 
 metic, only for the conveniency of the decimal notation ; we may therefore 
 use any multipliers whatsoever, providing we use the same in both cases. 
 Hence, we have this definition of proportionals, When there are four mag- 
 nitudes, and any multiple whatsoever of the first, when divided by the 
 second, gives the same quotient with the like multiple of the third, when 
 divided by the fourth, the four magnitudes are proportionals, or the first 
 has the same ratio to the second that the third bas to the fourth. 
 
 We are now arrived very nearly at Euclid's definition ; for, let A, B, C, 
 D be four proportionals, according to the definition just given, and m any 
 number ; and let the multiple of A by m, that is mA, be divided by B ; and 
 first, let the quotient be the number n exactly, then also, when mC is divided 
 by D, the quotient will be n exactly. But when mA divided by B gives n 
 for the quotient, mA=nB by the nature of division, so that when mA=nB, 
 mC=/iD, which is one of the conditions of Euclid's definition. 
 
 Again, when mA is divided by B, let the division not be exactly perform- 
 ed, but let n be a whole number less than the exact quotient, then nB /_ 
 mA, or mAynB ; and, for the same reason, mC"/nD, which is another of 
 the conditions of Euclid's definition. 
 
 Lastly, when mA is divided by B,let nbea whole number greater than 
 the exact quotient, then mA^nB, and because n is also greater than the 
 quotient of mC divided by D, (which is the same with the other quotient), 
 therefore mC^/«D. 
 
 Therefore, uniting all these three conditions, we call A, B, C, D, propor- 
 tionals, when they are such, that if mA/wB, mC/nD ; if ?nA=nB, mC = 
 nD ; and if mA/_nB, mC/nD, m and n being any numbers whatsoever. 
 Now, this is exactly the criterion of proportionality established by Euclid in 
 the 5th definition, and is derived here by generalizing the common and most 
 familiar idea of proportion. 
 
 It appears from this, that the condition of mA containing B, whether 
 with or without a remainder, as often as mC contains D, with or without a 
 remainder, and of this being the case whatever value be assigned to the 
 number m, includes in it all the three conditions that are mentioned in Eu- 
 clid's definition ; and hence, that definition may be expressed a little more 
 simply by saying, that four magnitudes are proportionals, when any multiple of 
 the first contains the second, {with or without remainder,) as oft as the same mul- 
 tiple of the third contains the fourth. But, though this definition is certainly, 
 in the expression, more simple than Euclid's, it is not, as will be found on 
 trial, so easily applied to the purpose of demonstration. The three conditions 
 which Euclid brings together in his definition, though they somewhat em- 
 barrass the expression of it, have the advantage of rendering the demon- 
 strations more simple ihan they would otherwise be, by avoiding all discus- 
 sion about the magnitude of the remainder left, after B is taken out of mA aa 
 oft as it can be found. All the attempts, indeed, that have been made to de- 
 monstrate the properties of proportionals rigorously, by means of other defini 
 tions than Euclid's, only serve to evince the excellence of the method follow 
 ed by the Greek Geometer, and his singular address in the application of \t
 
 NOTES. 303 
 
 The great objection to the other methods is, that if they are meant to be 
 rigorous, they require two demonstrations to every proposition, one when 
 the division of mk into parts equal to B can be exactly performed, the otner 
 when it cannot be exactly performed whatever value be p.ssjgned to m, or 
 when A and B are what is called incommensurable ; and this last case will 
 generally be found to require an indirect demonstration, or a reductio adab- 
 surdum. 
 
 M. D'Alembert, speaking of the doctrine of proportion, in a discourse 
 that contains many excellent observations, but in which he has overlooked 
 Euclid's manner of treating this subject entirely, has the following remark : 
 " On ne peut demontrer que de cette manicre, (la reduction a absurde,) la 
 •' plupart des propositions qui regardent les incommensurables. L'id6e de 
 " l'infini entre au moins implicitemens dans la notion de ces sortes de quan- 
 " tites ; et comme nous n'avons qu'une id6c negative de Pinfini, on ne peut 
 " demontrer directement, et a priori, tout ce qui concerne l'infini mathema- 
 " tique." (Encyclopidie, mot GeomHrie.) 
 
 This remark sets in a strong and just light the difficulty of demonstrating 
 the propositions that regard the proportion of incommensurable magnitudes, 
 without having recourse to the reductio ad absurdum : but it is surprising, 
 that M. D'Alembert, a geometer no les? learned than profound, should 
 have neglected to make mention of Euclid's method, the only one in which 
 the difficulty he states is completely overcome. It is overcome by the in- 
 troduction of the idea of indefinitude, (if I may be permitted to use the word), 
 instead of the idea of infinity ; for m and n, the multipliers employed, are 
 supposed to be indefinite, or to admit of all possible values, and it is by the 
 skilful use of this condition that the necessity of indirect demonstrations is 
 avoided. In the whole of geometry, I know not that any happier invention 
 is to be found ; and it is worth remarking, that Euclid appears in another 
 of his works to have availed himself of the idea of indefinitude with the 
 same success, viz. in his books of Porisms, which have been restored by 
 Dr. Simson.and in which the whole analysis turned on that idea, as I have 
 shown at length in the Third Volume of the Transactions of the Royal So- 
 ciety of Edinburgh. The investigations of these propositions were founded 
 entirely on the principle of certain magnitudes admitting of innumerable 
 values ; and the methods of reasoning concerning them seem to have been 
 extremely similar to those employed in the fifth of the Elements. It is 
 curious to remark this analogy between the different works of the same 
 author ; and to consider, that the skill, in the conduct of this very refined 
 and ingenious artifice, acquired in treating the properties of proportionals, 
 may have enabled Euclid to succeed so well in treating the still more dif- 
 ficult subject of Porisms. 
 
 Viewing in this light Euclid's manner of treating proportion, I had no 
 desire to change any thing in the principle of his demonstrations. I have 
 jnly sought to improve the language of them, by introducing a concise 
 mode of expression, of the same nature with that which we use in arith- 
 metic, and in algebra. Ordinary language conveys the ideas of the diffe- 
 rent operations supposed to be performed in these demonstrations so slowly, 
 and breaks them down into so many parts, that they make not a sufficient 
 impression on the understanding. This indeed will generally happen when 
 the things treated of are not represented to the senses by Diagrams, nv
 
 304 NOTES. 
 
 they cannot be when we reason concerning magnitude in general, as in this 
 part of the Elements. Here we ought certainly to adopt the language of 
 arithmetic or algebra, which by its shortness, and the rapidity with which 
 it places objects before us, makes up in the best manner possible for being 
 merejy a conventional language, and using symbols that have no resem- 
 blance to the things expressed by them. Such a language, therefore, I 
 have endeavoured to introduce here ; and I am convinced, that if it shall 
 be found an improvement, it is the only one of which the fifth of Euclid will 
 admit. In other respects I have followed Dr. Simson's edition to the accu- 
 racy of which it would be difficult to make any addition 
 
 In one thing I must observe, that the doctrine of proportion, as laid down 
 here, is meant to be more general than in Euclid's Elements. It is intended 
 to include the properties of proportional numbers as well as of all magni- 
 tudes. Euclid has not this design, for he has given a definition of propor- 
 tional numbers in the seventh Book, very different from that of proportional 
 magnitudes in the fifth; and it is not easy to justify the logic of this man 
 ner of proceeding ; for we can never speak of two numbers and two magni- 
 tudes both having the same ratios, unless the word ratio have in both cases 
 the same signification. All the propositions about proportionals here 
 given are therefore understood to be applicable to numbers ; and accord- 
 ingly, in the eighth Book, the proposition that proves equiangular parallelo- 
 grams to be in a ratio compounded of the ratios of the numbers proportional 
 to their sides, is demonstrated by help of the propositions of the fifth Book. 
 
 On account of this, the word quantity, rather than magnitude, ought in strict- 
 ness to have been used in the enunciation of these propositions, because we 
 employ the word Quantity to denote not only things extended, to which 
 alone we give the name of Magnitude, but also numbers. It will be suffi- 
 cient, however, to remark, that all the propositions respecting the ratios of 
 magnitudes relate equally to all things of which multiples can be taken, that 
 is, to all that is usually expressed by the word Quantity in its most extend- 
 ed signification, taking care always to observe, that ratio takes place only 
 among like quantities, (See Def. 4.) 
 
 DEF. X. 
 
 The definition of compound ratio was first given accurately by Dr. Simson , 
 for, though Euclid used the term, he did so without defining it. I have 
 placed this definition before those of duplicate and triplicate ratio, as it is in 
 fact more general, and as the relation of al' 'he three definitions is best seen 
 when they are ranged in this order. It is en plain, that two equal ratios 
 compound a ratio duplicate of either of th /n ; three equal ratios, a ratio 
 triplicate of either of them, &c. 
 
 It was justly observed by Dr. Simson, that the expression, compound ratio, 
 is introduced merely to prevent circumlocution, and for the sake principally 
 of enunciating those propositions with conciseness that are demonstrated by 
 reasoning ex a>quo, that is, by reasoning from the 22d or 23d of this Book 
 This will be evident to any one who considers carefully the Prop. F. of this, 
 or the 23d of the 6th Book 
 
 An objection which naturally occurs to the use of the term compound ratw. 
 arises from its not being evident how the ratios described in the definition
 
 NOTES. 305 
 
 determine in any way the ratio which they are said to compound, since the 
 magnitudes compounding them are assumed at pleasure. It may be of use 
 for removing this difficulty, to state the matter as follows : if there be any 
 number of ratios (among magnitudes of the same kind) such that tho con- 
 sequent of any of them is the antecedent of that which immediately fol 
 lows, the first of the antecedents has to the last of the consequents a ratio 
 which evidently depends on the- intermediate ratios, because if they are de- 
 termined, it is determined also ; and this dependence of one ratio on all the 
 other ratios, is expressed by saying that it is compounded of them. Thus, 
 
 if — , — -, — , -ft, be any series of ratios, such as described above, the ratio 
 13 C L) iii 
 
 A A R 
 
 -=-, or of A to E, is said to be compounded of the ratios -=-, -~, &c. The ratio 
 
 A A B 
 
 tt, is evidently determined by the ratios -=-, — , &c. because if each of tho 
 
 latter is fixed and invariable, the former cannot change. The exact nature 
 of this dependence, and how the one thing is determined by the other, it is 
 not the business of the definition to explain, but merely to give a name to 
 a relation which it may be of importance to consider more attentively 
 
 BOOK VI. 
 
 DEFINITION II. 
 
 This definition is changed from that of reciprocal figures, which was ol no 
 use, to one that corresponds to the language used in the 14th and 15th 
 propositions, and in other parts of geometry. 
 
 PROP. A, B, C, &c. 
 
 Nine propositions are added to this Book on account of their utility and 
 their connection with this part of the Elements. The first four of them are 
 in Dr. Simson's edition, and among these Prop. A is given immediately 
 after the third, being, in fact, a second case of that proposition, and capable 
 of being included with it, in one enunciation. Prop. D is remarkable for 
 being a theorem of Ptolemy the Astronomer, in his Meyalrj Hvyia^ig, and the 
 foundation of the construction of his trigonometrical tables. Prop. E is the 
 simplest case of the former ; it is also useful in trigonometry, and, under 
 another form, was the 97th, or, in some editions, the 94th of Euclid's Data. 
 The propositions F and G are very useful properties of the circle, and are 
 taken from the Loci Plani of Apollonius. Prop. H is a very remarkable pro- 
 perty of the triangle ; and K is a proposition which, though it has been 
 hitherto considered as belonging particularly to trigonometry, is so often of 
 use in other parts of the mathematics, that it may be properly ranked among 
 elementary theorems of Geometry. 
 
 39
 
 SUPPLEMENT. 
 
 BOOK I. 
 
 PROP. V. and VI, &c. 
 
 The demonstrations of the 5th and 6th propositions require the method 
 of exhaustions, that is to say, they prove a certain property to belong to the 
 circle, because it belongs to the rectilineal figures inscribed in it, or described 
 about it according to a certain law, in the case when those figures ap- 
 proach to the circles so nearly as not to fall short of it or to exeeed it, by 
 any assignable difference. This principle is general, and is the only one 
 by which we can possibly compare curvilineal with rectilineal spaces, or the 
 length of curve lines with the length of straight lines, whether we follow 
 the methods of the ancient or of the modern geometers. It is therefore a 
 great injustice to the latter methods to represent them as standing on a foun- 
 dation less secure than the former ; they stand in reality on the same, and 
 the only difference is, that the application of the principle, common to them 
 both, is more general and expeditious in the one case than in the other. 
 This identity of principle, and affinity of the methods used in the elementary 
 and the higher mathematics, it seems the most necessary to observe, that 
 6ome learned mathematicians have appeared not to be sufficiently aware of 
 it, and have even endeavoured to demonstrate the contrary. An instance 
 of this is to be met with in the preface of the valuable edition of the works 
 of Archimedes, lately printed at Oxford. In that preface, Torelli, the learn- 
 ed commentator, whose labours have done so much to elucidate the writ- 
 ings of the Greek Geometer, but who is so unwilling to acknowledge the 
 merit of the modern analysis, undertakes to prove, that it is impossible, from 
 the relation which the rectilineal figures inscribed in, and circumscribed 
 about, a given curve have to one another, to conclude any thing concerning 
 the properties of the curvilineal space itself, except in certain circumstances 
 which he has not precisely described. With this view he attempts to show, 
 that if we are to reason from the relation which certain <;ctilineal figure* 
 belonging to the circle have to one another, notwithstanding that those 
 figures may approach so near to the circular spaces within which they are 
 inscribed, as not to differ from them by any assignable magnitude, we shall 
 be led into error, and shall seem to prove, that the circle is to the square of 
 its diameter exactly as 3 to 4. Now, as this is a conclusion which the dis- 
 coveries of Archimedes himself prove so clearly to be false, Torelli argues, 
 that the principle from which it is deduced must be false also ; and in this 
 he would no doubt be right, if his former conclusion had been fairly drawn. 
 But the truth is, that a very gross paralogism is to be found in that part of
 
 NOTES. SUPPL. BOOK I. 307 
 
 his reasoning, where he makes a transition from the ratios of the small rect. 
 angles, inscribed in »he circular spaces, to the ratios of the sums of thoso 
 rectangles, or of the whole rectilineal figures. In doing this, he takes fo. 
 granted a proposition, which, it is wonderful, that one who had studie* 
 geometry in the school of Archimedes, should for a moment have suppos 
 ed to be true. The proposition is this : If A, B, C, D, E, F, be any num- 
 ber of magnitudes, and a, b, t, d, e,f, as many others ; and if 
 A : B : : a : b, 
 C : D : : c : d, 
 
 E : F : : e : /, then the sum of A, C and E will be to the sum of B, D and 
 F, as the sum of a, c and e, to the sum of b, d and/, or A+C+E : B+D 
 +F : : a+c-\-e : b-{-d-\-f. Now, this proposition, which Torelli supposes 
 to be perfectly general, is not true, except in two cases, viz. either first, 
 when A : C : : a : c, and 
 
 A : E : : a : e ; and consequently, 
 B : D : : b : d t and 
 
 B : F : : b : f\ or, secondly, when all the ratios of A to B, C to D, E 
 to F, &c. are equal to one another. To demonstrate this, let us suppose 
 that there are four magnitudes, and four others, 
 
 thus A : B : : a : b, and 
 
 C : D : : c : d, then we cannot have 
 A+C : B+D : : a+c : &+^> unless either A : C : : a : c, and B : D : : b : 
 d; or A : C : : b : d, and consequently a : b : : c : d. 
 
 Take a magnitude K, such that a : c : : A : K, and another L, such that 
 b : d : : 1 1 : L ; and suppose it true, that A + C : B + D : : 
 a+c : b+d. Then, because by inversion ; K : A : : c : a, |~k, A, B, L, I 
 and, by hypothesis, A : B : : a : b, and also B : L : : b : d, Cj flj £ j\ 
 
 ex aequo, K : L : : c : d ; and consequently, K : L : : ' 
 
 C : D. 
 
 Again, because A : K : : a : c, by addition, 
 
 A+K : K : : a-\-c : c ; and for the same reason, 
 B-f-L : L : : b+d : d, or, by inversion, 
 L : B+L : : d : b-\-d. And, since it has been shewn, that 
 K : L : : c : d ; therefore, ex aequo, 
 
 A+K,K,L,B+L, 
 a-j-e, c, d, b-\-d. 
 
 A+K : B + L : : a-f c : b-\-d; but by hypothesis, 
 
 A+C : B+D : : a+c : b+d, therefore 
 
 A+K: A+C :: B+L: B + D. 
 Now, first, let K and C be supposed equal, then it is evident that L and 
 D are also equal ; and therefore, since by construction a : c : : A : K, we 
 have also a : c : : A : C ; and, for the same reason, b : d : : B : D, and 
 ihese analogies from the first of the two conditions, of which one is affirmed 
 above to be always essential to the truth of Torelli's proposition 
 Naxt, if K be greater than C, then, since 
 
 A+K : A+C : : B + L : B+D, by division, 
 
 A+K: K— C :: B + L : L— D. But, as was shewn, 
 
 K : L : : C : D, by conversion and alternation, 
 
 K— C : K : : L— D : L, therefore, ex aequo,
 
 308 NOTES. SUPPL. BOOK IT. 
 
 A -f-K : K : : B + L : L, and lastly, by division, 
 A : K : : B : L, or A : B : : K : L, that is, 
 A : B • : C : D. 
 
 Wherefore, in this case the ratio of A to B is equal to that of C to D 
 and consequently, the ratio of a to b equal to that of c to d. The same 
 may be shewn, if K is less than C ; therefore in every case there are con- 
 ditions necessary to the truth of Torelli's proposition, which he does not 
 take into account, and which, as is easily shewn, do not belong to the mag 
 nitudes to which he applies it. 
 
 In consequence of this, the conclusion which he meant to establish re 
 specting the circle, falls entirely to the ground, and with it the general in- 
 ference aimed against the modern analysis. 
 
 It will not, I hope, be imagined, that I have taken notice of these cir- 
 cumstances with any design to lessen the reputation of the learned Italian, 
 who has in so many respects deserved well of the mathematical sciences, 
 or to detract from the value of a posthumous work, which by its elegance 
 and correctness, does so much honour to the English editors. But I would 
 warn the student against that narrow spirit which seeks to insinuate itself 
 even into the abstractions of geometry, and would persuade us, that ele- 
 gance, nay, truth itself, are possessed exclusively by the ancient methods 
 of demonstration. The high tone in which Torelli censures the modern ma- 
 thematics is imposing, as it is assumed by one who had studied the writings 
 of Archimedes with uncommon diligence. His errors are on that account 
 the more dangerous, and require to be the more carefully pointed out. 
 
 PROP. IX. 
 
 This enunciation is the same with that of the third of the Dimensio Cir- 
 culi of Archimedes ; but the demonstration is different, though it proceeds 
 like that of the Greek Geometer, by the continual bisection of the 6th part 
 of the circumference. 
 
 The limits of the circumference are thus assigned; and the method of 
 bringing it about, notwithstanding many quantities are neglected in the arith- 
 metical operations, that the errors shall in one case be all on the side of de- 
 fect, and in another all on the side of excess (in which I have followed Ar- 
 chimedes,) deserves particularly to be observed, as affording a good intro 
 duction to the general methods of approximation. 
 
 BOOK II. 
 
 DEF. VIII. and PROP. XX 
 
 Solid angles, which are defined here in the same manner as in Euclid, 
 are magnitudes of a very peculiar kind, and are particularly to be remarked 
 for not admitting of that accurate comparison, one with another, which is
 
 NOTES. SUPPL. BOOK II. 309 
 
 common in the other subjects of geometrical investigation. It cannot, for 
 example, be said of one solid angle, that it is the half, or the double of an- 
 other solid angle ; nor did any geometer ever think of proposing the pro 
 blem of bisecting a given solid angle. In a word, no multiple or sub-mul 
 tiple of such an angle can be taken, and we have no way of expounding 
 even to tho simplest cases, the ratio which one of them bears to another 
 
 In this respect, therefore, a solid angle differs from every other magni 
 tudc that is the subject of mathematical reasoning, all of which have this 
 common property, that multiples and sub-multiples of them may be found. 
 It is not our business here to inquire into the reason of this anomaly, but it 
 is plain, that on account of it, our knowledge of the nature and the proper- 
 ties of such angles can never be very far extended, and that our reason- 
 ings concerning them must be chiefly confined to the relations of the plane 
 angles', by which they are contained. One of the most remarkable of those 
 relations is that which is demonstrated in the 21st of this Book, and which 
 is, that all the plane angles which contain any solid angle must together 
 be less thin lour right angles. This proposition is the 21st of the 11th of 
 Euclid. 
 
 This proposition, however, is subject to a restriction in certain cases, 
 which, 1 believe, was first observed by M. le Sage of Geneva, in a com- 
 munication to the Academy of Sciences of Paris in 1756. When the sec- 
 tion of the pyramid formed by the planes that contain the solid angle is a 
 figure that has none of its angles exterior, such as a triangle, a parallelo- 
 gram, «fcc. the truth of the proposition just enunciated cannot be question 
 ed. But, when the aforesaid section is a figure like that which is annexed, 
 viz. ABC1), having some angles such 
 as B DC, exterior, or, as they are some- 
 times called, re-entering angles, the 
 proposition is not necessarily true , 
 and it is plain, that in such cases the 
 demonstration which we have given, 
 and which is the same with Euclid's, 
 will no longer apply. Indeed, it were 
 easy to show, that on bases of this 
 kind, by multiplying the number of 
 sides, soiid angles maybe formed, such 
 that the plane angles which contain them shnll exceed four right ang! 
 any quantity assigned. An illustration of this from the properties of the 
 sphere is perhaps the simplest of all others. Suppose that on the surface 
 of a hemisphere there is described a figure bounded by any number of arcs 
 of great circles making angles with one another, on opposite sides alter- 
 nately, the plane angles at the centre of the sphere that stand on these arcs 
 may evidently exceed four right angles, and that too, by multiplying and 
 extending the arcs in any assigned ratio. Now, these plane angles con- 
 tain a solid angle at the centre of the sphere, according to the definition of 
 •i. solid angle. 
 
 We are to understand the proposition in the text, therefore, to be tru« 
 only of those solid angles in which the inclination of the plane angli 
 all the same way, or all directed toward the interior of the figure. To dis- 
 tinguish this class of solid angles from that to which the proposition aoer
 
 110 NOTES. SUPPL. BOOK II. 
 
 not apply it is perhaps best to make use of this criterion, that they are such 
 that when any two points whatsoever are taken in the planes that contain 
 the Bolid angle, the straight line, joining those points, falls wholly within 
 the sobd angle : or thus, they are such, that a straight line cannot meet the 
 planes which contain them in more than two points. It is thus, too, that I 
 would distinguish a plane figure that has none of its angles exterior, by 
 saying, that it is a rectilineal figure, such that a straight line cannot meet 
 the boundary of it in more than two points. 
 
 We, therefore, distinguish solid angles into two species : one in which 
 the bounding planes can be intersected by a straight line only in two 
 points ; and another where the bounding planes may be intersected by a 
 straight line in more than two points : to the first of these the proposition 
 in the text applies, to the second it does not. 
 
 Whether Euclid meant entirely to exclude the consideration of figuies 
 of the latter kind, in all that he has said of solids, and of solid angles, it is 
 not now easy to determine : it is certain, that his definitions involve no 
 such exclusion ; and as the introduction of any limitation would conside- 
 rably embarrass these definitions, and render them difficult to be understood 
 by a beginner, I have left it out, reserving to this place a fuller explanation 
 of the difficulty. I cannot conclude this note without remarking, with the 
 historian of the Academy, that it is extremely singular, that not one of all 
 those who had read or explained Euclid before M. le Sage, appears to 
 have been sensible of this mistake. (Memoires ds VAcad. des Sciences, 
 1756, Hist. p. 77.) A circumstance that renders this still more singular 
 is, that another mistake of Euclid on the same subject, and perhaps of all 
 other geometers, escaped M. le Sage also, and was first discovered bv 
 Dr. Simson, as will presently appear. 
 
 PROP. IV. 
 
 This very elegant demonstration is from Legendre, and is much easiei 
 than that of Euclid. 
 
 The demonstration given here of the 6th is also greatly simpler that 
 that of Euclid. It has even an advantage that does not belong to Legen 
 dre's, that of requiring no particular construction or determination of any 
 one of the lines, but reasoning from properties common to every part o? 
 them. The simplification, when it can be introduced, which, however 
 does not appear to be always possible, is, perhaps, the greatest improve 
 ment that can be made on an elementary demonstration. 
 
 PROP. XIX. 
 
 The problem contained in this proposition, of drawing a straight line per 
 pendicular to two straight lines not in the same plane, is certainly to be ac- 
 counted elementary, although not given in any book of elementary geome- 
 try that I know of before that of Legendre. The solution given here is 
 more simple than his, or than any other that I have yet met with : it also 
 leads more easily, if it be required, to a trigonometrical computation.
 
 NOTES. SUPPL. BOOK III. 3.M 
 
 BOOK III. 
 
 DEF. II. and PROP. I. 
 
 These relate to similar and equal solids, a subject on which mistakes hav# 
 prevailed not unlike to that which has just been mentioned. The equality 
 of solids, it is natural to expect, must be proved like the equality of plane 
 figures, by showing that they may be made to coincide, or to occupy the 
 same space. But, though it be true that all solids which can be shewn to 
 coincide are equal and similar, yet it does not hold conversely, that all solids 
 which are equal and similar can be made to coincide. Though this asser- 
 tion may appear somewhat paradoxical, yet the proof of it is extremely 
 simple. 
 
 Let ABC be an isosceles triangle, of which the equal sides are AB and 
 AC ; from A draw A E perpendicular to the base BC, and BC will be bisected 
 in E. From E draw ED perpendicular to the 
 plane ABC, and from D, any point in it, draw 
 DA, DB, DC to the three angles of the tri- 
 angle ABC. The pyramid DABC is divided 
 into two pyramids DABE, DACE, which, 
 though their equality will not be disputed, 
 cannot be so applied to one another as to coin- 
 cide. For, though the triangles ABE, ACE 
 are equal, BE being equal to CE, EA common 
 to both, and the angles AEB, AEC equal, be- 
 cause they are right angles, yet if these two 
 triangles be applied to one another, so as to 
 coincide, the solid DACE will nevertheless, 
 as is evident, fall without the solid DABE, for the two solids will be on the 
 opposite sides of the plane ABE. In the same way, though all the planes 
 of the pyramid DABE may easily be shewn to be equal to those of the py- 
 ramid DACE, each to each ; yet will the pyramids themselves never coin- 
 cide, though the equal planes be applied to one another, because they are 
 on the opposite sides of those planes. 
 
 It may be said, then, on what ground do we conclude the pyramids to 
 be equal 1 The answer is, because their construction is entirely the same, 
 and the conditions that determine the magnitude of the one identical with 
 those that determine the magnitude of the other. For the magnitude of 
 the pyramid DABE is determined by the magnitude of the triangle ABE, 
 the length of the line ED, and the position of ED, in respect of the plane 
 ABE ; three circumstances that are precisely the same in the two pyra- 
 mids, so that there is nothing that can determine one of them to be greater 
 than another 
 
 This reasoning appears perfectly conclusive and satisfactory ; and it 
 aeems also very certain, that there is no other principle equally simple, on 
 which the relation of the solids DABE, DACE to one another can be de- 
 termined. Neither is this a case that occurs rarely ; it is one, that, in the 
 comparison of magnitudes having three dimensions, presents itself cunti
 
 312 NOTES. SUPPL. BOOK III. 
 
 mially ; for, (.hough two plane figures that are equal and similar can always 
 be made to < oincide, yet, with regard to solids that are equal and similar, L 
 they have not a certain similarity in their position, there will be found iust 
 as many cases in which they cannot, as in which they can coincide. Even 
 figure." described on surfaces, if they are not plane surfaces, may be equal 
 and similar without the possibility of coinciding. Thus, in the figure de- 
 scribed on the surface of a sphere, called a spherical triangle, if we suppose 
 it to be isosceles, and a perpendicular to be drawn from the vertex on the 
 base, it will not be doubted, that it is thus divided into two right angled 
 spherical triangles equal and similar to one another, and which, neverthe- 
 less, cannot be so laid on one another as to agree. The same holds in in- 
 numerable other instances, and therefore it is evident, that a principle, more 
 general and fundamental than that of the equality of coinciding figures, 
 ought to be introduced into Geometry. What this principle is has also ap- 
 peared very clearly in the course of these remarks ; and it is indeed no 
 other than the principle so celebrated in the philosophy of Leibnitz, under 
 the name of the sufficient reason. For it was shewn, that the pyra- 
 mids DABE and DACE are concluded to be equal, because each of them 
 is determined to be of a certain magnitude, rather than of any other, by 
 conditions that are the same in both, so that there is no reason for the one 
 oeing greater than the other. This Axiom may be rendered general by 
 saying, That things of which the magnitude is determined by conditions 
 ihat are exactly the same, are equal to one another ; or, it might be ex- 
 pressed thus ; Two magnitudes A and B are equal, when there is no rea- 
 son that A should exceed B, rather than that B should exceed A. Either 
 of these will serve as the fundamental principle for comparing geometrical 
 magnitudes of every kind ; they will apply in those cases where the coin- 
 cidence of magnitudes with one another has no place ; and they will apply 
 with great readiness to the cases in which a coincidence may take place, 
 such as in the 4th, the 8th, or the 26th of the First Book of the Ele- 
 ments. 
 
 The only objection to this Axiom is, that it is somewhat of a metaphy- 
 sical kind, and belongs to the doctrine of the sufficient reason, which is looked 
 on with a suspicious eye by some philosophers. But this is no solid objec- 
 tion ; for such reasoning may be applied with the greatest safety to those 
 objects with the nature of which we are perfectly acquainted, and of which 
 we have complete definitions, as in pure mathematics. In physical ques 
 tions, the same principle cannot be applied with equal safety, because iu 
 such cases we have seldom a complete definition of the thing we reason 
 about, or one that includes all its properties. Thus, when Archimedes prov- 
 ed the spherical figure of the earth, by reasoning on a principle of this sort, 
 he was led to a false conclusion, because he knew nothing of the rotation of 
 the earth on its axis, which places the particles of that body, though at 
 equal distances from the centre, in circumstances very different froa one 
 another. But, concerning those things that are the creatures of the mind 
 altogether, like the objects of mathematical investigation, there can be no 
 danger of being misled by the principle of the sufficient reason, which at the 
 same time furnishes us with the only single Axiom, by help of which we 
 can compare together geometrical quantities, whether they be of one, o! 
 two or of three dimensions.
 
 NOTES. SUPPL. BOOK III. 313 
 
 Legendre in his Elements has made the same remark ttiat nas been just 
 ■tated, that there are solids and other Geometrical Magnitudes, which 
 though similar and equal, cannot be brought to coincide with one another 
 and he has distinguished them by the name of Symmetrical Magnitudes. H« 
 has also given a very satisfactory and ingenious demonstration of the equa- 
 lity of certain solids of that sort, though not so concise as the nature of a 
 simple and elementary truth would seem to require, and consequently not 
 *uch as to render the axiom proposed above altogether unnecessary 
 
 But a circumstance for which I cannot very well account is, that Legen- 
 dre, and after him Lacroix, ascribe to Simson the first mention of such solid* 
 as we are here considering. Now I must be permitted to say, that no re- 
 mark to this purpose is to be found in any of the writings of Simson, whicb 
 have come to my knowledge. He has indeed made an observation concerning 
 the Geometry of Solids, which was both new and important, viz. that solids 
 may have the condition which Euclid thought sufficient to determine their 
 quality, and may nevertheless be unequal ; whereas the observation made 
 here is, that solids may be equal and similar, and may yet want the condition 
 of being able to coincide with one another. These propositions are widely 
 different ; and how so accurate a writer as Legendre should have mistaken 
 the one for the other, is not easy to be explained. It must be observed, 
 that he does not seem in the least aware of the observation which Simson 
 has really made. Perhaps having himself made the remark we now spean 
 of, and on looking slightly into Simson, having found a limitation of th* 
 usual description of equal solids, he had without much inquiry, set it dowt 
 as the same with his own notion ; and so, with a great deal of candour 
 and some precipitation, he has ascribed to Simson a discovery which reallj 
 belonged to himself. This at least seems to be the most probable solution 
 of the difficulty. 
 
 I have entered into a fuller discussion of Legendre's mistake than 1 
 should otherwise have done, from having said, in the first edition of these 
 elements, in 1795, that I believed the non-coincidence of similar and equal 
 solids in certain circumstances, was then made for the first time. This it 
 is evident would have been a pretension as ridiculous as ill-founded, if the 
 same observation had been made in a book like Simson's, which in this 
 country was in every body's hands, and which I had myself professedly 
 studied with attention. As I have not seen any edition of Legendre's Ele- 
 ments earlier than that published in 1802, I am ignorant whether he or 1 
 was the first in making the remark here referred to. That circumstance 
 is, however, immaterial ; for I am not interested about the originality of the 
 remark, though very much interested to show that I had no intenton of ap- 
 propriating to myself a discovery made by another. 
 
 Another observation on the subject of those solids, which, with Legendre, 
 we shall call Symmetrical, has occurred to me, which I did not at first 
 «.hink of, viz. that Euclid himself certainly had these solids in view when he 
 formed his definition (as he very improperly calls it) of equal and similar solids. 
 He says that those solids are equal and similar, which are contained under 
 in; same number of equal and similar planes. But this is not true, as Di. 
 Simson has shewn in a passage just about to be quoted, because two solids 
 may easily be assigned, bounded by the same numbe; of equal and similar 
 planes, which are obvioi sly unequal, the one being contained within th* 
 
 40
 
 314 NOTES. SUPPL. BOOK III. 
 
 othef. Simson observes, that Euclid needed only to have added that the 
 equal wid similar planes must be similarly situated, to have made his des- 
 cription exact. Now, it is true, that this addition would have made it exact 
 in one respect, but would have rendered it imperfect in another ; for though 
 all the solids having the conditions here enumerated, are equal and similar, 
 many others are equal and similar which have not those conditions, that is, 
 though bounded by the same equal number of similar planes, those planes 
 are not similarly situated. The symmetrical solids have not their equal 
 and similar planes similarly situated, but in an order and position directly con- 
 trary. Euclid, it is probable, was aware of this, and by seeking to render 
 the description of equal and similar solids so general, as to comprehend so- 
 lids of both kinds, has stript it of an essential condition, so that solids ob- 
 viously unequal are included in it, and has also been led into a very illogical 
 proceeding, that of defining the equality of solids, instead of proving it, as if 
 he had been at liberty to fix a new idea to the word equal every time that 
 he applied it to a new kind of magnitude. The nature of the difficulty he 
 had to contend with, will perhaps be the more readily admitted as an apo- 
 logy for this error, when it is considered that Simson, who had studied the 
 matter so carefully, as to set Euclid right in one particular, was himselt 
 wrong in another, and has treated of equal and similar solids, so as to ex- 
 clude the symmetrical altogether, to which indeed he seems not to have at 
 all adverted. 
 
 I must, therefore, again repeat, that I do not think that this matter can 
 be treated in a way quite simple and elementary, and at the same time 
 general, without introducing the principle of the sufficient reason as stated 
 above. It may then be demonstrated, that similar and equal solids are 
 those contained by the same number of equal and similar planes, either with 
 similar or contrary situations. If the word contrary is properly understood, 
 this description seems to be quite general. 
 
 Simson's remark, that solids may be unequal, though contained by the 
 same number of equal and similar planes, extends also to solid angles 
 which may be unequal, though contained by the same number of equal 
 plane angles. These remarks he published in the first edition of his Eu- 
 clid in 1756, the very same year that M. le Sage communicated to the 
 Academy of Sciences the observation on the subject of solid angles, men- 
 tioned in a former note ; and it is singular, that these two Geometers, with- 
 out any communication with one another, should almost at the same time 
 have made two discoveries very nearly connected, yet neither of them com- 
 prehending the whole truth, so that each is imperfect without the other. 
 
 Dr. Simson has shewn the truth of his remark, by the following reason- 
 ing. 
 
 " Let there be any plane rectilineal figure, as the triangle ABC, and from 
 a point D within it, draw the straight line DE at right angles to the plane 
 ABC ; in DE take DE, DF equal to one another, upon the opposite sides 
 of the plane, and let G be any point in EF ; join DA , DB, DC , EA, EB, 
 EC ; FA. FB, FC ; GA, GB, GC : Because the straight line EDF is at 
 right angles to the plane ABC, it makes right angles with DA, DB, DC, 
 which \t meets in that plane ; and in the triangles EDB, FDB, ED and 
 DB are equal to FD, and DB, each to each, and they contain right angles ; 
 therefore the base EB is equal to the base FB ; in the same manner EA is
 
 NOTES. SUPPL. BOOK III. 315 
 
 equal to FA, and EC to FC : and in the triangles EBA, FBA, EB, BA art 
 equal to FB, BA, and the base EA is equal to the base FA ; wherefore 
 the angle EBA is equal to the angle FBA, and the triangle EBA equai 
 to the triangle FBA, and the other angles equal to the other angles ; there- 
 fore these triangles are similar: In the same manner the triangle EBC is 
 shnilar to the triangle FBC, and the triangle E AC to FAC ; therefore there 
 are two solid figures, each of which is contained by six triangles, one of them 
 by three triangles, the common vertex of which is the point G, and their 
 bases the straight lines AB, BC, CA, and by three other triangles the com- 
 mon vertex of which is the point E, and their bases the same lines AB, BC, 
 CA. The other solid is contained by the same three triangles, the common 
 vertex of which is G, and their bases AB, BC, CA ; and by three other tri- 
 angles, of which the common vertex is the point F, and their bases the same 
 straight lines AB, BC, CA : Now, the three triangles GAB, GBC, GCA 
 are common to both solids, and the three others EAB, EBC, ECA, of the 
 first solid have been shown to be equal and similar to the three others. 
 FAB, FBC, FCA of the other solid, each to each ; therefore, these two 
 solids are contained by the same number of equal and similar planes : But 
 that they are not equal is manifest, because the first of them is contained in 
 the other : Therefore it is not universally true, that solids are equal which 
 are contained by the same number of equal and similar planes." 
 
 " Cor. From this it appears, that two unequal solid angles may be con- 
 tained by the same number of equal plane* angles." 
 
 " For the solid angle at B, which is contained by the four plane angles 
 EBA, EBC, GBA, GBC is not equal to the solid angle at the same point 
 B, which is contained by the four plane angles FBA, FBC, GBA, GBC ; 
 for the last contains the other. And each of them is contained by four 
 plane angles, which are equal to one another, each to each, or are the self- 
 same, as has been proved :' And indeed, there may be innumerable solid 
 angles all unequal to one another, which aro each of them contained by 
 plane angles that are equal to one another, each to each. It is likewise 
 manifest, that the before-mentioned solids are not similar, since their solid 
 angles are not all equal."
 
 PLANE TRIGONOMETRY. 
 
 DEFINITIONS, &c. 
 
 Trigonometry is defined in the text to be the application of Number 
 to express the relations of the sides and angles of triangles. It depends 
 therefore, on the 47th of the first of Euclid, and on the 7th of the first of the 
 Supplement, the two propositions which do most immediately connect 
 together the sciences of Arithmetic and Geometry. 
 
 The sine of an angle is defined above in the usual way, viz. the perpen- 
 dicular drawn from one extremity of tne arc, which measures the angle on 
 the radius passing through the other ■ out in strictness the sine is not the 
 perpendicular itself, but the ratio of tnat perpendicular to the radius, for it 
 is this ratio which remains constant, wnile the angle continues the same, 
 though the radius vary. It might be convenient, therefore, to define the 
 sine to be the quotient which arises from dividing the perpendicular just 
 described by the radius of the circle. 
 
 So also, if one of the sides of a rigQt angled triangle about the right an- 
 gle be divided by the other, the quotient is the tangent of the angle op- 
 posite to the first-mentioned side, &c. But though this is certainly the 
 rigorous way of conceiving the sines, tangents, &c. of angles, which are 
 in reality not magnitudes, but the ratios of magnitudes ; yet as this idea is 
 a little more abstract than the common one, and would also involve some 
 change in the language of Trigonometry, at the same time that it would 
 in the end lead to nothing that is not attained by making the radius eoual 
 to unity, I have adhered to the common method, though I have thought 
 it right to point out that which should in strictness be pursued. 
 
 A proposition is left out in the Plane Trigonometry, which the astro- 
 nomers make use of in order, when two sides of a triangle, and the angle 
 contained by them, are given, to find the angles at the base, without 
 making use of the sum or difference of the sides, which, in some cases, 
 when only the Logarithms of the sides are given, cannot be conveniently 
 found.
 
 NOTES PL TRKIONOMETR*. J17 
 
 THEOREM. 
 
 If, as the greater of any two sides of a triangle to the less, so the radius to the 
 tangent of a certain angle ; then will tlie radius be to the tangent of the diffe- 
 rence between that angle and half a right angle, as the tangent of half the 
 sum of the angles, at the base of the triangle to the tangent of half their 
 difference. 
 
 Let ABC be a triangle, the sides of 
 which are BC and CA, and the base 
 AB, and let BC be greater than CA. 
 Let DC be drawn at right angles to 
 BC, and equal to AC ; join BD, and 
 because (Prop. 1.) in the right angled 
 triangle BCD, BC : CD : : R : tan 
 CBD, CBD is the angle of which the 
 tangent is to the radius as CD to BC 
 that is, as CA to BC, or as the least 
 of the two sides of the triangle to the 
 greatest. 
 
 But BC + CD : BC-CD : : tan }(CDB+CBD) : 
 tan I (CDB-CBD) (Prop. 5.) ; 
 
 and also, BC+CA : BC— CA : : tan } (CAB+CBA) : 
 tan I (CAB— CBA). Therefore, since CD=CA, 
 tan X (CDB+CBD) : tan £ (CDB— CBD) : : 
 tan { (CAB + CBA) : tan \ (CAB— CBA). But because the 
 angles CDB + CBD=90°, tan \ (CDB+CBD) : 
 tan \ (CDB-CBD) : : R : tan (45°— CBD), (2 Cor. Prop. 3.) , 
 therefore, R : tan (45°— CBD) : : tan I (CAB + CBA) : 
 tan 1(CAB — CBA); and CBD was already shewn to be such an angle 
 that BC : CA : : R : tan CBD. 
 
 Cor. If BC, CA, and the angle C are given to find the angles A and B ; 
 find an angle E such, that BC : CA : : R : tan E ; then R : tan (45° — E, 
 : : tan \ (A+B) : tan } (A— B). Thus J (A— B) is found, and \ (A+B) 
 being given, A and B are each of them known. Lem. 2. 
 
 In reading the elements of Plane Trigonometry, it may be of use to ob- 
 serve, that the first five propositions contain all the rules absolutely neces- 
 sary for solving the different cases of plane triangles. The learner, when 
 he studies Trigonometry for the first time, may satisfy himself with these 
 propositions, but should by no means neglect the others in a subsequen* 
 perusal. 
 
 PROP. VII. and VIII. 
 
 I have changed the demonstration which I gave of these propositions in 
 the first edition, for two others considerably simpler and more concise, given 
 me by Mr. Jardine, teacher of the Mathematics in Edinburgh, formerly 
 one of my pupils, to whose ingenuity and skill I am very glad to bear this 
 public testimony.
 
 SPHERICAL 
 TRIGONOMETRY. 
 
 PROP V 
 
 The angles at the base of an isosceles spherical triangle are symmetrical 
 Magnitudes, not admitting of being laid on one another, nor of coinciding, 
 notwithstanding their equality. It might be considered as a sufficient 
 proof that they are equal, to observe that they are each determined to be 
 of a certain magnitude rather than any other, by conditions which are pre • 
 cisely the same, so that there is no reason why one of them should be 
 greater than another. For the sake of those to whom this reasoning may 
 not prove satisfactory, the demonstration in the text is given, which is 
 strictly geometrical. • 
 
 For the demonstrations of the two propositions that are given in the end 
 of the Appendix to the Spherical Trigonometry, see Elementa Sphaericorum, 
 Theor. 66. apud Wolfii Opera Math. torn, iii.; Trigonometrie par Cagnoli 
 h 463 : Trigonom&Tia Spherique par Mauduit, § 165. 
 
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