I DearCs Stereotype Edition. ELEMENTS OF GEOMETRY: CONTAINING THE FIRST SIX BOOKS OF EUCLID,' WITH A SUPPLEMENT ON THE QUADRATURE OF THE CIRCLE, AND THE GEOMETRY OF SOLIDS: TO WHICH AR£ ADDED, ELEMENTS OF PLANE AND SPHERICAL TRIGONOMETRY. BY JOHN PLAYFAIR, F.R.S. Lond.> Edin. PBOFSSSOB OF NATI7BAL PHILOSOPHY, FORMERLY OF MATHBUITICS, ZIf THE VNITEBSITY OF EDINBT7BOH. FROM THE LAST LONDON EDITION, ENLARGED. NEW YORK : '^'^S^- W. E. DEAN, PRINTER AND PUBLISHER, 2 ANN STREET. 1849. ^s-J I'^'i-^ Entered according to the Act of Congress, in the year One Thousand Eight Hundred and Forty-five, hy W. E. Dean, in the Clerk's Of- fice of the Southern District of New-York. PREFACE. It is a remarkable fact in the history of science, that the oldest book of 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. Simson, 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 Eleiiients 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 his object to restore the writings of Euclid to their original perfection, 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 modems. But, after all this was accomplished, something still remained to be done, since, notwithstanding the acknowledged excellence of Euclid's Ele- ments, it could not be doubted that some alterations might be 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 they originally had, as that which may at present render them most useful. PREFACE. One ot 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 great advantages accompanied with considerable defects ; of 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 forni 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. . vivFi^ni^ THE PRINCIPLES. - EXPLANATION OF TERMS AND SIGNS. 1. Geometry is a science which has for its object the measurement of mag- nitudes. 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- liums, 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 definition 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. 5. 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 the method of collecting the several arguments and proofs, and laying them together in proper order, by nveans of which the truth of the proposition becomes evident, is called a Demonstration. 6. A Direct Demonstration is that which concludes with the direct and cer- tain proof of the proposition in hand. It is also called Positive or Affirmative, and sometimes an Ostensive Dc' monstration, because it is most satisfactory to the mind. 6 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 Ahsurdum ; 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. 11 . A Corollary/, or Consectary, is a consequence drawn immediately from some proposition or other premises. 12. 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 J 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. 17. 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 ^hat which is commonly used in Geometry. 18. 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 / B is used. And to signify that A is less than B, the expression A/ B is used. OF GEOMETRY. BOOK I. 7 20. The sign of Addition is an erect cross ; thus A+B implies the sum of 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 remaining, when a part equal to' B has been taken away from it. In hke 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, AxB, 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 AxB+C — D, re- presents the product of A by the quantity B-f C — D. In like manner, (A4-B)x(A — B+C), indicates the product of A-f-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 ; JAB 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 B 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^, the cube of the line AB is designated by AB^ ; 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 ■v/5 ; -^/(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 yAxExCl or (A X B X C)^, and so on. 28. Numbers in a parenthesis, such as (15. 1.), refers back to the number 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. In like manner, (3. Ax.) designates the third axiom ; (2. Post.) the second postulate ; (Def. 3.) the third definition, 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 io he proportional: thus, if — = k 5 then, A is to B as C to D. 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." 4. 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, the 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. \ 13 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, * CB, is named the angle ABC, or CBA ; that which is contained by AB, * The definitions marked with inverted commas are djflferent from those of Euclid. OF GEOMETRY. BOOK I. 9 ' BD, is named the angle ABD, or DBA ; and that which is contained by * BD, CB, is called the angle PBC, 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 the ' angle at E.* 7. 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 figure, without any reference to the nature of the line or lines which hound it. 11. A circle is a plane figure contained by one line, w^hich 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. ■m ':^^^^k 12. And this point is called the centre of the circle. 13. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. 14. A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter. 2 1^ ELEMENTS 15. Rectilineal figures are those which are contained by straight lines. 16. Trilateral figures, or triangles, by three*straight lines. 17. Quadrilateral, by four straight lines. 18. Multilateral figures, or polygons, by more than four strstight lines. 19. Of three sided figures, an equilateral triangle is that which has three 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 wliich 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 its sides equal. 27. A rhombus is that which has all its sides equal, but its angles are not right angles. ( OF GEOMETRY. BOOK I. jl 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 pomt 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 one 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. S. 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- " rallel to the same straight line." 12 ELEMENTS PROPOSITION I. PROBLEM. To describe an equilateral triangle upon 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- tre 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 has been drawn equal to the given straight line BC. PROP. III. PROB. From the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the tv^o 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.) 0, 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 opposite* 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 AG to DF ; and let the angle BAG be also equal to the angle EDF: then shall the 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 equal sides are opposite, shall .^ — .^— — be equal, each to each, '** C/ H d 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 saying, that the triangUs are every way equal. 14 ELEMENTS coinciding with^DE, AC shall coincide with DF, because the angle BAG is equal to the angle EDF ; wherefore also the point C shall coincide with the 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 which 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 he 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. L) to AF, the less, and join FC, GB. Because AF is equal to AG, and AB 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, -D/ \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 GOB, which are the angles upon the other side of the base. Corollary. Hence every equilateral triangle is also equiangular PROP. VI. THEOR. If two angles of a triangle he 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. VH. THEOR. 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 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 each 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 eqtial to DB, the angle BDC is equal (5. L) to the angle BCD ; A^ i ^B 16 ELEMENTS but it has been demonstrated to be greater than it ; which is impossi- ble. But if one of the vertices, as D, . Tg be within the " other triangle ACB ; / ^^ produce AC, AD to E, F ; therefore, ^a 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 ^ P 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 other, each to each, and have likewise their bases equal ; the angle which is contain-- ed hy the two sides of the one shall he equal to the angle contained by the two 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 ; B 1? E F and also the base BC equal to the base EF. The angle BAG is equal to the 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. 17 tion, 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 BAG coincides with the angle EDF, and is equal (8. Ax.) to it. I PROP. IX. PROB. To bisect a given rectilineal angle, that is, to divide it into two equal angles. Let BAG be the given rectUineal angle, it is required to bisect it. Take any point D in AB, and from AG cut (3. 1.) off AE equal to AD ; join DE, and upon it describe (!• 1-) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the angle BAG. 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 BAG is bisect- ed by the straight line AF. SGHOLIUM. By the same construction, each of the halves BAF, GAF, 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 oh. 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 ABG, and bisect (9. 1.) the angle ACB by the straight line GD. AB is cut into two equal parts in the point D. Because AG is equal to GB, and GD common to the two triangles AGD, BGD : the two sides AG, GD, are equal to the two BG, GD, each to each ; but the angle AGD is also equal to the an- gle BGD ; 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. 3 0f 18 ELEMENTS PROP. XL PROB. To^raw 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 jp 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, ECF, 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. XIL PROB. To draw a straight line perpendicular to a given straight line, of an unlimited length, from a given point without it. I^et 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 x\B,and from the centre C, at the distance CD, describe (3. Post.) the circle EOF 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 straight line AB. OF GEOMETRY. BOOK I. 19 PROP. XIII. THEOR. The angles which one straight line makes with another upon one side of 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.) B A B S B 5 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 AB, make the adja- cent angles ABC, ABD equal togethe* to two right angles. BD is in the same straight line with CB. For if BD be not in the same straight line with CB, let BE be in the same straight line with it ; therefore, because the straight line AB makes angles with the straight line CBE, upon one side of 20 ELEMENTS it, t'ae angles 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 lines meeting in one point, are together equal to four right angles. PROP. XVL THEOR. If one side of a triangle he 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 EF 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 angle CEF, because they are vertical angles ; therefore the base AB OF GEOMETRY. BOOK I. 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 ACD, 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 the angles ACD, ACB are greaiter than the angles ABC, ACB ; but ACD, ACB are to- gether equal (13. 1.) to two right an- gles : therefore the angles AB'C, 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 oflf(3. 1.) AD equal to AB, and join BD : and because ADB is the exterior angle of the triangle B DC, it is greater (16. 1.) than the interior and opposite 2« ELEMENTS angle DCB ; but ADB is equal (5. 1.) to ABD, because the side AB is 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, or has 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 -o 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, the line BC, for example, is the shortest distance from B to C ; therefore BC is less than BA+AC or BA+AC>BC. OF GEOMETRY. BOOK I. 23 PROP. XXI. THEOR. If from the ends of one side of a triangle, there he drawn two straight lines to a point within the triangle, these two lines shall be less than the 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 of 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, AE 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 (3. 1.) DF equal to A, FG to B, and pH equal to C ; and from 34 ELEMENTS the centre F, at 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 third. 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 ^ A given rectilineal angle DOE. Take in CD, CE any points D, E, and join DE ; and make (22. 1.) the triangle AFG, 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, hut the angle contained by the two sides of the one greater than the angle contained hy 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. 25 DF ; but the angle BAG greater than the angle EDF ; the base BC is 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 BAG : and make DG equal (3. 1.) to AC or DF, and join EG, GF. Because AB is equal to DE, and AG to DG, the two sides BA, AG are equal to the two ED, DG, each to each, and the angle BAG is equal to the angle EDG, therefore the base BG is equal (4.1.) ^ D to the base EG ; and be- cause DG is equal to DF, the angle DFG is equal (5. 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 (19. l.)side is opposite to the greater angle, the side EG is greater than the side EF ; but EG is equal to BG ; 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, hut 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, AG, 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 BAG 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 BAG is not equal to the angle EDF, because then the base BC would be equal (4. 1 .) to EF ; but it is not ; therefore the angle BAG is not equal to the angle EDF ; neither "is it less ; because then the base BG would be less (24. 1.) than the bas'^ EF ; but it is not ; therefore the an- gle BAG is not less than the angle EDF : and it was shewn that it is nor equal to it : therefore the angle BAG is greater than the angle EDF. aft* ELEMENTS PROP. XXVI. THEOR. If two triangles 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 he 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. AB to DE, and AC to DF ; and B 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 join 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 ABH to the trian- gle DEF, and the other angles are equal, each to each, to which the equal 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 equal 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. 1.) ; 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 equal to the base DF, and the third angle BAG to the third angle EDF. PROP. XXVn. THEOR. If a straight line falling upon two other straight lines makes the alternate angles equal to one another j 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, D 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. XXVni. 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 GHD, the interior and oppo- site angle upon 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 28 ELEMENTS angle AGH, tlie angle AGH is equal to the angle GHD ; and they are thu alternate angles ; therefore AB is parallel (27. 1.) to CD. Again, because 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 ; 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, ii GHC together less than two right angles, KG and CH will meet on the side I 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 two, KGH, CHG, are by supposition less than OF GEOMETRY. BOOK I. 29 two right angles ; therefore the other two, HGL, GHD are greater than 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 perpendicular 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 ftcute 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 one 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. To 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 to draw a straight line through the point A, parallel to the straight line 60. aO ELEMENTS In BC take any point D, and join _, -? AD ; and at the point A, in the ^ / E* straight line AD, make (23. 1.) the / angle DAE equal to the angle ADC ; •. — ^- . and produce the straight line E A to F. D D 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 is 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 he 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. 1.) 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 tlie 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. ST 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. 1.) 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 J of one right angle. CoR. 8. The sum of the angles in a quadrilateral is equal to two right 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 oart 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 right angles. 32 ELEMENTS And, by joinu^g 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 xAB, 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 "D the angle BCD ; therefore the base AC is equal (4. L) 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. rSlVERBfTY ; 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, BC A 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 AC 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, its 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. PROP. XXXV. THEOR. Parallelograms upon the same base and between the same parallels, are equal 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. / 34 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 ABCDjEBCF, 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 DF ; now AB is also equal to DC ; therefore the two E A, 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 parallels, are equal to one another. Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same parallels AH, - r% XT' tt BG ; the parallelogram ABCD A D E H 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. 1.) 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 parallelogram ABCD is equal to EFGH. B O OF GEOMETRY. BOOK I. 35 PROP. XXXVII. THEOR. '^ Triangles upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBG be upon the same base BC, and between the same parallels, AD, BC : The triangle ABC is equal to the trian- gle DBC. 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 of 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. Prod\ice 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, f^ A Fl TT 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 ABCisthehalf(34. l.)ofthe ^ p p „ parallelogram GBCA, because ■" C Hj r the 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 up i e.vi,q &di t 46 ELEMENTS The same process may be applied to every other polygon ; for, 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 follo:ws that any polygon may he reduced to an equivalent rectangle. PROP. H. PROB. Tojind 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 siun 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. L 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 other 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 anglei^ ; therefore, the square described upon DC is equivalent to the squares constructed upon AD and AC : hence (Cor. 1. 47. 1.), AD2=CD2-AC2. PROP. K. PROB. A rectangle being given, to construct an equivalent one, having a side of a given length. Let AEFHbe the given rectangle, and produce one of its sides, as AH, till OF GEOMETRY. BOOK I. 47 HB be the givem length, and draw BFD meeting the prolongation of AE in D j then produce EF till FG is equal to HB : draw BGO, 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 AEFH (43. 1.) Coft. A polygon may he converted into an equivalent rectangle^ having one of its sides of a given length, ^ ELEMENTS OF GEOMETRY. BOOK IL 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. " 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 lectangle contained by AD and DC, we' shall simply say " the rectangle AD . DC, placing a point 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. arel, 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 " complements AF, FC, is the gno- " mon of the parallelogram AC. This *^ gnomon may also, for the sake of * brevity, be called the gnomon AGK •orEHC." OF GEOMETRY. BOOK 11. 40 PROP. I. THEOR. ' If there he 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, Qnd the several parts of the divided line. Let A and BC be two straiglit lines ; and let BC be divided into any parts in the points D, E ; the rectangle A.BC is egual to the several rect- angles 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.DE4-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-\-kc-\-kd. PROP. 11. 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 whole line. 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=AB2. . 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=AB2. There- fore AB.AC + AB.BC =AB2. ^HOLIUM. This property is evident from Algebra : let AB be denoted by a, and the segments AC, CB, by 5 and d, respectively; then, a=&-|-c?; therefore, multiplying both members of this equality by a, we shall have a^=:ab-\-ad 7 F E 50 ELEMENTS PROP. III. THEOR. If a straight line he divided into any two parts, the rectangle contained hy the whole and one of the parts, is equal to the rectangle contained hy the two 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^. 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=BC2; there- fore AB.BC=AC.CB-|-BC2. SCHOLIUM. In this proposition let AB be denoted by a, and the segments AC and CB, by h and c ; then a=.h-{-c'. therefore, multiplying both members of this equality by c, we shall have ac=:.hc-\-c^, PROP. IV. THEOR. If a straight line he 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, AB2=AC2+CB2+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 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 lis 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. 1.) Wherefore CGKB is a square, and it is upon the side CB. For the same A. CI H Ct / T /^ Jl I ) ] F • E OF GEOMETRY. BOOK II. 51 reason HF also is a square, and it is upon the side HG, which is equal to AC : therefore HF, CK are the squares of AC, CB. And because the 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=AC2 and CK=CB2; therefore, HF+CK+AG + GE=AC2+CB2+2AC.CB. But HF+CK+AG+GE=the figure AE, or AB^; therefore AB^zs AC2+CB2+2AC.CB. CoR. From the demonstration, it is manifest that the parallelograms 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 h ; then, {a-\-})f =a24.2a54-62. PROP. V. THEOR. If a straight line be dividedinto two equal parts, and also into two ufiequal parts ; the rectangle contained hy the unequal parts, together with the square of 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 +CD2= CB2. 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 CL or 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 fCor. 4. 2.) ; therefore gnomon CMG =AD.DB. To each add LG=CD2, then, gnomon CMG4-LG=AD.DB + CD2. But CMG+LG=BC2; therefore AD.DB + CD2=BC2. " Cor. From this proposition it is manifest, that the difference of the *' squares of two unequal lines, AC, CD, is equal to the rectangle contain- *' ed by their sum and difference, or that AC2— CD2=(AC+CD) (AC— ** CD)." SCHOLIUM." In this proposition, let AC be denoted by a, and CD by b ; then, AD=s a-^-b, and DB=a — b-, therefore, by Algebra, (a-{-b)x{a — b)=a^ — &2. that is, the product of the sum and difference of two quantities, is equivalent to the difference of their squares 52 ELEMENTS G F PROP. VI THEOR. If a straight line be 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+CB2. But CMG+LG=CF =CD2, therefore AD.DB4-CB2=CD2. SCHOLIUM. This property is evinced algebraically ; thus, let AB be denoted by 2a, and BD by & ; then, AD=2a+6, and CD=a+i. Now by multiplication, b(2a'\-b)=2ab-\-b'^ ; therefore, by adding a^ to each member of the equality, we shall have, b(2a-)rb)-^a^=a'^-\-2ab-^b'^ ; .-. 5(2a+6)4-a2=(a+ J)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^+BC^ =2AB.BC+AC2. Upon AB describe (Prop. 46. 1.) the square ADEB, and construct the figure as in the pre- ceding propositions : Because AG=GE, AG -fCK = GE4-CK, that is, AK = CE, and therefore AK+CE=2AK. But AK+CE =gnomon AKF+CK ; and therefore AKF OF GEOMETRY. BOOK II. &3 +CK=2AK = 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+HF=AE=AB2, AB2+CK=2AB.BC+HF, that is, (since CK=CB2, and HF=AC2,) AB24-CB2=2AB.BC+AC2. " Cor. Hence, the sum of the squares of any two lines is equal to " twice the rectangle contained by the lines together with the square of " 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 a2=52-f2Jc+c2; adding c^ to each member of this equality, we shall have, a2+c2 = 62 + 2Jc + 2c2; .•.a2^c2=524-2c(5+c), or a2_j_c2=c2ac+^^. 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 respectivelfj minus twice the rectangle contained by the lines. For a — c=ib ; therefore, by involution, a^ — 2ac-\-c^=b'^. This may be also derived from the above algebraical equality, by transposition. PROP. VIH. THEOR. If a 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 ; aild congtruct 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.) to RN, because they are the complements of 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-}- BN + GR + RN = 4CK. Again, be- cause CB is equal to BD, and BD equal C B B M E G E / P / R / • K li F 54 ELEMENTS (Cor. 4. 2.) to BK, that is, to CG ; and CB equal to GK, that is, to GP ; therefore CG is 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 CK4-BN+GR4-RN=4CK ; wherefore, adding equals to equals, the whole gnomon A0H=4AK. Now AK=AB.BK=AB.BC, and 4AK=4AB.BC ; therefore, gnomon A0H=4AB.BC ; and adding XH, or (Cor. 4. 2.) AC^, to both, gnomon AOH+XH=4AB.BC+AC2. But A0H+XH=AF = AD2; therefore AD2=4AB.BC+A.C2. " 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 h ; then AD=c+2^. Now, since c=5-i-c, multiplying both members by 4&, we shall have 4ai=462+4k; and adding c^ to each member of this equality, we shall have, 4a5+c2=c2+46c+462, or 4a&4-c2=(c+2&)2. PROP. IX. THEOR. If a straight line he divided into two equal, and also into two unequal parts , the squares of the two unequal parts are together double of the square of half 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 aligle (Cor. 4. 32. 1.) ; and they are equal to one ano- ther ; 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-s half a right angle ; and therefore the whole 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 a 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 is equal to the angle BFD, and the side DF to (6. 1.) the side DB. Now, be- cause AC=CE, AG2=CE2, and AC24-CE2=2AC2. But (47. 1.) AE2= AC2-f CE2 ; therefore AE2=2AC2. Again, becauseEG=GF, EG2=GF2, and EG2+GF2=2GF2. But EF2=EG2+GF2 ; therefore, EF2=:2GF2 =2CD2, because (34. 1.) CD=GF. And it was shown that AE2=2AC2 ; therefore AE2+EF2=2AC2+2CD2. But (47. 1.) AF2=AE2+EF2, and AD2+DF2= AF2, or AD2+DB2=AF2 ; therefore, also, AD2+DB2= 2AC2+2CD2. SCHOLIUM. This property is evident from the algebraical expression, (a+Z>)2+(a— 5)2=2a24-2Z>2 ; where a denotes AC, and h denotes CD ; hence, a-\-'b =AD, a — J=DB. PROP. X. THEOR. If a straight line hebisected, 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 line 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 56 ELEMENTS right angle, 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 half 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^ + CA^ == 2CA2. Now AE2z= (47. 1.) AC2 4- CE2; therefore, AE2= 2AC2. Again, be- cause EF^FG, EF2=FG2, andEF2+FG2=2EF2. ButEG2r=: (47. 1.) EF2-I-FG2; therefore EG2=2EF2; and since EF=CD, EG2z=z2CD2. And it was demonstrated, that AE2=2AC2 ; therefore, AE2-f.EG2=2AG2 +2CD2. Now, AG2=:AE2-f-EG2, wherefore AG2=2AC2+2CD2. But AG2(47. l.) = AD24-DG2=AD2-fDB2, because DG=DB : Therefore, AD2-1-DB2=2AC2+2CD2. SCHOLIUM. Let AC be denoted by a, and BD, the part produced, by b ; then AD=: 2a-\-b, and C\)=a+b. Now, (2a+6)2+62_4a2+4a6_|.263; but 4024.406+262=202+2 (a+ 6)2; hence, (2o+6)2+62_2a2+2(a+ 6)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. m Let AB be the given straight line ; it is required to divide it intof 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 is equal to EB ; therefore the rectangle CF. FA, together with the square of AE, is OF GEOMETRY. BOOK II. 67 equal to the square of EB ; And the squares of BA, AE are equal (47. 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 equal 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 the 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.BH 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. XH. THEOR. In obtuse angled triangles, if a perpendicular he 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, when 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, an'd 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 liae BD is Hvided j^ into two parts in the point C, BD2 = (4. 2.) BC24-CD2-F2BC.CD; aSd AD^ to both: Then BD^+AD^ = BC2+ CD2+ AD2+ 2BCJiD. But AB2^BD2+AD2 (47. 1.), and AC2= CD2+%D2 (47. L); tfllefore, AB2=BC2+AC2+2BC.CD ; that is, AB2 is greater than BC2-i-AC2 by 2BC.CD. PROP. XHI. THEOR. In every triangle the square of the side subtending any of the acute angles, is less than the squares of the sides containi/ig 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 the angle B, is less than the squares of CB, BA by twice the Actangle CB.BD. 8 58 ELEMENTS First, 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^-j- BD2:=2BC.BD+CD2. Addtoeach AD2; thenBC2+BD2+AD2=2BC.BD + CD2+ AD2. But BD2+AD2=AB2, and CD2+ DA2r= AC2 (47. 1.) ; therefore BC2+AB2=: 2BC.BD+AC2 . that is, AC2 is less than BC2+AB2 by 2BC,BD. Secondly, let AD fall without the triangle ABC :* Then because the angle at D is a right angle, the angle AC3 is greater (16. 1.) than a right angle, and AB2= (12. 2.) AC2+BC2+2BC.CD. x\dd BC2 to each; then AB2+BC2=AC2 4-2BC24-2BC.CD. But because BD is divided into two parts in C, BC2+BC.CD=(3. 2.) BC.BD, and 2BC2+2BC.CD =2BC.BD: therefore AB2+ BC2=:AC2+ 2BC.BD ; and AC2 is less than AB2+BC2, by 2BD.BC. Lastly, let the side AC be perpendicular to BC ; then is BC the straight line between the perpendicular and the acute angle at B ; and it is manifest that (47. 1.) AB2+BC2= AC2+2BC2=AC2+2BC.BC. PROP. XIV. PROB. To describe a square that shall he 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 orte 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 the 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 : but GF is equal to GH ; therefore the rectangle BE, EF, together with 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. d9< HE and EG are equal (47. 1.) to th% 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 the 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 together double of the squares of BD and DA. From A draw AE perpendicular to BC, and because BEA is a right an- gle, AB2=(47. 1.) BE2+AE2 and AC2= CE2+ AE-^ ; wherefore AB2+AC2=BE2 J^ -f CE2+2AE2. But because the line BC ifs cut equally in D, and unequally in E, BE2 -f CE2 = (9. 2.) 2BD2 + 2DE2 ; therefore AB2 + AC2==2BD2 -f 2DE2.2AE2: Now DE24-AE2=(47. 1.) AD2, and 2DE2+2AE2=2AD2 ; wherefore AB2+ AC2=2BD2+2AD2. PROP. B. THEOR. The sum of the squares of the diameters of any parallelogram is equal 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 squares of AB, BC, CD, DA. Let AC and BD intersect one another in E : and because the vertical angles AED, CEB are equal (15. 1.), and also the alternate angles EAD, 60' ELEMENTS, &c. ECB (29. 1.), the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each ; but the sides AD and Bp, which are opposite to equal angles in these triangles, are also equal (34. 1.); therefore the other sides which are opposite to the eq^ual angles are also equal (26. 1.), viz. AE to EC, and ED to EB. Since, therefore, BD is bi- sected in E, AB2+AD2=:(A. 2.) 2BE2-f-2AE2; and for the same reason, CD^ + BC^ = 2BE2+2EC2=2BE2+2AE2, because EC = AE. Therefore AB^-f AD^ + DC2+BC2r=4BE2+4AE2. But 4BE2=BD2, and 4AE2=rAC2 (2. Cor. 8. 2.) because BD and AC are both bisected in E ; therefore AB2-[- AD2+CD2+BC2=BD2+AC2. Cor. From this demonstration, it is Bftanifest 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 rhom- bus 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 the 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 tvfi 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 ofT. 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 circle 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. i.) 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. S9 sible : Therefore G is not the centre of the circle ABC : In the same 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 is 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 he 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 ; the straight line drawn from A to B shall fall within the circle. Take any point in AB as E ; find D (1. 3.) the 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 pointy moreover ^ in the production of KB, is farther from ihe centre than the circumference. PROP. III. THEOR. // a straight line drawn through the centre of a circle bisect a straight line tn the circle, which does not pass through the centre, itfwill 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 stajading upon another makes the 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 AB, which does not pass through the centre, cuts AB at right 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 E A, 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 th^ 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 (26. 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 hy 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 loth 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 ^ot both pass through the centre : AC, BD do not bieect 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 eauEli 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. W 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 : iwt 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. 'm 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 €qual to FE : but CF was shown to be equal to FB ; therefore FE is equal to FB, the less to the greater, which is impossible ; Where- fore F is not the centre of the circles ABC, CDE. PROP. VII. THEOR. If any point he taken in the diameter of a circle which is not the centre, of all 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 one another, one upon each side of the shortest line. Let ABCD be a circle, a^d AD its diameter; in which let any point F be taken which is not the centre : let the centre be E ; of all the straight lines 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 greater (20. 1.) than the third, BE, EF are greater than BF ; but AE is equal to EB ; therefore AE and EF, that is, AF, is greater than BF : Again, be- cause 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 EF ; but the angle BEF 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. [ If any point he taken without a circle^ and straight lines he 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 he 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 j EM, MD OF GEOMETRY. BOOK III. 41 are equal to FM, MD ; but the angle EMD is greater than the a'ngle 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, KD 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 maimer, 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 he 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 impossible : Therefore E is not the centre of the 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 circle 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, 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 another, 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 their 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 to 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 their 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. 1 .), 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 touches 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 *be points B, D are in the circumference of each of the circle^ the straight line BD falls 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 therefore GH 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 the 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 : therefors, 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 U?ies in a circle are equally distant from the centre ; and those which are equally distant from the centre, are equal to ona 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 AB double of AF. For the same reason, CD is double of CG : But AB is equal to CD ; therefdre 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 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 ar« 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, that is, if FE be eqjial to EG, AB is equal to CD. For, the same coa- B D and, for the like reason, OF GEOMETRY. BOOK III. n struction being made, it may, as before, be demonstrated, that AB is double of AF, and CD double of CG, and that the squares of EF, FA are equal to the squares of EG, GO ; of which the square of FE is equal to the square of EG, because FE is equal to EG : therefore the remaining square of AF is equal to the remaining square of CG ; and the straight line AF is therefore equal to CG : But AB is double of AF, and CD double of CG ; wherefore AB is equal to CD. PROP. XV. THEOR. The diameter is the greatest straight line in a circle ; and of all others, that which is nearer to the centre is always greater than one more remote ; and the greater is nearer to the centre than the less. Let ABCD be a circle, of which the diame- ter is AD, and the centre E ; and let BC be near- er to the centre than FG ; AD is greater than any straight line BC which is not a diameter, and BC greater than FG. From the centre draw EH, EK perpendiculars to BC, FG, and join EB, EC, EF ; and because AE is equal to EB, and ED to EC, AD is equal to EB, EC : But EB, EC are greater (20. 1.) than BC ; wherefore, also, AD is greater than BC. And, because BC is nearer to the centre than FG, EH is less (4. Def. 3.) than EK ; But, as was demonstrated in the preceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares of EK, KF, of which the square of EH is less than the square of EK, because EH is less than EK ; therefore the square of BH is greater than the square of FK, and the straight line BH greater than FK : and 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 straight 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 draion between that straight line and the circumference, from the extremity of the diameter, so as not to cut the circle. Let ABC he a circle, the centre of which is D, and the diameter AB : and let AE be drawn from A perpendicular to AB, AE shall fall without the circle. 72 ELEMENTS In AE take any point F, join DF and let DF meet the circle in C. Because DAF is a right angle, it is greater than the angle AFD (32. 1.) ; but the greater angle of any triangle is subtended by the greater side (19. 1.), therefore DF is greater than DA : now Dxl is equal to DC, there- fore DF is greater than DC, and the point F is therefore without the circle. And F is any point whatever in the line AE, there- fore AE falls without the circle. Again, between the straight line AE and the circumference, no straight line can be drawn from the point A, which does not cut the circle. Let AG be drawn in the angle DAE : from D draw DH at right angles to AG; and because the angle DHA is a right angle, and the angle DAH less than a right angle, the side DH of the triangle DAH is less than the side DA (19. L). The point H, therefore, is within the cir- cle, and thef^ore thf*>il^raight line AG cuts the circle. ^ CoR. 1. From this it is manifest, that the jj j straight line which is drawn at right angles to the diameter' of a circle from the extremity of it, touches the circle ; and that it touches it only in one point ; because, if it did meet the circle in two, it would fall within it (2. 3.). Also it is evident that there can be but one straight line which touches the circle in the same point. CoR. 2. Hence, a perpendicular at the extremity of a diameter is a tan- gent to the 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 tljeir 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 EA, join EBF, and draw AB. AB touches the circle BCD. Because E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB ; therefore the two sides AE EB are equal to the OF GEOMETRY. BOOK III. W two FE, ED, and they contain the angle at E common to the two 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- O/ 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 point D, draw DE to the centre E, and DF at right angles to DE ; DF touches 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 point of intersection, and also A and the point where this last line will intersect 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 piferpendicular 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 cin be perpendicular to DE ; FC is there- fore perpendicular to DE, 10 74 ELEMENTS PROP. XJX. THEOR. If a straight lirw touch a circle, and from the paint 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 aright 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 douhle of the angle at the circumfer- ence, upon the same base, that is, upon the same part of the circumfer- 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, and 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. 4.) 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. 76 PROP. XXL THEOR. The angles 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 greater 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 segiicircle ; 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. XXIL THEOR. The opposite angles of any quadrilateral figure described in a circle, are 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 (2L 3.) to the angle 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. L) ; therefore also the angles 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 angles. 76 ELEMENTS Cor. 1. If any 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 : and 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 lines 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 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. B C J) coinciding with CD, the segment AEB must (23. 3.) coincide with the 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 (II. 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 III. 77 are all equal ; D is the centre of the circle (9. 3.) ; from the centre 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 A. JJ ^ E* A D ^C is described : and because the centre D is in AC, the segment ABC is a semicircle. Next, let the angles ABD, BAD be unequal ; at the point A, in 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 (6. 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 he at the centres or circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences : the arc BKC is equal to the arc ELF. 78 ELEMENTS Join BC, EF ; and because the circles ABC, DEF are equal, the straight 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 the 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. XXVIL THEOR. In equal circles, the angles which stand upon equal arcs are equal to one another, whether they he 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, l^EF stand upon the equal arcs BC, EF : the angle BGC is equal to the angle EHF, and the angle BAC 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. XXVHL 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 oflf the two greater arcs BAC, EDF, and the two less OF GEOMETRY. BOOK III. 79 BGC, EHF : the greater BAG is equal to the greater EDF, and the less BGC to the less EHF. Take (1. 3.) K, L, the centres of the circles, and join BK, KG, EL, LF ; and because the circles are equal, the straight lines from their centres are equal ; therefore BK, KG are equal to EL, LF ; but the base BG is also equal to the base EF ; therefore the angle BKG is equal (8. 1.) to the angle ELF : and equal angles stand upon equal (26. 3.) arcs, wljen they are at the centres ; therefore the arc BGG is equal to the arc EHF. But the whole circle ABG is equal to the whole EDF ; the remaining part, therefore, of the circumference viz. BAG, is equal to the remaining part EDF. PROP. XXIX. THEOR. In equal circles equal arcs are subtended by equal straight lines. Let ABG, DEF be equal circles, and let the arcs BGG, EHF also be equal ; and join BG, EF : the straight line BG is equal to the straight line EF. Take (1. 3.) K, L the centres of the circles, and join BK, KG, EL, LF : and because the arc BGG is equal to the arc EHF, the angle BKG is equal (27. 3.) to the angle ELF : also because the circles ABG, DEF are equal, their radii are equal : therefore BK, KG are equal to EL, LF : and c E they contain equal angles ; therefore the base BC is equal (4. 1.) to the base 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 ACD, 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 ^ C B 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. XXXL 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 then 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 Cxi 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. . m And because the two angles ABC, BAG of the triangle ABC are to- gether less (17. 1.) than two right angles, and BAG 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 ABCD is a quadrilateral figure in a circle, any two of its 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. If a straight line touch a circle^ and from the point of contact a straight line he drawn cutting the circle, the angles made hy this line with the line which touches the circle, shall he 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 straight 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 ABF 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 right angles ; therefore the angles DBF, DBE, being likewise equal (13. 1.) to 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 82 ELEMENTS PROP. XXXIII. PROS. 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. L) AE at right angles to AD ; bisect (10. L) 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.) ; but 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. ^ 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. 1.) 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 ; but 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 to 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- 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^ = FB-^ = AF^. But AF2 = AE2 + (47. 1.) EF2, therefore BE.ED + EF2, ^ 4lE2 + EF2, and taking EF2 from each, BE.ED=AE2=AE.EC. Next, let BD, which passes through the centre, cut the other AC, which does not pass through 84 ELEMENTS the centre, 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, jy and from F draw (12. 1.) FG perpendicular to AC ; therefore AG is equal (3. 3.) to GC ; where- fore AE.EC + (5. 2.) EG2 = AG^, and adding GF2 to both, AE.EC + EG2+GF2=:AG2+GF2. Now EG2+GF2 = EF2, and AG2+GF2=zAF2 ; ^ ^ ^ ,^ ,^ therefore AE.EC + EF2=:AF2=:FB2. But FB2 A\ ^"^TX /C = BE.ED + (5. 2.) EF2, therefore AE.EC + EF2 =BE.ED4-EF2, and taking EF^ 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. XXXVL 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 hy 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 DC A, DB two straight lines drawn from it, of which DC A 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 + EC2=ED2 (6. 2.). But EC = EB, therefore AD.DC + EB^ = ED^. Now ED2= (47. 1.) EB2-I- BD^, because EBD is a right angle ; therefore AD.DC + EB^rr EB2 4-BD2, and taking EB^ from each, AD.DC =BD2. 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 join EB, EC, ED ; and because the straight line EF, which passes through the centre, cuts OF GEOMETRY. BOOK III. 85 tlie straight line AC, which does not pass through the centre, at right angles, it likewise bisects it (3. 3.) ; therefore AF is equal to FC ; and because the straight line AC is biseclfcd in F, and produced to D (6. 2.), AD.DC + FC2= FD2; add FE2 to both, then AD.DC + FC^H- FE2-FD3 + FE2. But (47. 1.) EC2=:FC2+ FE2, and ED2=FD2+FE2, because DFE is a right angle; therefore AD.DC + EC^^ED^. Now, because EBD is a right angle, ED2 = EB2-1-BD2=EC2+BD2- and therefore, AD. DC + EC2=EC2-f BD2, and AD.DCz=BD2. 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 hy 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 of 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 cuts 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 the square of DB : therefore the square of DE is 86 ELEMENTS equal to 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 ; and the base FD is commonio 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. I 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 ; othier- 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 he 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 lines DB, CB, perpendicular to them would also.be parallel (Cor. 2. 29. 1.), or else form but one straight line : but they 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 pne centre, there can be but one circumference. PROP. C. THEOR. If two circles cut each other ^ the line which passes through their centres will he 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 td the two circles. And if a perpendicular be erected from the middle of this chord, it will pass (C(# 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 he 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 less 88 ELEMENTS than the sum of the smaller and the distance between the centres. For, CD is less (20. 1.) than CA-f AD, and for the same reason, AD/AC+ 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 + Cr) ; 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 each other. CoR. 2. Hence, also, if the distance between the centres be less than 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 of a triangle exceeds the difference between the other two. Hence, the tri- angle is impossible when the distance between the centres is less than the difference of the radii ; and consequently the two circles cannot cut each other. PROP. D. THEOR. In the same circle, equal angles at the centre are subtended hy 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 coii^ cide, inasmuch as they are all equally diP tant from the centre. Conversely. Let the arc AFD be equal to the arc BGD ; then the an- 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 is equal to the angle BCD (8. 1.). Cor. L It follows, moreover, that equal angles at the centre are sub- OF GEOMETRY. BOOK III. tended 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 arc and the chord which it subtends shall also be bisected. Cor. 4. It follows, likewise, that a perpendicular through the middle of 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 to HK. 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 9a ELEMENTS, &c. which 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 lines 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, andthe straight line BG will be the tangent required. For the angle CBD = CDB ; and there- fore the angle GBC (32. 3.) is also equal to CD B, an angle in the 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 rectilineal 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 the 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 the 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 the 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, hexch gons ; those of seven sides, heptagons ; those of eight sides, octagons i 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 three sides ; and the square is one of four sides. LEMMA. Any regular polygon may he inscribed in a circle, and circumscribed about one. Let ABODE, &c. be a regular polygon : describe a circle through the three points A, B, C, the centre being O, and OP the perpendicular let fall from it, to the middle point of BO : join AO and OD. If the quadrilateral OPOD be placed upon the quadrilateral OPBA, they will coincide ; for the side OP is common : the angle 0P0= OPB, being right ; hence the side PO will ap- ply to its equal PB, and the point will fall on B ; besides, from the nature of the polygon, the angle POD=PBA; hence OD will take the direction BA, and since CD =B A, the point D will fall on A, and the two quadrilaterals will entirely coincide. ^_^ _, The distance OD is therefore equal to AO ; a* and consequently the circle which passes through the three points A, B, 0, 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, 0, 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, 0, 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, BO, OD, &c. are equal chords ; they are therefore 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 BO, 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 he 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 froTn 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 0, 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 same 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 eqiMl to a given straight line, not greater than the diameter of the 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 CF ; 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 ; it 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 point 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 n 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 BAG 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 ABG PROP. III. PROB. About a given circle to descrj^ 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, KG, 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 j and because two of 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 ing 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. 1.) 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 straigjit lines DE, DF, DG, are equal to one another, and the circle described from the centre D, at the distance of any of them, will pass through the extremities of the other two, and will touch 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 about ABC. Bisect flO. 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, AE, 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. Join 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 = 0B, is proved as before. The joints between the arch-stones, or voussoirSf are only continuations of radii drawn from the centre 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. because EA. is at rigtit angles to BD, and common to the triangles ABE, ADE ; the base BA is equal (4. 1.) to the base AD ; and, for the same reason, BC, CD are each of them equal to B A or AD ; therefore the quad- rilateral figure ABCD is equilateral. It is also rectangular ; for the straight line BD be- ing a diameter of the circle ABCD, 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 it is inscribed in the circle ABCD. SCHOLIUM. Since the triangle AED is right angled and isosceles, we have (Cor. 2. 47. 1) AD : AE : : -v/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, G .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 ; fo^| GBEA being a parallelogram,' and AEB a right an- J"" 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 was demonstrated to be equilateral ; therefore it is a square ; and it is de- scribed about the circle ABCD. 13 B r ^ y D K 9S 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. L) 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, BAG, 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 BAG, 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^the half of ABC ; the angle EAB is equal to the angle EBR : and the side EA (6. 1.) to the side EB. In the same manner, it may be demonstrated, that the straight lines EC, ED a,re 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 pass I OF GEOMETRY. BOOK IV. 99 tlirough 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, BP 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 CD A ; 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 also 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 ADB is equal to twice the an- " gle BAD, they are together equal to four times BAD, and therefore all "the three angles ABD, ADB, BAD, taken together, are equal to fivo 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, musj; " 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. XL PROB. * To inscribe ari equilateral and equiangular pentagon in a given circle. Let ABODE be the given circle, it is required to inscribe an equilateral and equiangular pentagon in the circle ABODE. 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 ABODE inscribe (2. 4.) the triangle AOD equiangular to the triangle FGH, so that the angle OAD be equal to the angle at F, and each of the angles AOD, ODA equal to the angle at G or H : where- fore each of the angles AOD, ODA is double of the angle OAD. Bisect (9. 1.) the angles AOD, ODA by the straight lines OE, DB ; and join AB, BO,ED, EA. ABODE is the pentagon required. Because the angles AOD, ODA are each of them double of OAD, and are bisected by the straight lines OE, DB,the five angles DAO, AOE, EOD, ODB, BDA are equal to one another ; but equal angles stand upon equal arcs (26. 3.) ; therefore the five arcs AB, BO, 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, BO, OD, DE, E A are equal to one another. Where- fore the pentagon ABODE is equilateral. It is also equiangular ; be- cause the arc AB is equal to the arc DE ; if to each be added BOD, the whole ABOD is equal to the whole EDCB ; and the angle AED"* stands •on the arc ABOD, and the angle BAE on the arc EDOB : therefore the angle BAE is equal (27. 3.) to the angle AED : for the same reason, each of the angles ABO, BOD, ODE is equal to the angle BAE or AED: there- fore the pentagon ABODE 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. 101 "(11. 2.). Apply in the circle, on each side of a given point, a line *' equal to the greater of these parts ; then (2. Cor. 10^. 4.), each of the "' 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, bj'- the last pro-^'i position, be in the points A, B, C, D, E, so that the arcs AB, BO, 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 ABODE in the point 0, to which FC is drawn from the cen- tre F, FC is perpendicular (1.8. 3.) to KL ; therefore each of the angles at is aright 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, OK. For the same reason, the square of FK is equal to the squares of FB, BK : therefore the squares of FC, OK 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 OK 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 OF, 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, arid OLD 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 CFD double of CFL ; therefore the angle KFC is equal to the angle CFL : now the right angle FCK is equal to the right angle FOL ; and therefore, in the two triangles FKC, FLO, 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'^ 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 FLO : and because KC is equal to CL, KL is double of KC ; in the same manner, it may be shewn that HK is double of BK ; and because BK is 102 -"^^ ELEMENTS equal to KC, as was demonstrated, and KL is double of KC, and HK double of BK, HK is equal to'KL ; in like manner, it may be shewn that GH, GM, ML are 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 KI^M 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 equiangUMr 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 s^e 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 drawn from the extremity of the diameter of a circle at right angles to it, OF GEOMETRY. BOOK IV. 108 touches ^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 ABODE. PROP. XIV. PROB. To describe a circle about a given equilateral and equiangular pentagon. Let ABODE be the given equilateral and equiangular pentagon ; it is required to describe a circle about it. Bisect (9. 1.) the angles BCD, ODE by the straight lines OF, 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 CBA, 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 tlie 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 ABODE. 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 equal 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 angleg ; therefore the angle EGD is the third part of two right an- 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 GO makes witt EB the adjacent angles EGO, 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; and also the angles vertical to them, BGA, AGF, FGE (15. 104 ELEMENTS 1.); therefore the six angles EGD, DGC, CGB, BGA, 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, BO, 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 FABCD 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 qmndecagon 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 AB, 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. 105 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 is plain that each polygon will exceed the preceding in surface or area. It is obvious that any regular polygon whatever might be inscribed in a 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 Gottingen, at length proved, in a tvork entitled Disquisitiones Arithmeticm, 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"+l> provided it be a prime number, that 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=z2. 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 depends upon the the- ory of algebraical equations, the student may consult Barlow's Theory of Numbers. 14 ELEMENTS OF 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 w», 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, 3 A 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 ; ?w(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 mnA. is A mul- " tiplied by the product of m into n. DEFINITIONS. 1. A less magnitude is said to be apart 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. lOT 4. Magnitudes are said to be of the same kind, when the less can be mul- tiphed so as to exceed the greater ; and it is only such magnitudes that 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. 6. Magnitudes are said to be proportionals, when the first has the same ratio to the second that the third has to th^ourth ; 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 has 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 of the ratios which *e 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. IJ. 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 to 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- " ties ; a ratio compounded of four equal ratios quadruplicate," &c. 13. 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. 18. Convertendo, by conversion ; when there are four proportionals, and it is inferred, that the first is to its excess above ^e second, as the third to its excess above the fourth. Prop. D. Book 5. r OF GEOMETRY. BOOK V. 109 19. Ex aequali (sc. distanlia), or ex asquo, 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 ane 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 pf 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 sequali, 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 <4f 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 ant/ 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+B+C is the same multiple of D-f- E+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 no ^ ELEMENTS Then, because A contains D three times, A=D4-D+D. For the same reason, B=E-i-E4-E ; And also, C=F+F+F. Therefore, adding equals to equals (Ax. 2. 1.), A+B+C is equal to D+E+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. CoR. Hence, if m be any number, 77jD+'mE+^F=m(D+E+F). For 7?iD, mE, and wiF are multiples of D, E, and F by m, therefore their sum is also a multiple of D+E+F by w. PROP. II. THEOR. If to a multiple of a magnitude hy any number^ a multiple of the same mag' nitude hy any number he added, the sum will he the same multiple of that magnitude thai the sum of the two numbers is of unity. Let A=mC, and B=nC ; A+B=(m+n)C. For, since A=7nC, A^C+C + C+&c. C being repeated m times. For the same reason, B=C+C+&c. C being repeated n times. Therefore, adding equals to equals, A+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=mE, B=7iE, C=j?E, it is shown, that A+B + C=(m+» +;>)E. CoR. 2. Hence also, since A+B+C=(m+n+jj)E,andsinceA=mE, B=nE, andC=pE, mE+nE+pE=(m+n+;))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=mB, and B=nC ; then A=mnC. Since B=nC, mB=7iC+nC + &c. repeated m times. But nC+nC, &c. repeated m times is equal to C (2. Cor. 2. 5.), multiplied by n+n+&c. n being added to itself m times ; but n added to itself m times, is n multi- plied by m, or mn. Therefore nC+nC+&c. repeated m times=mnC; whence also ffiB=»inC, and by hypothesis A=mB, therefore A=m7iC OF GEOMETRY. BOOK V. - HI PROP. IV. THEOR. If the first of four magnitudes has the same ratio to the second which the third has to the fourth, and if any equimultiple whatever he 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 of 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 j9, 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 pmA 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, y'lz.pmA and;?»iC, and of B and D, any equimultiples ^nB, qnD, ii pmA be greater than qnB,pmC must be greater than^nD (def. 5. 5.) ; if equal, equal ; and if less, less. But pmk, pmC are also equimultiples of »iA and mC, and qnB, qnJ) are equimultiples of nB and nD, therefore (def. 5. 5.), mA : nB :: mC :nJ). 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 mC, 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 luhole 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=D4-B. Therefore (1. 5.) mA=7/iD+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 and n, and let m be greater than n ; mA— nA contains A as oft as m — n con- tains imity, or mA— nA=(m— n)A. 1 12 ELEMENTS Let m—n=zq\ then ^=71+5^. Therefore (2. 5.) mk=:nk-\-qk ; take nk. from both, and mh. — nkz=qA.. Therefore mk — nk contains A as oft as there are units in q, that is, in wi— n, or mk—nk=^{m—n)k. Cor. When the difference (jf the two numbers is equal to unity or m—> n=l, then mk — nA=A. PROP. A. THEOR. If four magnitudes he proportionals, they are proportionals also when taken inversely. If A : B : : C : D, then also B : A : ; D : C. Let mk and mO be any equimultiples of A and C ; nB and nD any equi- multiples of B and D. Then, because A : B : : C ; D, if mk be less than nB, mC will be less than nJ) (def. 5. 5.), that is, if nB be greater than mk, nD will be greater than mC For the same reason, if nB=niA, nD=niC, and if nB /^ mk, nD / mC. But nB, nD are any equimultiples of B and D, and mk, mC any equimultiples of A and C, therefore (def. 5. 5.), B": A • . D : C. PROP. B. THEOR. If the first he 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 mk, mB be equimultiples of the magnitudes A and B, mk : A : : mB : B. Take of mk and mB equimultiples by any number n ; and of A and B equimultiples by any number p ; these will be nmk (3. b.),pk, nmB (3. 5.), pB. Now, if nmA be greater than pk, nm is also greater than p ; and if nm is greater thanp, nmQ is greater than pB, therefore, when nmk is great- er than pk, nmB is greater than pB. In the same manner, if nmk=pk, nmB=pB, and if nmk/_pk, nmB/_p^- Now, nmA, nmB are any equi- multiples of mk and mB ; and pk, pB are any equimultiples of A and B, therefore mk : A : : rnB : 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 C : : B : D and inversely (A. 5.) C : A : : D : B. PROP. C. THEOR. If the first he to the second as the third to the fourth; and if the first he 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 is the same multiple of D, that is, if A=mB, C=nip. Take of A and C equimultiples by any number as 2, viz. 2A and 2C j and of B and D, take equimultiples by the number 2m, viz. 2mB, 27nD (3. OF GEOMETRY. BOOK T. 113 5.) ; then,because A=OTB,2A=2mB ; and since A : B : : C : D, and since 2A=2mB, therefore 2C=2mD (def. 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 : 0. Let mA, wiB, be any equimultiples of A and B ; and nC any multiple ofC. Because A=B, mA=mB (Ax. 1.5.); wherefore, if mA be greater than nC, mQ is greater than nC ; and if mA=nC, »iB=7iC ; or, iimK./^nC, wB /_nC But wiA and mB are any equimultiples of A and B, and /»C is any multiple of C, therefore (def. 5. 5.) A : C : : B : 0. Again, if A=B, C : A : : C : B ; for, as has been proved, A : C : : B : C, and inversely (A. 5.), : 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 4- 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+B. Let m be such a number that mk. and mQ are each of them greater than C ; and let nO be the least multiple of C that exceeds mA+mB ; then nC — C, that is {n—\)C (1. 5.) will be less than mA+mB, or mA+^B, that is, m(A-l-B) is greater than (n — 1)C. But because nC is greater than wiA+TwB, and C less than /nE, nQ — C is greater than mA, or mA is less than nC— C, that is, than (n— 1)0. Therefore the multiple of A+B by m exceeds the multiple of by n — 1, but the multiple of A by m does not exceed the multiple of C by ra— 1 ; therefore A+B has a greater ratio to C than A has to C (def. 7. 5.). Again, because the multiple of C by n — 1, exceeds the muWple of A by m, but does not exceed the multiple of A+B by m, C has a greater ratio to A 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 equal ^ to one another. '■•] If A: C :: B : C, A=B. For if riot, let A be greater than B ; then because A is 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 ifmA exceed 71C, 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 jGlrst case, A=B. PROP. X. THEOR. That magnitude, which has a greater ratio than 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 mAynC, and mB/nC (def. 7. 5.). Therefore also mA/mB, and A/B (Ax. 4. 5.). Again, let : B/C : A; B/A. For two numbers, m andn maybe found, such that mC^nB, and mCj/nA (def. 7. 5.). Therefore, since nB is less, and nA greater than the same magnitude mC, nB /_ nA, and there- fore B/ 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, nJ), nF, any equimultiples of B, D, and F. Because A : B : : C : D, if mAynB, mC/nD (def. 5. 5.) ; butifmC/nD, mE7nF(def. 5. 5.), because C : D : : E : F ; therefore if mAynB, mE ynF. In the same manner, if mA=: nB,mE=n'^; and if wiA/nB, wiE/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. IU PROP. XII. THEOR. // any number of magnitudes he proportionals, as one of the antecedents is to its consequent, so are all the antecedents, taken together, to all the conse- quents. If A : B : : C : D, and C : D : : E : F ; then also, A : B : : A+C+E : B+D+F. Take wiA; mC, twE 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 mK y nBjWiC/nD (def. 5. 5.) ; and when mCynD, mE/nF, because C : D : : E : F. Therefore, if mA/nBjmA+mC+mE/nB+wD-i-wF: In the same manner, if mA=nB, mk-\-mC-\-m^z=nB-\-nD-\-nF ; and if mA./^ nB, mA+mC+mE/nB + wD+nF. Now, mA+mC4-mE=m(A+CH- E) (Cor. 1. 5.), so that mA and mA+mC+mE are any. equimultiples of A, and of A+C+E. And for the same reason nB, and wB+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 the fourth, hut 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 : : : D ; but C : D7E : F ; then also, A : B/E : F. Because C : D /E : F, there are" two numbers m and n, such that mC y nD, but 7nE/_nF (def. 7. 6.). Now, if mC ynJ), mkynB, because A : B : : C : D. Therefore mkynB, and mE/_nF, wherefore, A : B/E : 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 he greater than the third, the second shall he greater than the fourth; if equal, equal ; and if less, less. . If A : B :: : D; then if A/C, B/D ; if A=C,B=D; andif A/ C, B/D. . First, let A7C ; then A : B /C : B (8. 5.), but A : B : : C : D, there- fore C : D/C : B (13. 5.), and therefore B/D (10. 5.). In the same manner, it is proved, that if A=C, B=D ; and if A/C, B/D. • PROP. XV. THEOR. Magnitudes have the same ratio to one another which their equimultiples have. If A and B be two magnitudes, and m any number, A : B . : mk : mB, Because A : B : : A : B (7. 5.) ; A : B : : A+A : B+B (12, 5.), or A: 116 ELEMENTS B : : 2A : 2B. And in tlie 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 he proportionals, they will also be pro- portionals when taken alternately. If A : B : ; C : D, then alternately, A : C . : B : D. . Take mA, mB any equimuhiples of A and B, and nC, nJ) any equimul tiples of C and D. Then (15. 5.) A : B : : mA : wB ; now A : B : : C : D, therefore (11. 5.) C : D : : mA : mB. ButC : D : : nC : wD (15. 5.) ; therefore mA : mB : : nC : nD (jl. 5.) : wherefore if mA/nC, mB'/nD (14. 5.) ; if mA=nC, mB=wD, or if mA/wC, mB/nD ; therefore (def. 5. 5.) A ; C : : B : D. PROP. XVII. THEOR. If magnitudes J taken jointly, he proportionals, they will also he 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 third has to the fourth. If A+B : B : : C+D : D, then by division A : B : : C : D. Take mA and nB any multiples of A and B, by the numbers m and n ; and first, let mA ynB : to each of them add mB, then mK-^-mB ymB-\-nB, But mA+mB=m(A4-B) (Cor. 1. 5.), andmB+;iB=(m+n)B (2. Cor. 2. 5.), therefore m(A+B)7(m-hw)B. And because A+B : B :: C+D : D, if m(A+B)7(m+w)B, m(C+D) /(m-j-wjD, or mC+mD/mD+yiD, that is, taking mD from both, mC/ nD. Therefore, when mA is greater than nB, mC is greater than nD. In like manner it is demonstrated, that if mA=nB, mC=nD, and if mA^wB, that mDZnB; therefore A : B : : C : D (def. 5. 5.). PROP. XVIII. THEOR. If magnitudes, taken separately, he proportionals, they will also he proportion- als when taken jointly, that is, if the first he to the second as the third to the fourth, the first and second together will he to the second as the third and fourth together to the fourth. If A : B : : C : D, then, by composition, A+B : B : : C+D : D. Take m(A+B), and wB 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)7''iB. For the same reason, m(C+D)7'wD. In this case, therefore, that is, when my n, m^+B) is greater than nB, and m(C+D) is greater than nD. And in the same manner it may be proved, that when «i=n, m(A+B) is greater than nB, and m(C+D) greater than nD. OF GEOMETRY. BOOK V. 117 Next, let m/n, or n/m, thenm(A+B) maybe greater than nB, or may- be equal to it, or may be less ; first, let m(A+B) be greater than nB ; then also, mA-f-mB/nB ; take mB, which is less than wB, from both, and wiA 7nB— »iB,or mA'/(n^m)B (6. 5.). But if mA7(n— m)B,mC7(n— m) D, because A : B : : : D. Now, (n— w)D=7iD— mt) (6. 5.), therefore mCynB—mJ), and adding mD to both, mC+TziD/fiD, that is (1. 5.), ^(C+D)7nD. If, therefore, 77i(A+B)7nB,m(C+D)7nD, In the same manner it will be proved, that if w(A4-B)=nB, m{C+'D) =nD; and if m(A+B)ZnB, m{C-{-D)ZnD ; therefore (def. 5.5.),A+ 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 he to the remainder as the whole to the whole. If A ; B : : C : D, and if be less than A, A— : B— D : : A : B. Because A : B : : C : D, alternately (16. 5.), A : C : : B: D ; and there- fore by division (17. 5.) A— : : : B— D : D. Wherefore, again alter- nately, A— C : B— D : : C : D ; but A : B : : : D, therefore (11. 5.) A -C: B-D:: A: D. CoR. A-C : B-D : : C : D. PROP. D. THEOR. If four magnitudes he proportionals, they are also proportionals hy conversion, that is, the first is to its excess above the second, as the third to its excess above the fourth. If A : B : : : D, by conversion, A : A— B : : C : C— D. For, since A : B : : G ; D, by division (17. 5.), A— B : B : : G~D : D, and inversely (A. 5.) B : A— B : : D : 0— D ; therefore, by composition (18. 5.), A : A-B :: : C-D. CoR. In the same way, it may be proved that A : A+B : : C : C-j-D. PROP. XX. THEOR. If there be three magnitudes, and other three, which taken two and two, have the same ratio ; if the first be greater than the third, the fourth is greater than the sixth ; if equal, equal ; and if less, lees. If there be three magnitudes, A, B, and C, and other three D, E, and F ; and 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 ZF. A, B, C, D, E, F. First,letA7C; thenA : B7C : B (8. 5.). ButA : B : : D : E, there- fore also D : E7C : E (13. 5.). Now B : C : ; E : F, and inversely (A. A, B, D, E, c, F. but A : B : :E:F, 118 ELEMENTS . 5.), C : B : : F : E ; and it has been shewn that D : E/C : B, therefore D : E/F: E (13. 5.), and consequently D/F (10. 5.). i Next, let A=C; then A : B :: C : B (7. 5.), but A : B :: D : E ; there- 1 fore, C : B : : D : E, but C : B : : F : E, therefore, D : E : : F : E (11." 5.), and D=F (9. b.). Lastly, let A/ C. Then C / 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, F 7 D, that is, if A / C, D ^ F, PROP. XXI. THEOR. If there he three magnitudes, and other three, which have the same ratio taken two and two, hut in a cross order; if the first magnitude he 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; ifA7C,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 : F7C : B (13.5.). Now, B : C : : D : E, and inversely, C : B : : E : D ; there- fore,E : F7E : D (13. 5.), wherefore, D 7F (10. 5.). Next, let A=C. Then (7. 5.) A : B : : C : B ; 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 C/A, 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 he any numher 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 : : : D : F. Take of A and D any equimultiples whatever, mk, mD ; and of B and D any whatever, nB, nF : and of C and F any whatever, ^^C, qF. Because A : B : : D : E, TwA : nB : : TwD : nE (4. 5.) ; and for the same reason, nB : ^C : : nE : ^F. Therefore (20. 5.) according as mk is«greater than ^C, equal to it, or less, mD is greater Ihan* ^F, equal to it, or less ; but mk, mD are any equimultiples of A and D ; and ^C, ^F are any equimultiples of G and F ; therefore (def. 5. 5.), A : C : : D : F. Again, let there be four magnitudes, and other four which, taken two * N. B. This proposition is usually cited by the words " ex aequali," or " ex aequo." A, B, C, D, E, F, mk. nB, qO, mD, nE, qY. OF GEOMETRY. BOOK V. It9 A, E, B, and two in order, have the same ratio, viz. A:B:: E:F; B:C G; C : D :: G : H,thenA: 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. In the same manner is the demonstration extended to any num- ber of magnitudes. C, G, PROP. XXIII. THEOR. If there he 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 : 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 : : tjE : «F ; therefore, mA : mB : : nE ~ because B : : : D : E. mB : nC : : mD : tiE (4. 5.) ; and it has been just shewn that mA : mB : : nE :nF; therefore, if mA7nC,mD /nF (21.5.) ; if mA=nC, mD=nF ; and if mA^nC, mD/nF. Now, mA and mD 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 : D : : E : F, then, A : D : : E : H. Fcf, since A, B, C, are three magnitudes, and F, G, H, other 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. nF(ll .5.). Again, A, B, c, D, E, F, mA, mB, nC, mD, nB, nF. A, E, B, F, C, G, D, H. PROP. XXIV. THEOR. If the first has to the second the same ratio which the third has to the 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 : D. * N. B. This proposition is usually cited by the words " ex aequali in proportione pertur bata:" or, " ex »quo 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 A/B, A4-B : A-B :: C+D : G-D; orifA^B A+B ; B-A:: C+D : D-C. For, if A 7B, 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 another. 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 dift'erence 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 similar. For, let A=mC, and B=nC ; A+B = (m+n)C, and A— B== (m— n)C. CoR. IfC 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 diflference of B and A+B is also A. A, B, C, D, E, F. ELEMENTS OF 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 fibres, the sides which lie fdjacent to equal angles, are called homologous sides. Those angles themselves are called homo- logous angles. In different circles, similar arcs^ sectors, and segments, are those of which the arcs subtend equal angles at the centre. Two equal figures are always similar ; but two similar figures may be vewjr unequaf 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 th€ 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 parallelogram is the perpendicu- lar which measures the distance of two oppo- site sides, taken as bases. And the altitude of a trapezoid is the perpendicular drawn 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 the triangle ACD, and Uie 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 IjC is of the base CD, the same mul- tiple is the triangle ALC of the triangle ADC. But if E A V the base HC be equal to the base CL, the triangle AHC is also equal to the triangle ALC (38. 1.): and if the base HC be greater ihan the base CL, likewise the trian- gle AHC is greater than the triangle ALC ; and if less, hess. 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 whateverhave 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 tnangle ALC ; and since it has been shewn, that if the base HC be greater than 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. 1.), 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 the parallelogram CF. And because it has been shewn, that, as the base BC is to the base CD, so is the triangle ASC 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 (XL 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. i OF GEOMETRY. BOOK VI. izs PROP. II. THEOR. If a straight line he drawn parallel to one of the sides of a triangle^ it will 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 he parallel to the remaining side of the tri- 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 toEA(ll. 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 ADE (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. ELEMENTS PROP. III. THEOR. If the angle of a triangle he bisected hy a straight line which also cuts the hose ; 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 BAG, 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 inE. 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 constructioir 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.) : C()nsequently AC is equal to AE (9. o.), 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 Vl. |5Mf PROP. A. THEOR. If the exterior angle of a triangle he bisected hy a straight line which also cuts the base produced ; the segments between thebisecting line and the extremities of the base have the sarhe ratio which the other sides of the triangles have to 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 afngle DAE. The same construction being jmade, 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 sides about the equal angles of equiangular triangles are proportionals ; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequent^ 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 ABC, 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 n tp (34. 1.) : And because AC is parallel to FE, -K O lii one of the sides of the triangle FBE, BA : AF : : BC : CE (2. 6.) : but AF is equal to CD ; therefore (7. 5.) BA : CD : : BC : CE ; and alter- nately, BA : BC : : DC : CE (16. 5.) : Again, because CD is parallel to BF, BC : CE : : FD : 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 two triangles, about each of their angles, he proportionals, the triangles shall be equiangular, and ha.v€u 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 ; th« 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 EOF (4. Cor. 32. 1.), and the trian- gle ABC is therefore equiangular to the triangle GEF ; and consequently they have their sides opposite to the equal angles proportionals (4. 6.). Wherefore, 4 AB : BC : : GE : EF ; but by supposition, AB : BC : : DE : EF, therefore, DE:EF:: GE : EF. Therefore (11. 5.) DE and GE have OF GEOMETRY. BOOK VI. 127 the 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 G*:iF, 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 he equian- gular, and shall have those angles equal which are opposite to the homolo' gous sides. Let the triangles ABC, 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, make (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. 1.), 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 angle EDF (Hyp.) ; wherefore also the re- maining angle at B is equal to the remaining angle at E. Therefore tho triangle ABC is equiangular to the triangle DEF. 128 ELEMENTS PROP. VII. THEOR. If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals y then, if each of the remaining angles he either less, or not less, than a right angle, the triangles shall be equiangular, and have those angles equal about which the sides are propor- tionals. Let the two triangles ABC, DEF have on€ angle in the one equal to one angle in the other, viz. the angle BAG 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 aright 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 bp 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 angle : Where- OF GEOMETRY. 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 right angled triangle if a perpendicular he drawn from the right angle to the base; the triangles on each side of it are similar to the whole triangle ^ 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 pfoportionals (4. 6.) ; wherefore the triangles are similar (def 1. 6.). In like manner, it may be demonstrated, that the triangle ADC is equiangular 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 tjie base ; and also that each of the sides is a mean proportional between the base, and its segment adjacent to that side. For in the triangles BDA, ADC, BD 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 which 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 take AC such that it shall contain AD, as oft as AB 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. Because ED is parallel to one of the sides of the 17 tan 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 muUiple 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. l.).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 BG : GF ; therefore the given straight line AB is divided similarly PROP. XL PROB. Tojind 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 straight lines AB, AC a third proportional, CE is found. OF GEOMETRY. BOOK VI. m PROP. XII. PROB. Tojlnd a fdurth 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 hav- 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 tie triangle DEF, DG : GE : : DH : HF (2. 6.) ; but DG=A, GE=B, and DH=C ; and therefore A : B : : : HF. Wherefore to the three given straight lines, A, B, C, a fourth proportional HF is found. PROP. XIIL PROB. To find a mean proportional befuoeen 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 . 1.) 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 proportional between AB, BC, the seg- ments of the base (Cor. 8. 6.) ; therefore between the two given straight lines AB, BC, a mean proportional DB is found. m r\df) ELEMENTS PROP. XIV. PROB. Equal parallelograms which have one angle of the one equal to one angle of the other f 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^ ore 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 sif^es 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 BAG 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 line ; 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 EAD OF GEOMETRY. BOOK VI. 133 : triangle BAD ; but CAB : BAD : : CA ; AD, and E AD : BAD : ; EA : AB ; therefore CA: AD::EA: AB(11.5), 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. C A 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 is equal (9. 5.) to the triangle EAD. PROP. XVI. THEOR. to If four straight lines he proportionals, the rectangle contained hy the extremes is equal to the rectangle contained hy the means; And if the rectangle contained by the extremes he equal to the rectangle contained hy 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 : ]g| and the parallelogram BG is contain- ed by the straight lines AB, F ; be- F— ■ cause AG is equal to F ; and the pa- rallelogram DH is contained by CD Q. 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 the straight lines AB, F be equal to that which is contained by CD, E | these fom: lines are proportionals, viz. AB is to CD, as E to F. if BOD m ■' ELEMENTS 0/ 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. XVn. THEOR. If three straight lines he proportionals, the rectangle contained by the extremes ts equal to the square of the mean : And if the rectangle contained hy the ex- tremes he equal to the square of the mean, the three straight lines are propor- 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 equal 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- ^ ' ~" angl^ B.D is equal to the square of B, because B= :r ~ 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 B ; 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 ; wherefore A ; B : : B : C. PROR XVHL PROB. Upon a given straight line to describe a rectilineal fgure 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 describe a rectiline&l 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 SAG equal to the angle atC, and the angle ABG equal to the angle CDlf* ; therefore the remaining angle CFD is equal 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 the 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 : I F for the same reason, the angle ABH is equal to the angle ODE ; 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 figurts have their sides about the equal angles proportionals : for the triangles GAB, FCD being equiangular, BA : AG : : DC : OF (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 «quali (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 iimilarly 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. 6.) ; therefore, ex aequali (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.) DKto KE, because the triangles BLH, DKE 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- 1^ ELEMENTS tilineal figure of 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 oneanother (15. 6.): therefore -A. the triangle ABG is equal to thetriangle DEF; and because x/ \ .*v that BC is to EF, as EF to / / \ 1> 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 IE 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 triangle upon the first to a similar, and similarly described triangle upon the second. OF GEOMETRY. BOOK VI. 137 PROP. XX. THEOR. Similar polygons may he 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 ABODE, FGHKL, be similar polygons, and let AB be the homo- logous side to FG: the polygons ABODE,* 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 ABODE has to the poly- gon FGHKL a ratio duplicate of that which the side AB has to the side FG. Join BE, EC, GL, LH : and because the polygon ABODE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL (def. L 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 ABO is equal (def. 1. 6.) to the whole angle FGH ; therefore the remaining angle EBO 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 : BO : : FG : GH (det 1.6.); therefore, ex sequali (22. 5.) EB : BO : : LG : GH, that is, the sides about the equal angles EBO, LGH are proportionals ; therefore (6. 6.) the triangle EBO is equiangular to the triangle LGH, and similar to it (4. 6.). For the same reason, the triangle EOD is likewise similar to the triangle LHK ; therefore the similar polygons ABODE, 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, EBO, EOD, and the consequents FGL, LGH, LHK : and the polygon ABODE 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 the triangle FGL, ABE has to FGL the duplicate ratio (19. 6.) of that which the side BE has to the side 18 r t . I ' n^ ELEMENTS GL : for 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 (11 . 5.) is the triangle BEC to the triangle GLH. Again, because the triangle EBG 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 triangle I 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 to 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 the squares of their homologous sides. ill* * OF GEOMETRY. BOOK VI. 139 SCHOLIUM. If two 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:, 4fec. 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 j are also similar to one another. Let each of the rectilineal figures A, B be simikr 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, they 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 (1 1. 5.). Therefore A is similar (def. 1. 6.) to B. PROP. XXII THEOR. If four straight lines le proportionals, the similar rectilineal figures similarly described upon them shall also he 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. To AB, CD take a third proportional (11.6.) X; and to EF, GH,ft third proportional ; and because , . .- - ** 140 ELEMENTS AB : CD ; : EF : GH, and CD : X : : GH : (11. 5.) 0, ex aequaU (22. 5.) AB : X : : EF : 0. But AB : X (2. Cor. 20. 6.) : : KAB : LCD ; and EF : : : (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 'the 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 ]P G HO P R 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 similay, 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. XXHL THEOR. Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. Let AC, CF 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 ratios 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 ta 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 • t I 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 (11. 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 hy their altitudes. SCHOLIUM. Hence the product of the base by the altitude may be assumed as the measure 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 3x 10, 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 1^ 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 : the number 30, for instance, by which the rectangle A was measured, now represents 30 superficial units, or 30 of those squares, which have each of their sides equal to unity, _,^^ 142 ELEMENTS Cor. 1. Hence, the area of any parallelogram is equal to the product of its base 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 J 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 BAG, 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 GA : 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 rectilineal figures which are similar to the same rectilineal figure, are also similar to 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.orl4.1.), 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 1 OF GEOMETRY. BOOK VI. 143 GH to CF, and if three straight lines be proportionals, as the first is to the 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 is 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 B E ; 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 I>; 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. XXf I. THEOR. If two similar parallelograms have a common angle, and he 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 «0 V/ 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 sam'e diameter ; wherefore ABCD and AEFG must be about the same diameter. A G^ I 144 ELEMENTS PROP. XXVII. THEOR. Of all the rectangles contained hy 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. XXyill. 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 thai#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 is divided equally in D, and unequally in G, AG.GB4-DG2=(5.2.) DB2=: EG2. But (47. 1.) ED2+DG2=EG2; therefore, AG. GB+DG2=ED2 +DG2, and taking away DG^, AG.GB=ED2. 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. ^B PROP. XXIX. PROB. To produce a given straight line, so that the rectangle contained by the segments 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 AB 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 equal 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+DB2= DG2=DE2. But (47. 1.) DE2=DB2+BE2, there- fore AG.GB + DB2 = DB2 + BE2, and AG.GB=BE2. 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. 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 and mean ratio. Upon AB describe (Prop. 46. 1.) the square BC, and produce CK 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 extreme 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 ]^ q g 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 U6 ELEMENTS PROP, XXXI. THEOR. i In 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 be a right angled triangle, having the right angle BAG : 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 ABD, 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. XXXIL THEOR. . If two triangles, which have two sides of the one proportional to two sides oj the other, be pined 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 to 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. B A to AC, as CD to DEj 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 BAG was proved to be equal to ACD : Therefore the whole angle ACE is equal to the two angles ABC, BAG ; add the common angle AGB, then the angles ACE, ACB are equal to the angles ABC, BAG, ACB : But ABC, BAG, 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, AGB equal to two right angles ; therefore (14. 1.) BC and 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 BAG, EDF at their circumferences ; as the arc BC to the arc EF, so is the angle BGC to the angle EHF, and the angle BAG 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, GK, KL are all equal, the angles BGC, GGK, 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 EHN of 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 has been proved, that if the arc BL be greater than EN, the angle BGL is greater than EHN ; and if equal, equal ; and if less, less ; As therefore, the arc BC to the arc EF, so (def. 5. 5.) is the angle BGC to the angle 148 ELEMENTS lo m 1 1 EHF : But as the angle BGC is to the angle EHF, so is (15. 5.) the an- gle BAG to the angle EDF, for each is double of each (20. 3.) : Therefore, as the circumference BG is to EF, so is the angle BGC to the angle EHF, and the angle BAG to the angle EDF. Also, as the arc BG to ^F, so is the sector BGG to the sector EHF. Join BG, GK, and in the arcs BG, GK take any points X, 0, and join BX, XG, GO, OK : Then, because in the triangles GBG, GGK, the two sides BG, GG are equal to the two GG, GK, and also contain equal angles ; the base BG is equal (4. 1.) to the base GK, and the triangle GBG to the tri- angle GGK : And because the arc BG is equal to the arc GK, the remain- ing part of the whole circumference of the circle ABG is equal to the re- maining part of the whole circumference of the same circle : Wherefore the angle BXG is equal to the angle GOK (27. 3.) ; and the segment BXG is therefore similar to the segment GOK (def. 9. 3.) ; and they are upon equal straight lines BG, GK : But similar segments of circles upon equal straight lines are equal (24. 3.) to one another : Therefore the seg- ment BXG is equal to the segment GOK : And the triangle BGG is equal to the triangle GGK ; therefore the whole, the sector BGG is equal to the whole, the sector GGK : For the same reason, the sector KGL is equal lo each of the sectors BGG, GGK; 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 BG, the same multiple is the sec- tor BGL of the sector BGG. For the same reason, whatever multiple the arc EN is of EF, the same multiple is the sector EHN of the sector JEHF ; 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 BG, EF, and the two sectors BGG, EHF, and of the arc BG, 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- tor EHN ; if equal, equal; and if less, less ; therefore (def. 5. 5.) as the arc BG, is to the arc EF, so is the sector BGG to the sector EHF. OF GEOMETRY. BOOK VI. 149 PROP. B. THEOR. If an angle of a triangle be bisected by a straight line, which likewise cuts the 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 BAG 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 : : EA : (4. 6.) AC, and consequently, BA.AC = (16. 6.) AD.AE=ED.DA(3. 2.) +DA2. But ED. DA=BD.DC, therefore BA.AC = BD.DC +DA2. PROP. C. THEOR. If from any angUof 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 ; B^ause the right angle BDA is equal to the angle ECA 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 the rectangle EA.AD. 150 ELEMENTS^ PROP. D. THEOR. The rectangle contained by the diagonals of a quadrilateral inscribed in a oircle, is equal to both the rectangles^ contained by its 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 : And 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 B DC, 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 BCD A + BA.DC = BD.CE4- 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. !F]or since ACBD is a quadrilateral in- scribed in a circle, of which the diagonals are AB and CD, AD.CBH-DB.AC (D 6.) = AB.CD : but AD.CB+DB.AC = AD.AC -I- 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. 151 PROP. F. THEOR. If two points he 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: and if from 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 the 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 AD, 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 ; therefore 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 : DIG. 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 angle 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 aequo, FA : AE : : FC : CE. But FB : 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 he drawn tn 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 hy the straight line 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 hy the diameter and the segment of it cut offhy 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 equal 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 BAG 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, AEF : then, because AEF is a right angle, the circle described about the triangle AEF will have AF for 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 same circle. OF GEOMETRY. BOOK VI. 15a If to the angle DFC (15. 1.), and also ■ the angle BEF to the angle CDF, being both right angles, the triangles BEF, and CDF are equiangular, and therefore BF : EF : : OF : 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 abou^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 f 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 vtie base BC, so that BD, DC are the segments of the base ; (AC-j-AB) \C-AB)=(CD-f DB) (CD-DB.) 154 ELEMENTS From A as a centre with the radius AC, the greater of the two 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-1-DC= the sum of the segments. But when AD falls without the triangle, BG=DG+DB=CD-fDB= the sum of the segments of the base, and BC=CD— DB= the diff*erence 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 he 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. l.)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^=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 lerefore 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 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 contained 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 he 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^ (36. 3.) ; hence that rectangle is equivalent to the given square C. PROP. 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 parallQj|to the diameter, at a distance AD from it, equal to the side of the 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 the square 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. O. PROB. To construct a square that shall he 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 dravs^ the perpendicular KL. Through 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 .*— *^ .j— ^_. to GH, LM : LN : : LG : M N LH ; consequently, LM^ : LN^ : : LG^ : LH^ (22. 6.) ; but, since the trian- gle LGH is right angled, we have LG2 : LH2 : : GK : KH ; hence LM^ ; LN2 : : 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 hy a line from the vertex of one of its angles^ so that the parts may he 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 propor- tional to given lines, by dividing the base in the same proportion. OF GEOMETRY. BOOK VI. w PROP. Q. PROB. To divide a triangle into two 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 AB2 to AD2 (Prob. 4.) ; Draw DE parallel to BC, and the triangle is divided as required. For the triangles ABC, ADE being similar, ABC : ADE : : AB^ : AD2 ; but M+N : N : : AB2 ; AD2 ; 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 ACEP : PEB : : M : N B M : N ; 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 triangle into equal parts. I5d ELEMENTS PROP. S. PROB. To divide a triangle into three equivalent parts hy lines drawn from the vet' tices of the angles to the same point within the triangle. Make BD equal to a third part of BC, and draw DE parallel to BA, the side to which BD is adjacent. From F, the 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 third 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, FBO, FCA, are each equal to a third part of ABC. PROP. T.' PROB. To divide a trtangle into three equivalent parts y by lines drawn pom 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 ACG the quadrilateral ACGP will be equivalent to the triangle ACE ; but this triangle is one third of ABC ; hence the quadrilateral ACGP is one third of the triangle ABC : consequently, the spaces ABFP, ACPG, PFG are each equal to a third part of the triangle ABC. OF GEOMETRY. BOOK VI 159 PROP. U. PROB. To divide 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 to 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 ^in 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 th^ 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. f PROP. W. PROB. To divide a quadrilateral into two parts by a line parallel to one of its sides, so that these parts may be to each other as the line M is to the tine N. Produce AD, BC tDl they meet in E ; draw the perpendicular EF and bisect it in G. Upon the side GF construct a rectangle equivalent to the 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^ to Ea^ ; draw ab paral- lel to AB, and it will divide the quad- rilateral into the required parts. For since the triangles EAB, Ea& are similar, we have the proportion EAB :Eab: : EA2 : Ea^; but by - -^ ^ _ ^ construction, EA2 : Ea^ : ; AB : "^ ^^ ^ *^ KB ; so that EAB : Eab : : AB : KB : : AB.GF : KB.GF ; and conse- quently, since by construction EAB=AB.GF, it follows that Eab=KB. GF, and therefore AK.GF=AJ, and since by construction AH. GF= AC, it follows that KH.GF=aC. Now AK.GF : KH.GF : ; AK : KH ; but AX : KH : : M : N ; consequently, A5 : aC : : M : N ; that is, the quadrilateral is divided, as required. 160 ELEMENTS, &c. PROP. X. PROB. To divide a quadrilateral into two parts hy a line drawn from a point in one of its sides, so that the parts may be to each other as a line M is to a line N. Draw PD, upon wliicli 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 k.a ; 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 + PD.;>5, that is, PD.DF is equivalent to twice the sum of the triangles APD, ;>PD , consequently, since DL is half DF, PD.DL=APpD ; and therefore PD. LK=PBCo ; but PD.DL : PD.LK : : DL : LK : : M : N ; consequently. APpD : PBCp : : M : N ; hence the quadrilateral is divided, as required. PROP. Y. PROB. To divide a quadrilateral hy 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 trH, then PQ will also divide the quadri- lateral AC as required. For, by construction DE.EF=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 ; that is, the quadrilateral is divided, as required. wg^atw I SUPPLEMENT TO THE ELEMENTS OP GEOMETRY. 21 ELEMENTS OF GEOMETRY. SUPPLEMENT, I BOOK I. OF THE QUADRATURE OF THE CIRCLE, LEMMA Any 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 APDB which envelopes it. We have already said that by the term convex line we understand a line, polygonal, 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, th^re 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 at least only touching, the line AMB. The straight line PQ is shorter than PCDEQ ; hence, if instead of the part PCDEQ, we substitute the straight line PQ, the enveloping line APQB will be shorter than APDQB. But, by hypothesis, this latter was shorter than any other ; hence that hypothesis was false ; hence all of the enveloping lines are longer than AMB 164 SUPPLEMENT TO THE ELEMENTS 1 Cor. 1. Hence the perimeter of any polygon inscribed in a circle is less than the circumference of the circle. Cor. 2. If from a point two straight lines be drawn, touching 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 is greater than the circumference of the circle. PROP. L THEOR. Ifjrom the greater of two unequal magnitudes there he 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 D 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 -j^i i™ 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 remainder AH, take HK equal to its half, and so on, until there be as many divisions in AB as there are 46 in 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 remainder HA. !B C iE! 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 sideSy inscribed in circles, are 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 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 andK. 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 all equal, and jthey OF GEOMETRY. BOOK I. IS^^ -^C H. W are equal in number to the others ; therefore, whatever part the arc AB is 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, that 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 NO, 01, it may be proved that the angles NBC, OHI are equal to one another, and to the angle ABN. Therefore the whole angle ABC 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- lar 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 square of GH ; but because the triangles ANB, GOH are equiangidar, the 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 the polygon GHIKLM 16¥ SUPPLEMENT TO THE ELEMENTS as the 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. in. PROB. The side oj 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. Let 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 KL 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 an- gles of the one equal to two angles of the other, and they have also the side GM common, therefore they are equal (26. 1 .),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 \ r OF GEOMETRY. BOOK I. 167 described about the circle, the sides of which will each be equal to KL, and will 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 same number of sides with the inscribed polygon ABGDEF. Cor. 1. Because GL, GK, GN, and the other straight lines drawn from the centre G to the angular points of the polygon described about the 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 ABGDEF. 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 hefound, 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 (1. 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 aro 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 remainder 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, 1«B SUPPLEMENT TO THE ELEMENTS they are as the squares of the homologous sides HL and AF (3. Corol. 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 tl^an 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 be described OF GEOMETRY. BOOK I. 169 r 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 hy 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 touching the circle, and meeting DA produced in E, and DB produced in Ku: L F ; EF will be the side of an equilateral polygon described about the cir- 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 EDF, DG is drawn perpendicular to the base, the triangle EDF 23 170 SUPPLEMENT TO THE ELEMENTS is equal to the rectangle contained by DG and the half of EF (41. 1.) ; and as the same is true of all the other equal triangles having their vertices ia 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 rect- 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. iVgain, 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 the circle ABC. But the same rectangle DA.AH has been proved to be less 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 semidiameter of the circle ABC, and AH the half of its circumference. Cor. 1. Because DA : AH : : DA2 : 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. And 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 less than the square of NP ; and therefore, since DA is greater than NP, HL is 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. VL THEOR. The areas of circles are to one another in the duplicate ratio, or as the squares f 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 space that the square of AD is to the square of GL as the circle ABD to the space Q. Because the polygons ABCDEF and GHKLMN are equi- lateral and of the same number of sides, they are similar (2. 1. Sup.), and \ 172 SUPPLEMENT TO THE ELEMENTS their areas are as the squares of the diameters of the circles in which they are inscribed. Therefore AD^ : GL^ : : polygon ABCDEF ; polygon GHKLMN; butAD2 : GL2 : : circle ABD : Q; and therefore, ABCDEF : GHKLMN: -.circle ABD: Q. Now, circle ABD /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 the 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. L 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 ABD, 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. I. Sup.), therefore AO.X : GP.Y : : AD2 : GL2 : : AO2 : GP2 ; and alternately, AO.X : AO2 : : GP.Y : GP2 ; whence, because rectangles that have equal altitudes are as their bases (1. 6.), X : A0-: : Y : GP, and again alternately, X : Y : : AO : GP : wherefore, taking the doubles of each, the circumference ABD is to the circumference GHL as 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 circle 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. VH. THEOR. Equiangular parallelograms are to one ahother as the products of the num. hers proportional to their sides. Let AC and DF be two equiangular parallelograms, and let M, N, P and Q be four numbers, such that AB : BC : : M : N ; AB : DE : : M : OF GEOMETRY. BOOK I. 173 P ; and AB : EF : : M : Q, and therefore ex aquali, BC : EF : : N : Q. The parallelogram AC is to the parallelogram DF as MN to PQ. Let NP be 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- A. B D K 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 MN to PQ 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 DF as 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 Q 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. : T) : : 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 tliird , that is, if A.C=B2, mXr=nXn, OT=n^. For by this Prop. A.C : B^ :: mXr : n^ ; hut A.C=B2, therefore mXr =n^. Nearly in the same way it may be demonstrated, that whatever is 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=B2, and if a number n be found such, that n'^z=mr, then A : B : : m : n. For let A : B : : m : 5', then since m, q, r are proportional to A, B, and C, and A.C=B2; therefore, as has just been proved, q^=m Xr ; but n'^=qxr, by hypothesis, therefore n^-^cf-^ and n-=.q ; wherefore A \B \\ m \ n. SCHOLIUM. In order to have numbers proportional to any set of magnitudes of the ^ \ 174 SUPPLEMENT TO THE ELEMENTS same kind, suppose one of them to be divided into any number m, of equal 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= a number m, it is understood that A=mx H, 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 the numerical fraction indicates. Thus, if A= 360.375, it is meant that there is a certain magnitude H, such that A=360xH-f' 375 XH, or that A is equal 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. VIIL 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 drawnfrom 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 OF and CG at right angles to DE and DB ; if OF 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 I OF, GEOMETRY. BOOK I. 175 is 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, CBG 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 ofCG. 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, AD2=4AH.AF. Therefore also, because 4CG2=AD2, 4CG2=4AH.AF, and CG2=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 digmeter^ hy a line less than ten of the parts ^ of which the diameter contains seventy ^ hut greater 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 — , or 70' -, of AB, but greater than — - of AB. * In this proposition, the character -f" placed after a numb-jr, signliles that something is to be added to it ; and the character — , on the other hand, signifies tirat something is to be taken away 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, CG2=(17. 6.) AH. AF=500x 1500=750000; wherefore CG=866.0254+, because (866. 0254)2 is less than 750000. Hence also, AC-f CG=1866.0254-f . 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 P2=AH (AC+CG)= 500 X (1866.0254-f ) = 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 Q2=AH (AC+P)=500(1965.9258+)=982962. 9+ ; and therefore Q=991.4449+, because (991.4449)2 jg 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, S2=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, T2=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) =2000 X (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 sides, 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, that 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 viz. 6285.461—, then, because, 999.46458 : 1000 i : 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 of these proportionals is greater than the second, the third, viz. 6286 461 — 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 those 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 CG2=750000 ; wherefore 00=866.02545—, because (866.02545)2 is greater than 750000 ; therefore AC-f CG=1866. 02545—. Now, P being, as before, the perpendicular from the centre on the chord of one twelfth of the circumference, P2=AH (AC+CG) =r-500x(1866. 02545)— =933012.73— ; and P = 965.92585— , because (965.92585)2 is gxeater than 633012.73. Hence also, AC-t-P=1965.92585 — . 23 178 SUPPLEMENT TO THE ELEMENTS Next, as Q= the perpendicular drawn from the centre on the chord of one twenty-fourth of the circumference, Q2= 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, S2=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-f-, and the chord itself = 65.4377-f because (65.4377)2 is less than 4282.1 : Now the chord of one ninety-sixth part of the circumference being =:65.4377-|-, the perimeter of a polygon of ninety-six sides inscribed in the circle =(65.4377 -|-)96= 6282.01 9 -j-. But the circumference of the circle is greater than the pe- rimeter of the inscribed polygon ; therefore the circumference is greater than 6282.019, of those parts of which the radius contains 1000 ; or than 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-f -- to 1 ; therefore the circumference of the circle has a greater ratio to the diameter than 34- rrrhas to 1 ; that is, the excess of the circumference? 71 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. And if as 1 to 3 -|- — , or ^s 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 sam^ 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 carried 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 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 of the polygons differ by less than that quantity ; and consequently the cir- cumferenca of the circle, which is greater than the least, and less than the greatest of these numbers, is determined within less than the milliontli part of the radius. Hence also, if R be the radius of any circle, the circumference is greater than Rx 6.283185, or than 2Rx 3.141592, but less than 2Rx 3.141593 ; 180 SUPPLEMENT TO THE ELEMENTS, &c. and these numbers differ from one "another only by a millionth part of the radius. So also R2+3.141592 is less, and R^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 as- i^igned. ELEMENTS OF GEOMETRY. SUPPLEMENT. BOOK II. r ■ ■ ■ . ' / OF. THE INTERSECTION OF PLANES. DEFINITIONS. 1. 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 plane 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 drawn 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. 6. 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. PROP. I. THEOR. One part of a straight line cannot he in a plane and another part ahove 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 in 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 another are in one plane : Also, that any three points whatever are in one plane ill 0-, •; OF GEOMETRY. BOOK II. 183 PROP. III. THEOR. If two planes cut one another^ their common section is a straight line. Let two planes AB, BO 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 T) 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 BD is therefore common to the planes AB and BC, or it is the common section of these planes. PROP. IV. THEOR. 3 JT 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 line ; 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"=AB2 +AD2 (47. 1.); and for the same reason, BF2 = AB2+AF2, therefore BD2+BF2= 2AB2 + AD2 + AF2 ; and because DF is bisected in G (A. 2.), AD2+AF2=2AG2+ 2GF2, therefore BD2+BF2=2AB2+2AG2 +2GF2. But BD2 + BF2 2GF2=2AB2+2AG2+2GF2 ; and taking 2GF2 from both, 2BG2=2AB2 +2AG2, or BG2=AB2+AG2; whence BAG (48. 1.) is a right angle. Now AG is any straight line drawn in the plane of the lines AD, AF ; and when a straight line is at right angles to any straight line which it meets 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. (A. 2.) 2BG2+2GF2, therefore 2BG2+ 184 SUPPLEMENT TO THE ELEMENTS PROP. V. THEOR. If three straight lines meet all in one point, and a straight line stand at right 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 BO, 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, EC, 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. VL THEOR. 7\oo 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, AB24-BE2= (47. 1.) AE2, and because BDE is a right angle, BE2=BD2 + DE2; therefore AB24.BD2+DE2=AE2 ; now, AB2+BD2=AD2, because ABD is a right angle, therefore AD2+DE2=AE2, 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 of the angles ABD, BDC is a. right angle ; therefore AB is parallel (Cor. 28. l.)toCD. PROP. VII. THEOR. If two straight lines he parallel, and one of them at right angles to a plane ; the other is also at right angles to the same plane. Let AB, CD be two parallel straight lines, and let one of them AB be at ^ C Q. 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. 1.). 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 anajles 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. lei (6. 2. Sup.) to one another. Therefore AB is parallel to CD. PgOP. 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, and are not in the same plane with AB, 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 equal 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 the 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.) AD perpendicular 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 plane right point ►DE; D draw (Prop. 11. 1.), in the BH, the straight line DE at angles to BC ; and from the A draw AF perpendicular to 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, t)k. 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, Dx\, and is perpendicular (def. 1. 2. Sup.) to every straight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets it : Therefore GH is per- pendicular to AF, and consequently AF is perj^ndicular to GH ; and AF is also perpendicular to DE : Therefore AF is perpendicular to each of the straight lines GH, DE. But if a straight line stands at right angles to each of two straight lines in the pointof 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 AF is perpendicular 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 perpendifcular 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 AF it will be perpendicular to the same plane to which AF is perpendicular (7. 2. Sup.). PROP. XL 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 he 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 of 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 be 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 plane 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 a right angle ; wherefore the two angles ABK, BAK of the triangle ABK are equal to two right angles, which is impossible. 188 SUPPLEMENT TO THE ELEMENTS (17. L): Therefore the planes CD, EF, though produced, do not meet one another ; that is, they are parallel (def. 7. 2. Sup.). PROP. Xm. THEOR. If two straight lines meeting one another , he parallel to two straight lines which also meet one another, hut 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 Gdraw.GH parallel to ED (Prop. 31. 1.), and GK parallel to EF : And because BG is perpendicular to the plane through DE, EF, it must 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 a right 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 Bx\, 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. r OF GEOMETRY. BOOK II. PROP. XIV. THEOR. 189 If two parallel planes be cut by another plane, their common sections with it 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 [^: A>\ meet; therefore EF and GH are parallel (def. 30. 1.). r^vD e>^ 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 B \< a- V A X X c \ H 3) \ ^ I ^s. I K m: 19a 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 (4. def. 2. Sup.)- For the same reasoi\,the angle MKN is the mclma- tion of the plane AB to the plane EH. But because KN and LO are pa- rallel, being the common sections of the pa;:allel planes AB and CD witn 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. XVL 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 pomts 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 (IL 5.), as AE to EB, so is CF to FD. PROP. XVH. 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 first 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 the 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 OF GEOMETRY. BOOK II. 191 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.) ; and 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 OK ; 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. E 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 plane. B Let the two planes AB, BO be each of them perpendicular to a third plane, and BD be the common section of the first two ; BD is perpendicular to the plane ADC. From D 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 AB (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.). JlE 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. 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, the 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 he contained hy 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 BAG, 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. L) to the base BE. And because BD, DC are greater (20. L) 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. L) 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, DAG are together greater than BAE, EAC, that is, than the angle BAG. But BAG is not less than either of the angles DAB, DAG ; therefore BAG, with either of them, 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 cTf plane angles BAG, 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 BGDEF. And because the solid angle at B is contained by three plane angles GBA, ABF, FBG, of which any two are greater (20. 2. Sup.) than the third, the angles GBA, ABF are greater than the an- gle FBG : For the same reason, the two plane angles at eajh of the points G, 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 BGDEF : 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 BGDEF ; 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(l cr. 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 angles. Otherwise. Let the sum of all the angles at the bases of the triangles =S; the sum of all the angles of the rectilineal figure BGDEF==.2' ; the sum of the plane angles at A=X, and let R= a right angle. - Then, because S-l-X= twice (32. 1.) as many right angles as there are triangles, or as there are sides of the rectilineal figure BGDEF, and as .^-f-4R is also equal to twice as many right angles as there are sides of the same figure ; therefore S+X=^-j-4R. But because of the three plane angles which contain a solid angle, any two are greater than the third, 25 194 SUPPLEMENT TO THE ELEMENTS, &c. Sy^] 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 is 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 3.re 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 S is necessarily great- J> er than -2". This proposition, therefore, is subject to a limitation, which is farther explained in the notes on this Book. ELEMENTS OF GEOMETRY SUPPLEMENT. BOOK III. OF THE COMPARISON OF SOLIDS. DEFINITIONS. 1. A Solid is that which has length, breadth, and thickness. 2. Similar solid figures are such as are contained by the same number of similar 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 is a solid figure contained by six equal squares. ^ 7. A sphere is a solid figure described by the revolution of a semicircle about a diameter, which remains unmoved. 8. The axis of a sphere is the fixed straight line about which the semi- circle revolves. 9. The centre of a sphere is the same with that of the semicircle. 10. The diameter of a sphere is any straight line which passes through the centre, and is terminated both ways by the superficies of the sphere. 196 SUPPLEMENT TO THE ELEMENTS n. 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. 13. The base of a cone is the circle described by that side, containing the right angle, which revolves. 14 A cylinder is a solid figure described by the revolution of a right 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 It ^ V. a \j E \ IE \ \ P k M )Vr — ' \ \ \ 1_ r ] i J L 1 5 AC, which are equal and similar, may coincide (8. Ax. 1.), the point N coinciding with the poinffD, 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 AC, 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 other 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. H. THEOR. If a solid he contained hy 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,l)ecause 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 triangle 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. HI. THEOR. If a solid parallelopiped he cut hy a plane parallel to two of its opposite planis, it will he divided into two solids, which will he to one another as the hases. Let the solid parallelopiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides tfie 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, and take any number of straight lines HM, MN, each equal to EH, and any number AK, KL each equal to EA, and complete the parallelograms LO, KY, HQ, MS, and the solids LP KR, Idft SUPPLEMENT TO THE ELEMENTS HIT, MT ; then, because the straight lines LK, KA, AE are all equal, and also the 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 G K \p \l R. ^1\D \T7 X Z K A E H M NF \ \ \ \ \ \ \ O Y r O Q s (2. 3. Sup.) the parallelograms LZ, KP, AR, because they are opposite planes. FQ.r 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 LV 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 as 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. If a solid parallelopiped be cut hy a plane passing through the diagonals of two of the opposite planes, it will he cut into two equal prisms. Let AB be a solid parallelopiped, and DE, CF the diagonals of tlie 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 parallels 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^'s 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 following 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 of the opposite sides DH, EK ; wherefore DH is equal to EK : add, or take away the common part HE ; then DE is equal lo HK : Wherefore also the 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.), because 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 s,a%e solid there be taken the prism AGD, the remaining solid, viz. the parallelepiped AH is equal to the remaining parallelopiped AK. PROP. VL THEOR. Solid parallelopipeds upon the same base, and of tne same altitude, the «n- 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. 201 also is the figure BLPQ ; and because the plane ACDF is tljat in which are the parallels AC, FDOR, and in which also is the figure CAOR ; therefore the figures 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 parallelepiped. Now the solid parallelepiped CM is equal (5. 2. Sup.) to the solid parallelepiped CP, because they are upon the same base, and their insisting straight lines AF, AO, CD, CR ; 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 upon the same base ACBL, and their insisting straight lines AO, AG, LP, LN ; 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 are upon equal bases, and of the same altitude, are equal to one another. Let the solid parallelepipeds, 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 sides 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 straightline(14. 1.). Produce OD, HB, and let them meet in Q and complete the solid parallelepiped 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 parallelepiped 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 solid R o \ sr \m \ ►F* ^"^ v^ G- - n B K \ ^\ \ o L\ ^ \ 2 ^ s 3 I I 1 X 202 SUPPLEMENT TO THE ELEMENTS AE to the sojid 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, CD 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, DF, 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 soUd 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 another ; wherefore, the solids AE and CF are also equal. OF GEOMETRY. BOOK III. 263 PROP. VIII. THEOR. Solid parallelopipeds which have the same altitude, are to one another as their bases. « Let AB, CD b^ solid parallelopipeds of the same altitude ; they are to one another as their bases ; that is, as the base AE to the base CF, sols 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 \ / \ ^ T\ \ A M 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 parallelepiped 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 parallelepiped, which have the same alti- tude, g,re to one another as their bases ; that is, the prism BNM is to the parallelepiped 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 J^ME to the triangle CGF, and therefore the prism BNM is to twice the pnsm DPG as the triangle AME to twice the triangle CGF (4. 5.) ; that is, the prism BNM is to the parallelepiped CD as the triangle AME to the paral- lelogram LG. 304 SUPPLEMENT TO THE ELEMENTS PROP. IX. THEOR. Solid parallelopipeds are to one another in the ratio that is compounded of the ratios of the areas of their bases, and of their altitudes. Let AF and GO be two solid parallelopipeds, of whiah the bases are the ^rallelograms 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, aiid 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 -e \ Tl \ \ N \ vTMC \ B & IS K 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 of 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 thaaltitude 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 i 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- tude GM. Case 2. When the insisting lines are not perpendicular to the bases. 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 parallelepipeds Y and Z on these bases. Then, if the upright parallelepipeds AF and GO be constituted on the bases AG and GK, with the altitudes AE and GM, they will be equal to the parallelepipeds Y and Z (7. 3. Sup.). Now, the solids AF arid 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 parallelepipeds 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 AE 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 AY 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 soHd 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 parallelepipeds, and if to AB, to the perpendicular from A upon DC, and to the altitude of the parallelepiped 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 parallelepiped GO, the numbers L, Z, m, n, be proportional ; the solid AF is to the solid GO as LxMxN to ZXmXn. • For it may be proved, as in the 7th of the 1st of the Sup. that L X Mx N is to IXmXnm the ratio compounded of the ratio of hxMtolxm, and of the ratio of N to n. Now the ratio of L X M to Ixmis that of the area of the parallelogram AC to that of the parallelogram GK ; and the ratio of N to 71 is the ratio of the altitudes of the parallelepipeds, by hypothesis, therefore, the ratio ofLxMxNto IXmXn is compounded of the ratio of the areas of the bases, and of the ratio of the altitudes of the parallelepipeds 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 IxmXn. 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 parallelepiped of the same altitude with it, and of an equal base (2. Cor. 8. 3. Sup.). f PROP. X. THEOR. Solid parallelopipeds, which have their bases and altitudes reciprocally propoT' tionalj 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 are 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 AE (11. 5.) ; wherefore AF is equal to S (9. 5.). But the solid AG is I ■ ( J J El ( ^ \ e; X V \ p \ V D c M \ \ \ \ \ ^YL A B \ \ S K L 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 x\C 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, audit will be shewn, ^s 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. XL THEOR. Similar solid parallelopipeds are to one another in the triplicate ratio of theif 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 of AC to KM, and of AE to KO. Now, the ratio of AC to KM, because they are equiangular parallelograms, is compounded (23. 6.) of the ratios of AB to KL, and of AD to KN. Wherefore, the ratio of AG to KQ is OF GEOMETRY. BOOK III. 207 11 I— t ■» \ E \ F [x N y n "W m: I> k 1 N \ A L B ^ jj compounded of the three ratios of AB to KL, AD to KN, and AE to KQ ; 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 of the ratio of AB to KL. Cor. 1. If as AB to KL, so KL to m, and as KL torn, so ism to n, 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. THEQR. If two triangylar pyramids^ wfUch have equal bases and altitudes j be cut by planes that are parallel to the bases, and at equal distances from them, the sections are equal to one another. Let ABCD and 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 planes 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- son, 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 are 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 208 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 hasieen shewn that the triangle KLM is similar to the base BCD ; therefore, any section of j^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. XHL THEOR. A series of prisms of the same altitude may be circumscribed about any pyramid, 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 solid Z is not represented in the figure of this, or the following Proposition. I OF GEOMETRY. BOOK III. 209 Let Z be equal to a prism standing on the same base with the pyramid, viz. the triangle BCD, and having for its altitude the perpendicular drawn 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 piany equal parts as there are units in OT,*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 qlJ 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 ABCD 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 equal altitudes, viz. the perpendiculars drawn JfVom 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 (13. 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^X, V*Ybe made parallel to the base FGH. The section NQL is equal to the section WZT (12. 3. Sup.) ; as also ORI to X-aU, and PSM to Y(2>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, EFGH* therefore, are not unequal, that is, they are equal to one another. OF GEOMETRY. BOOK III. 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 let DEF be the triangle opposite the base : The prism ABCDEF may be divided into three equal pyramids having triangular bases. Join AE, EC, CD ; 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 ADEC, 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 ; for if the base of the prism be any other figure than a triangle, it maybe 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, DFG 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 ; €ind let AGHD be the rectangle (14. def. 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, the 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 t'hat 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. XVH. THEOR. A cylinder and a parallelopiped having equal bases and altitudes j 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 ABCD is equal to the parallelopiped EF. I If not, let them be unequal ; and first, let the cylinder be less than the parallelopiped EF ; and from the parallelopiped EF let there be cut oflf a I OF GEOMETRY. BOOK. III. 213 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. If 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 may be inscribed in BCD, that shall differ from it less than LIM does (4. 1. 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 Because the polygon BECFD is greater than the circle LIM, the prism BCFKHG is greater than the cylinder LMNO, for they have the same altitude, but the prism has the greater base. But the pyramid ABECFD is the third part of the prism (15. 3. Supt) 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 is 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 he 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 solid 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 T>K 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 Jess 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 thus 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 than OE GEOMETRY. BOOK III. D •215 I o ± the 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 cylinders 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 solid 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 CGL 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. Cor. 6. 1 Sup.) or GO ; now, CG is equal to GR, 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 GH, are equal to the fli6 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 shcAvn 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 hemispnere. PROP. XXL THEOR. Evert/ sphere is two-thirds of the circumscribing cylinder. 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 described by BD, let it be greater by the solid W. Then, as the cone de- scribed 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 the 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 OP THE SUPPLEMENT 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 of 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 AG 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 four times the angle ABD as the arc AC to four times the arc AD (4. 5.). But ABD is a right angle, and there- fore four times the arc AD is equal to 28 218 PLANE TRIGONOMETRY. the whole circumference AGF ; therefore the angle ABC is to four right angles as the arc AG to the whole circumference AGF. 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, B^t the centre of the circles AGE, GHK, stand on the arcs AC, GH, AC is to the whole circumference of the circle ACE, as the angle ABC to four right angles ; and 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, the arc 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, estch 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 (rf 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 ^l^ PLANE TRIGONOMETRY. m 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. Cor. 1 . The sine of a quadrant, 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 of tA 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 sine. AE the CMD 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 : : rad. 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 220 PLANE TRIGONOMETRY. therefore, if tables be constructed, exhibiting in numbers the sines, tan- gents, secants, and versed sines of certain angles to a given radius, they 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=R2. 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 B A 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=R2. 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 tangent of the arc DE, or of the angle C. PLANE TRIGONOMETRY. 221 ■ "The two triangles DFC, BAG, are equiangular, because the angles DFC, BAG are right angles, and the angle at G is common. Therefore, CB : BA : : GD : DF ; but GD is the radius, and DF the sine of the angle G, (Def. 4.) ; therefore GB : BA : : R : sin. G. Also, because EG touches the cir- cle in E, GEO is a right angle, and therefore equal to the angle BAG ; and since the angle at G is common to the triangles CBA, GGE, these triangles are equiangular, wherefore GA : AB : : GE : EG ; but GE is the radius, and EG the tangent of the •^ngle G ; therefore, GA : AB : : R : tan. G. GoR. 1. As the radius to the secant of the angle G, so is the side adja- cent to that angle to the hypotenuse. For GG is the secant of the angle G (def. 7.), and the triangles GGE, GBA being equiangular, GA : GB : : GE : GG, that is, GA : GB : : R : sec. G. GoR. 2. If the analogies in this proposition, and in the above corollary- be arithmetically expressed, making the radius = 1, they give sin. G = - ; tj^n. G == T^j sec. G = -77^. Also, since sin. G=co8. B, because B BG AG' AG AB is the complement of G, cos. B =^57^, and for the same reason, cos. G = AC BG* GoR. 3. In every triangle, if a perpendicular be drawn from any of the 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 ABG, AD be drawn perpendicular to the base BG, each of the triangles GAD, ABD being right angled, AD : DG : : R : tan. GAD, and AD : DB : : R : tan. DAB ; therefore, ex »quo, DG : DB : : tan. GAD : tan. BAD. SGHOLIUM. The proposition, just demonstrated, is most easily remembered, by stating it 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. 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 angled at D, AB : AD : : R : sin. B ; "and for the same reason, AG : AD : : R : sin. G, and inversely, AD : AG : : sin. C : R ; therefore, ex aequo inversely, AB : AG : : sin. C : sin. B. In the same manner it may be demonstrated, that AB : 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 ofhcuf their difference. Let AB, AG be two arcs of a circle ABGD ; let E be the centre, and AEG the diameter which passes through A ; sin. AG + sin. AB : sin. AG -sin. AB : : tan. J (AG+AB) : tan. \ (AC-AB). Draw BF parallel to AG, meeting the circle again in F. Draw BH and GL perpendicular to AE, and they will be the sines of the arcs AB and AG ; produce GL till it meet the circle again in D ; join DF, FG, DE, EB, EG, DB. Now, since EL from the centre is perpendicular to CD, it bisects the line CD in L and the arc GAD in A : DL is therefore equal to LG, 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 AG and AB, because 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 : KG : : tan. DFK : tan. CFK ; but tan. DFK=tan. J 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. GFK=tan. J arc. BC ; and consequently, DK : KG : : tan. \ arc. BD : tan. J arc. BC. But DK is the sum of the sines of the arcs AB and AG ; and KG is the difference of the sines ; also BD is the sum of the arcs AB and AG, and BC the dijffe- rence of thos« arcs I 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. ^ arc. DB : tan. J 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. J (AB+AC), therefore cos. AB+cos. AC : sin. AB+sin. AC : : R : tan.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. i(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 sura 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 of half their difference. Let ABC be any plane triangle ; CA+AB : CA-AB : : tan. J(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, (11. 5.) CA+AB : CA-AB : : tan. J (B + C) : tan. J (B-C). 224 PLANE TRIGONOMETRY. Otherwise, without the 3d. *»«*>♦*« Let 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 of half their difference. About the centre A with 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, orBFG, 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 the 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 BCis parallel to FG, CE : CF : : BE : BG, (2. 6.) that is, the sum of the two sides of the triangle ABC is to their difference as the tangent of half the sum of the angles opposite to those sides to the tangent of half their difference. PLANE TRIGONOMETRY. 22S^' PROP. V. THEOR. If a perpendicular he drawn from any angle of a triangle to the opposite side, or base ; the sum of the segments of the base is to the sum of the other 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 is to the dif- ference between the sum of the squares of those sides, and the square of the base J 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^+BC^ and AC^ 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. 15 Ji C BD : : R : cos. B. Now 2BC.BD is the difference between ABHBC" and AC^, therefore twice the rectangle AB.BC is to the difference between jf^ AB24-BC2, and AC2 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.BAXC0S. B = BC2H-BA2 — AC2, and adding AC2 to both; AC^ + 2 COS. B X BC.BA = BC24- BA2 ; and taking 2 cos. Bx BC.BA from both, AC2=BC2— 2 cos. Bx BC.BA +BA2. Wherefore AC= V(BC2— 2 cos. B X BC.BA4-BA2). If B is an obtuse angle, it is shewn in the same way that AGs V(BC2+2 COS. BXBC.BA+BA2). 29 PLANE TRIGONOMETRY. PROP. YII. THEOR. "Four times the rectangle contained hy any two sides of a triangle^ is to the rectangle contained hy two straight lines, of which one is the base or third side of the triangle increased by the difference of the two sides, and the 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)) : : R2 : (sin. I BACp. Produce the side AG to D, so that AD=AB ; join BD, and draw AE, A\'^. '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^ : : EF.ED : ED2 (1. 6.), and therefore 4AC.AD : AD2 : f 4EF.ED : ED^, or alternately, 4AC.AD : 4EF.ED : : AD^ : ED^. But since 4EF=2BK, 4EF.ED=2BK.ED=2ED.BK=DB.BK= HB.BG ; therefore 4AC.AD : DB.BK : : AD2 : ED2. Now AD : ED : : R : sin. EAC=sin. -J BAC (1. Trig.) and AD2 : ED^ : : R2 : (sin. J BAC)2 : therefore, (11. 5.) 4AC.AD : HB.BG : : R2 : (sin. ^ BAC)2, or since AB =AD, 4AC.AB : HB.BG : : R2 : (sin. J BAC)2. Now 4AC.AB is four times the rectangle coatained by the sides of the triangle ; HB.BG is that contained by BC4-(AB-AC) and BC-(AB— AC). CoR. Hence 2 ^AC.AD : ^HB.BG : : R : sin J BAG. PLANE TRIGONOMETRY. m PROP. VIII. THEOR. Four times the rectangle contained by any two sides of a triangle, ts 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 of 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 of the other two sides, 4AB.AC : (AB-f AC+BC) (AB+AC-BC) : : R2; (cos. J RAC)2. From the centre C, with the radius CB, describe the circle BLM, meet- ing AC, produced, in L and M. Produce AL to 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-AC4-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=2B0 =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 bases (1. 6.), ACAD : AD2 : : AR.AE : AE2,and alternately ACAD : AR.AE : : AD2 : AE2, and 4 AC AD : 4 AR.AE : : AD2 : AE2. But 4AR.AE=: 2ARx2AE=:NP.NB=MN.NL ; therefore 4AC.AD : MN.NL : : AD^ : AE2. But AD : AE : : R : cos. DAE (1) =cos. l (BAG): Wherefore 4ACAD : MN.NL : : R2 : (cos. J BAC)2 228 PLANE TRIGONOMETRY. Now 4AC.AD is four times tlie 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. 1. Hence 2 ^ACAB : V MN.NL : : R : cos. ^ BAG. Cor. 2. Since by Prop. 7. 4AC.AB : (BC+(AB-AC)) (BC— (AB — BC)) : : R2 : (sin. J BAC)^; and as has been now proved 4AC.AB : (AB+AC+BC) (AB+AC-BC) : : R2 : (cos. -J- BAC)2; therefore, ex aequo, (AB + AC + BC) (AB+AC-BC) : (BC + (AB-AC)) (BC- (AB-AC)) : : (cos. -J BAC)2 : (sin. J BAC)^. But the cosine of any arc is to the sine, as the radius to the tangent of the same arc ; therefore, ( AB 4-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. J BAG. LEMMA IL If there he two unequal magnitudes, half their difference added to half their 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 X E D B C 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 difierence, and it will give the less. i SCHOLIUM. This property is evident from the algebraical sum and difference of the two quantities a and 6, of which a is the greater ; let their sum be denoted by Sf and their difference by d : then, a-\-h=s > a-h=dS /.by addition, 2o=^+ J; By subtraction, 25=^— (? ; and.-.i=i— i. ,^j^^^. PLANE TRIGONOMETRY. Ml SECTION II. OP THE RULES OF TRIGONOMETRICAL CALCULATION. The General Problem which Trigonometry proposes to resolve is: In 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 beiHg 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 angle. PROBLEM I. 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 a right angled triangle 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 the 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. PLANE TRIGONOMETRY. ^ GIVEN. SOOGHT. SOLUTION. CB and B, the hypotenuse and angle. AC. AB. R : sin B : : CB : AC. R : cos B : : CB : AB. I AC and C, a side and one of the acute angles. BC. AB. Cos C : R : : AC : BC. R : tan C : : AC : AB. 3 4 CB and BA, the hypotenuse and 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 madeR : sec. C : : AC : CB, CB will be found. In the third case, when the hypotenuse BC and the side AB are given to find AC, thib may be done either as directed in the Table, or by the 47th of the first ; for since AC^ = BC2 - BA2, AC = ^BC^ — BA2. This value of AC will be easy to calculate by logarithms, if the quantity BC^ — B A^ be separated into two multipliers, which may be done ; because (C or. 5. 2. ), BC2-BA2 =(BC + BA) . (BC-BA). Therefore AC = V(BC-hBA) (BC-BA). When AC and AB are given, BC may be found from the 47th, as in the preceding instance, for BC=r ^BAH^AC^. But BA2+AC2 cannot be separated into two multipliers ; and therefore, when B A 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 anaJogy in the Table, AC : AB : : R : tan. C ; but if C itself is not required, it is sufl[icient, having found tan. C by this proportion, to take from the Trigonometric PLANE TRIGONOMETRY. Tables the cosine that corresponds to tan. C, and then to compute CB from the proportion cos. C : R : : AC : CB. PROBLEM IL In an oblique angled triangle, of the three sides and three angles, any three being given, and one of these three being a side, it is required to find the other three. This problem has four cases, in each of which the solution depends on some of the foregoing propositions. CASE L Two angles A and B, and one side AB, of a triangle ABC, being given, to find the other sides. SOLUTION. Becailse 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. B O : CASE n. 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=180O— 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 AB in the one, being the supplement of that 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 832 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=ACC, 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 to 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. 4 « SOLUTION. First, make AB+AC : AB— AC : : tan. | (C+B) : tan. J (C— B). Then, since J (C + B) and J (C— B) are both given, p and C may be found. For B=} (C+B)+i (C-B), and C=J (C+B)-J (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 ; for BC=: VAB2-.2 cos. AXAB.AC+AC2, Prop. 6. PLANE TRIGONOMETRY. 23i This method of finding BC is extremely useful in many geometrical in- vestigations, but it is not very w^l adapted for computation by logarithms, because the quantity under the radical sign cannot be separated into sim- ple multipliers. Therefore, when AB and AC are expressed by large numbers, 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-fAC : : 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 difl^erence 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. H C B SOLUTION II. C D Let D be the diff'eren ce of the sides AB, AC. Then (Cor. 7.) 2 ^AB-AC V(BC+D) (BC-D) : : R : sin. J BAC. SOLUTION III. Let She the sum of t he sides BA and AC. Then (1. Cor. 8.) 2 -/AB.AC V(S+BC) (S-BC) : : R : cos. J BAC. SOLUTION IV. S and D retaining the significations above, (2.Cor.8.) ^(S+BC)(S— BC) : V(BC+D) (BC-D) : ; R : tan. J 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 is preferable to the second, when the angle sought is greater than a right 30 234 PLANE TRIGONOMETRY. angle ; and in extreme cases, that is, when the angle sought is very acute or very obtuse, this distinction is very flfciterial 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 of 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 I", 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 CG is 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 l-fcos. BD. Or, for the greater generality, supposing A = any arc, cos. -J A is a mean proportional PLANE TRIGONOMETRY. 235 between J and l+cos. A, and therefore (cos. J Af^l (i+cos. A) or cos. 5A = Vi(l+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- V3 fore COS. J BD, or cos. 30°= \/i(l+i)= Vi=^^' ^^ *^® s^"^® "^^^" ner, cos. 15°= Vi(l+cos.30o), and cos.7o, 30'= Vi(l+cos.l5o),&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^. 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, ihe remainder is the square of the sine, and its square root is the sine itself. Thus the sine of 52". 44"'. 03'"'. 45^ 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 tbe 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 iwo arcs are both very small, the ifirst 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^. being found, the sine of 1' tm PLANE TRIGONOMETRY. becomes known , for, as 52". 44'". 03"". 45^. to 1, so is the sine of the former arc to the sine of the latter. Thus the sine of 1' is found = 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 aiid 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= i (BF+DH) ; therefore R : cos. BC : : sin. AC : ^ (sin. AB+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 of any arc as the sine of the arc is to half the sine of twice the arc ; or if any arc= A, ^ sin. 2A=sin. A x cos. A, or sin. 2A=2 sin. A x cos A. Therefore also, sin. 2'=2' sin. 1' X cos. 1' : so that from the sine and 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 : J (sin. I'+sin. 3'), or sin. I'+sin. 3'=2 cos. I'-f-sin. 2', and taking sin. 1' from both, sin. 3'=2 cos. I'xsin. 2'— sin. L In like manner, sin. 4'=2' cos. I'xsin. 3'— sin. 2, sin. 5'=2' COS. I'xsin. 4'— sin. 3, sin. G'=2' COS. I'Xsin. 5'— sin. 4, &c. Thus a table containing the sines for every minute of the quadrant may be computed ; and as the multiplier, cos. 1' remains always the sa?ne, the calculation is easy. For computing the sines of arcs that differ by more than 1', the method is the same. Let A, A+B, A4-2B be three such arcs, then, by this the- orem, R : cos.B : : sin. (A-}-B) : J (sin. A-fsin. (A-f 2B)) ; and therefore making the radius 1, PLANE TRIGONOMETRY. sin. A+sin. (A+2B)=5 cos. Bxsin. (A+B), or sin. (A-f2B)=2 cos. Bxsin. (A+B)— sin. A. By means of these theorems, a table of the sines, and consequently also of the cosines, of arcs of any number of degrees and minutes, from to 90, may be constructed. Then, because tan. 'k= — '—j, the table of tangents ■^ cos. A 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, if the difierence between any arc and 45° be called D, that tan. (45°— D) : 1 : : 1 : tan. (45°-f D), so that tan. (450+D)= ^^^ (450-,D) - Lastly, the secants are calculated from (Cor. 2. def. 9.) where it is 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. x\. 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. \mo, If in the last Theorem, the arc AC=A, the arc BC=B, and the radius EC=I, then AD=A-1-B, and AB=A— B ; and by what has just been demonstrated, 1 : COS. JB : : sin. A : J sin. (A-f B)+J sin. (A— B), and therefore, sin. Ax COS. B=isin. (A4-B)+i (A— B). 2(^0, 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 EGG, EKI, EC : EI : : GE : EK ; now, GE is the cosine of AC, therefore, R : cos. BC : : cos. AC : J cos. AD4-^ cos. AB, or 1 : cos. B : : cos. A : \ cos. (A4-B)4- J cos. (A— B) ; and therefore, cos. Ax cos. B=J COS. (A+B)+i cos. (A— B) ; 2tio, Again, the triangles IDM, CEG are equiangular, for the angles KIM, EID 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 EGG ; and the angles DMI, CGE are also equal, being both right angles, and therefore the triangles IDM, CGE have the sides about their equal angles proportionals, and consequently, EC : CG : : DI ; IM ; now, IM is half the difference of the cosines FE and EH, therefore, R : sin. AC : : sin. BC : \ cos. AB— J cos. AD, or 1 : sin. A : : sin. B : J cos. (A— B)— ^ cos, (A+B) ; 238 PLANE TRIGONOMETRY. and also, sin. Ax sin. B=i cos. (A— B)— ^ cos. (A+B). 4^0, Lastly, in the same triangles EGG, 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— J sin. AB, or 1 : COS. A : : sin. B : J sin. (A+B)— ^ sin. (A+B) ; and therefore, COS. Ax sin. B=J sin. (A+B)— J sin. (A— B). 6. If therefore A and. B be any two arcs whatsoever, the radius being supposed 1 ; I. sin. Axcos B=Jsin. (A+B)+isin. (A— B). IL cos.Axcos. B=Jcos.(A— B)+icos.(A+B) III. sin. Axsin. B=|cos.(A- B)-|cos. (A+B). ly. COS. Axsin. B=|sin. (A+B)— Jsin. (A B). ^ From these four Theorems are also deduced other four. For adding the first and fourth together, sin. Ax cos. 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=Jsin. (A+B)+^ sin.(A— B),if A+B=S, and A— B=D, 1. /T ON A S+D ^^ S-D , . . S+D then (Lem. 2.) A= — - — , and B= — - — ; wherefore sm. — - — X cos. g r) — - — =^sin. S+JD. But as S and D maybe 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 , . , , , . T) sm. — - — Xcos. — - — •=z\ sin. A+ J sm. B, or „ . A+B A-B . , . . ^ 2 sm. — - — xcos. — - — :i=sm. A+sm. B. z z In the same manner, from Theor. 2 is derived, 2 COS. xcos. — - — =cos. B+cos. A. From the 3d, 2 Z 2 sin. — - — xsin. — - — =cos. B— cos. A ; and from the 4th, A+B . A-B . , . „ 2 cos. — - — xsm. — - — =sm. A— sm. B. z z In all these Theorems, the arc B is supposed less than A. 8. Theorems of the same kind with respect to the tangents of arcs may be deduced from the preceding. Because the tangent of any arc is equal to the sine of the arc divided by its cosine, . PLANE TRIGONOMETRY. 239 tan. (A4-B)= ^^ — ^- — ^r^. But it has just been shewn, that ^ ^ '' cos. (A+B) •' sin. (A4-B)=sin. Ax cos. B+cos. AXsin. B, and that cos. (A-j-B)=cos. Axcos. B— sin. Axsin. B ; therefore, tan. (A+B) = sin. Axcos. B+cos. AXsin. B ,,..,., ,,.i , i •, 7 rr : 1 -. — r7, auQ dividinffboth the numerator and d^no- cos. Axcos. B— sm. AXsin. B ° minator of this fraction by cos. Ax cos. B, tan. (A4-B)=- '-^ '—^. i. tan. A X tan. x) T i-u / A Ti\ tan. A tan. B In like manner, tan. (A— B)= — ; ; =. ^ ' 1-ftan. Axtan. B 9. If the Theorem demonstrated in Prop. 3, be expressed in the same manner with those above, it gives sin. A+si n. B _ tan. J (A+B) sin. A — sin. B ~ tan. ^ (A— B)' Also by Cor. 1, to the 3d, COS. A+cos. B _ cot. i (A+B) cos. A— COS. B """ tan. ^ (A— B)' And by Cor. 2, to the same proposition, sin. A+sin. B tan. i (A+B) . ^ . , —j — = £— or since K is here supposed = 1, cos. A+cos. BR' x-i f sin. A+sin. B ^ i / » , r»\ COS. A+cos. 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 are 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 X cos. B=sin. (xA.+B)+sin. (iV— 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. Itthenbecomes 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 li7iear 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 Arithmetic of Sines. ELEMENTS OF SPHERICAL TRIGONOMETRY. PROP. I. If a sphere he 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 super^cies 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 its centre, is called a great circle of the sphere. Cor. All great circles 6f 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 ^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 ABC 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, therefore 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 he the same with the intersection of other two great 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, ADC are right angbs : therefore (4. def. 2. Sup.), the angle CDB is the inclination of the planes of 31 242 SPHERICAL TRIGONOMETRY. the circles AB, AC, and is (def. 3.) equal to the spherical angle BAC ; 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 to one 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 circle 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 poles of the ♦ When in any reference no mention is made of a Book, or of the Plane Trigonometry, the Spherical Trigonometry is meant. SPHERICAL TRIGONOMETRY. ^m 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 be 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 their poles in the circumference of a circle, the plane of which is perpendicular to thatrii^Tieter. PROP. V. J" 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 equal, viz. AE to AF(26. 1.), and DE to DF. Again, because the angles AGE, AGF are right angles, the sauares 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 ihe 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 (4. def. 2. Sup.), because FA and FG, which are drawn in these planes, are at right angles to DF, the common section of the planes. The angle AFG (3. def.) is therefore equal to the 244 SPHERICAL TRIGONOMETRY. spherical angle ACB ; and, for the same reason, the angle AEG is equal to the spherical angle ABC. But the angles AFG, AEG are equal. Therefore the spherical angles ACB, ABC are also equal. ii J i.. PROP. VI. If the angles at the base of a spherical triangle be equal, the triangU ''^ "sosceles. Let ABC be a spherical triangle having the angles ^3 equal to one another ; the sides AC and AB are also equal. Let D be the centre of the sphere ; join DB, DC, Da froni A on the straight lines DB, DC, draw the perpendiculars AE, ; and from the points E and F, draw in the plane DBC a 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 ^y — "D make with the plane DBC. But because ^ Jij JJ 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. I^et ABC be a spherical triangle, any two sides AB, BC are greater than the third side AC. SPHERICAL TRIGONOIVIETRY. 245 Let D be the centre of the sphere ; join DA, DB, 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 arcs AB, AC, BC, which measure these angles, aFrv\B and BC must also be greater than ^~ third AC. ^^ ' ' PROP. VIIL The three si.-..^ 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 (2L2. Sup.) thBrefore the sides AB, BO, AC, which are the measures of these anglps, are together less than four quadrants describ- ed with the radius AD, that is, than the circumference of a great circle. PROP.nX. 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 tp 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 B C. The converse is demonstrated as Prop. 19. L 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 of the two interior angles at the base greater than two right angles, equal to two right angles, or less than two right angles. Let ABC be a spherical triangle, of which the sides are AB and RC ; 246 SPHERICAL TRIGONOMETRY. produce any of the two sides as AB, and the base AC, till they meet again mp ; then, the arc ABD is a semicircle, and the spherical angles at A and D are equal, because each of them is the incHnation of the cirde ABD to the circle ACD. ■- 1. If AB, BC be equal to a semicircle, that is, to AD, 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. \y 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 is, 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, BOA are equal to two right angles, if the 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 he made the poles of three great circles, these three circles hy their intersections will form a triangle, which is said to he supplemental to the former; and the two triangles are suchj 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 the pole of AB. And since F, E are the poles of AL, AM, the arcs FL and EM (2.) are SPHERICAL TRIGONOMETRY. quadrant and FL, EM together, diat |s, F£ aad ML to a semi rcle. But since A is the pol© of ML, ML is angle BA (3.), consequently FE is the angle BA. In the same manner, ED, D" measuresf the angles AEG, BC A. , , , Since kewise CN, BH are quadrants, CN and BH t yUi, A*i », NH and C together, are equal to a semicirete ; tad nee D k tk» p«l» «f NH, NHs the measure of the angle FDE, tlmrefero the angle FDl is the supplement of the side BC. In the «i shewn tht the measures of the angles DEE, EFD an dM of the sids AB, x\C in the triangle ABC. PROP. XII. The three agles of a spherical triem^ tregntiUr tkm tvt, mi Int tkm , The misure of the angles A, B, C, in the triaa^ ABC, the three ides of the supplemental triangle DEF, mre (11.) semicircle ; but the three sides of the tiiangle FDE, ire (8.) len semicircle ; therefore the measures of the angles A, B, C,«rs a semicirie ; and hence the angles A, B, C sre greitac dia angles. And beause the interior angles of any triangle, toge&er wi& iks rior, are eual to six right angles, the interior aJoiie an less Asa mx, angles. PROP. XIIL If to the cKumference of a great ctrde,fram apeimi m the tmfm* tftkt Sflmm, which i&ot the poU of th!u circle, arcs rfgnatanUs he inmm; tkegrmUM of theseircs is that which passes through tktfok tf tks Jkstrmtmiimmi cv>- cle, andhe supplement of it is the least ; ami of the other ont^thM wkkkm nearer i the greatest is greater thmm thmt whiA is wmo i— ifi Let AB be the circumference of a great drde, of which the pole is H, and let C e any other point ; through C and H let the semicirde ACB W _ „ — ^j ^vxiv^i l^wiuk , Lxxxi^ugu \j ouu XA ic* use OCUUCUdC A\^0 0% drawn meting the circle^ ADB in A and B ; -and let the arcs CD, CE, CF also be dccribed. From C draw CG perpendicoiar to AB, and the*, he- cause the ircle AHCB which passes through Kthe pole of the circle ADB, is at right ngles to ADB, CG is per- pendicula to the plane ADB. Join GD,GE,JF,CA,CD,CE,CF, CB. Becaus AB is the diameter of the circle AB, and G a point in it, which IS not the.entre, (for the centre is in the pomt ^here the perpendicular from H meets B), therefore AG, the part ot the diaieter in which the centre is 248 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 AC2=AG2+GC2, and DC2=DG2+ GC2; but AG2+GC2 7DG2+GC2; because AG7DG; therefore AC2 7DC2, and AC /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 are 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 supplement 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 rigjit angled spherical triangle, the sides containing the right angle are 0, 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. Let 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 thap a right angle. But when AB is greater than AE, the angle ACB is gi'eater than ACE, or than a right an- gle. In the same way may the converse be demonstrated. SPHERICAL TRIGONOMETRY^ 24« PROP. XV. If the two sides of a right angled spherical triangle about the right angle he of 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 ztngled spherical triangle ; according as the two sides AB, AC are of the same or of different affection, the hypotenuse BC will 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 is 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 gi;eater (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- tion, the hypotenuse is less than a quadrant : and conversely. PROP. XVI. 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 perpe7idicularfall without the triangle, the angles at the base are of different affection. Let ABC be a spherical triangle, and let the arc CD be drawn from C perpendicular to the base AB. 1. 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 fl60 SPHERICAL TRIGONOMETRY. C 2. Let CD fall 'wjithout the triangle ; then (14.) the angle B is of the same 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 &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 jwid B-would be of the same affection, contrary to the supposition. PROP. XVH. If to the base of a spherical triangle a perpendicular he drawn from the opposite angle, which 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 Vnd 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 than CB is, CE is greatoT 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- position less than DB, AC is also less than CB (13.), and therefore in this case, viz. when the perpendicular falls within the triangle, and when the SPHERICAL TRIGONOMETRY. 251 sum of the sides is less than a semicircle, the least segment is adjacent 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 Ctf, EC is greater than CF, and EC is therefore nearer to the perpendicular CHG than CF is, wherefore EG is the least segment of the base, and is adjacent to tlb 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 ar^ 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 right angle, is to the radius of the sphere, as the tangent of the remaining side is to the tangent of the angle opposite to that side. Let ABC be a triangle, having the righl^ 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 TRIGONOxMETRY. 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. PL Tr.) ; but AF is the sine of 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. PL 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 either 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. PL 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 to the tangent of the angle ACB. SPHERICAL TRIGONOMETRY. 253 Describe the circle DE, of which B is the pole, and let it meet 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.). And since AC is perpendicular to BD, the plane of the circle AC is perpendi- 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 of 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 «quo, 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.): butsin. 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. 1. From the demonstration it is manifest, that the tangents of any two arcs AB, BC are reciprocally proportional to their cotangents. 254 SPHERICAL TRIGONOMETRY Cor. 2. Because cos. ABC : R : : tan. AB : tan. BC, and R : cos. BC : : tan. 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 : : sin. 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. EOF (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 ; therefore (19.), the sine of the hypotenuse BC is to the radius, (or the sine 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 tvyo 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 : sin. A ; wherefore, by equality inversely, sin. BC : sin. AC : : sin. A : sin. B. In the same manner, it may be proved that sin. BC : sin. AB : : sin. A : sin. C, J A I / (C-B) : : tan. ^ BC : tan. i (AB— AC) ; which is the first part of the tan. zi nns. Svsin T) fan. S proposition. Again, since sin. S X COS. D COS. Sx sin. D ; and since tan. ^ cos. Sxsin. D tan. 2 sin. S X cos. D tan. X sin. D X cos. D I or m Lversely tan. 2 tan. J ——-. — pr ^ ; therefore by multipli- tan. B sm. D X cos. S • "^ ^ tan. X tan. 2 (cos. t)f cation,^ :^X: -z=) ^. tan. B tan. ^ (cos. S)^' SPHERICAL TRIGONOMETRY. 261 ^ , •,■■■, 1 tan. X tan. ^X tan. ^ , - . But It was already shewn that =5= — =— — , wherefore also •^ tan. B (tan. B)^ tan. X tan. 2' _ (tan. 2f tan. B^tan. ^~(tan. B)2* __ tan. X tan. 2 (cos. Dp , . , , Now, r-x 7=7 T^s as has just been shewn. ' tan. B tan. ^ (cos. S)^' •' ^, ^ (cos D)2 (tan. 2f -, i cos. D tan. 2 Therefore ) ^=7; BT5, and consequently ^^=- ^, or cos. (cos. S)2 (tan. B)^' ^ •' cos. S tan. B S : COS. D : : tan. B : tan. -2", that is, cos. (C+B) : cos. (C— B) : ; tan. J 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 difference 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 cotahgent 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, ttj h, c the sides opposite to them, I. sin. i (A+B) : sin. J (A— B) : : tan. |-c : tan. J (a — h). II. cos. I (A-j-B) : cos. \ (A— B) : : tan. | c : tan. | (a+b). III. sin. I {a-\-b) : sin. i (a— ^>) : : tan. ^ C : tan. J (A—B). IV. COS. I (a+b) : cos. | (a—b) : : tan. ^ C : tan. | (A-f B). 262 SPHERICAL TRIGONOMETRY. PROBLEM I. In a right angled spherical triangle^ of the three sides and three ansles, am/ 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 an^ed at A. GIVEN. SOUGHT. SOLUTIOX. BO 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 0. 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 TABLE for detenmning 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- ^en^ (15.) 3 AB and C are of the same affection, (14.) 4 K AC and C are of the same affection, BC/ 90^ ; otherwise BCZ90^, (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 WTien B and C are of the same affection^ BC/90o. other- wise, BC79(P, (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, viz. that when the thing found by the rules in the first table is either a tangent or a cosine ; and tchen^ of the tangents or cosines employed in the computation cf 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 both 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 aifection with the opposite angles. Two rules are therefore sufficient to remove the ambiguity in all the cases of the 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; leib be the side opposite to B, and c the side opposite to C. GIVEN. SOUGHT. SOLUTION. a and B. b. c. C. sin & = sin a X sin B. tanc = tan a X cos B cotC = cos a X tan B. 1 2 3 b and C. c. a. B. tan c = sin 6 X tan C. tan 5 tan a == r-/ cosC cos B = cos J X sin C. 4 5 6 b and B. a, C. tan b sm c = - — ^, tanB sin b sm a = -^^5. sinB . ^ cosB smC= r. cos b 7 8 9 a and b. c. B. C. * . cos a 10 11 12 cos b . ^ sin 5 sm B = -. . sm a ^ tan & cos C = . tan a h and c. a. B. C. cos a = cos b X cos c. . r» tan 5 tan B = -, . sm c * /-. tan c tan C = -. — r. sm b 13 14 14 B and C. b a. cos C ' cos c = -, — - . sin B cosB cos b = -:— FT- sinC cotC cos a = : — 5. 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 SphericalTrigo- nometry. GIVEN, SOUGHT. SOLUTION. 1 Two sides AB, AC, and the in- 2 eluded 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. GIVEN. SOUGHT. SOLUTIOX. 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 90°, according as the angles A and BCD are of the same, or different affec- tion. The third angla 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. GIVEN. 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 -f BC is greater or less than ISQo, A-f-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 4- 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=ADiBD, wherefore AB is ambiguous. SPHERICAL TRIGONOMETRY. 269 TABLE continued. GIVEN. SOUGHT. SOLUTION. The side Sin B : sin A : : sin AC : sin BC, BC (24) ; the affection of BC is un- opposite certain, except when it can be de- 8 to the termined by this rule, that accord- other ing as A-f B is greater or less than given an- 180°, AC-fBC 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. 3.), cos A : cos B : : sin ACD : 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 i AB sides, : tan J (AC-fBC) : : tan J (AC- 11 One of the BC) : tan J E ; then, if AB be 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. SOLUTION. 12 The three angles A, B, C. One of the sides BC. Suppose the supplements of the three given angles, A, B, C, to be a, bj c, and to be the sides of a spherical triangle. Find, by the last case, the angle of this triangle, opposite to the side a, and it will be the supplement of 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. TJiis 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 with 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.). xm SPHERICAL TRIGONOMETRY. 271 The last table may also be conveniently expressed in the following manner, denoting the side opposite to the angle A, by a, to B by b, and to C by c ; and also the segments of the base, or of opposite angle, by x andy. 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 a;=tan b X cos A ; then _ sin ic X tan A tan B= — : — ; r-. sm (c — X) Find X, as above, , cos 5Xcos (c — x) then cos a= ^ -. Find X, so that cot a;=cos Z>xtan A ; then tan h X cos x tan a= ; r-. cos (c — X) Find 0?, as above. then cos B= cos A X sin (c— a;) sm X sin B: sin A X sin A sin a Find a?j so that cot a:=cos I X tan A ; then _, cos X X tan b cos G=- tan a Find Xj so that tana:=tan5xcos A; and find y, so that cos aXcos X cos y= cos b c=x^y. 272 SPHERICAL TRIGONOMETRY. TABLE continued. The angles AandB and tlie side b. 10 sma: sin 6 X sin A sin B Find Xj so that tan a;=tan b X cos A ; and y, so that sin irXtan A siny= tanB c=x^y. Find a:, so that cot a;=cos & X tan A ; and also y, so that sin irXcos B E2ny= cos A c=x^y. 11 a, 6, c. Let a-\-b-\-c=s. sin JA: ^sin (J^— Z>)xsin (\s—c) •/sin ^Xsin c or cos JA __ -v/sin \s X sin [\s--a) -v/sin ixsin c 12 A, B, C. Let A+B+C=S. . , J cos 4 S X cos (A S— A) sm \a=i— ^ „ '-A 3= ' Vsin Bxsin C ,. Vcos(lS— B)lcos S— C) or cos \a^— ^=^==41^=^ • -^/sin B X sin C \ ■b^l APPENDIX TO SPHERICAL TRIGONOMETRY, COTTTAINING 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 from the iSth to the 23d inclusive. They are perhaps the happiest example of 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 their 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 angled triangle ABC, A, being the right angle, AC, AB, 90°— B, 90O-BC, 90O-.C, are the circular parts, by Def. 1.; and if 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 for 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 BC=cot Bxcot C, and R x cos 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 = -^ — — ::,. Again, suppose that BC and C are given to find AC ; it is obvious that C is in the middle between the adjacent parts AC and (90°— BC), there- SPHERICAL TRIGONOMETRY. 275 fore R X cos C=tan AC X cot BC, or tan AC=-^^^7^=:cos C + tan BC -, cot JdU because, as has been shewn above, =77r=tan BC. cot BC In the same way may all the other cases be resolved. One or two tnais 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, even 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, 5, c, the sides opposite to them. I. Given two sides 5, c, and the angle A between them. To find the angles B and C. tan I (B— C)=cot ^ AX ^!" \ [^T"") . (31.) cor. 1. tanJ(B-{-C)=cotJAx^^^44TT4- (31.) cor. 1. ^ ' ' 2 cos J (6-t-c) ^ ' To find the third side a, sin B : sin A : : sin £ : sin a. II. Given the two sides 5, c, and the angle B opposite to one of them. To find C, and the angle opposite to the other side. sin 5 : sin c : : sin B : sin C. To find the contained angle A. cot \ A=tan \ (B-C) X ^^^ \ ^^^\ (31.) cor. 1. * ^ ^ ' sm J (o— c) ^ ' To find the third side a, sin B : sin A : : sin b : sin a. III. GiTen two angles A and B, and the side c between them. To find the other two sides a, b. 376 APPENDIX TO tauJ(Ha)=.anicx:^ii{|=|-;. (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 thcfm. To find bf the side opposite to the other. sin A : sin B : : sin a : sin b. To find c, the side between the given angles. «»4 — i(«-*)x^^|{^. (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 si3es 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, h, c, be the three given sides, to find the angle A, contained be- tween b and c. If Rad = 1, and a + i -|- c = j, . _ . ^sin (4 s-^b) X sin I (s^c) BUi^Af=»^ ^^ ^ -; or, ^sin 6 X sin c cos I A -^ V^s^^-(?^Xsini(^~fl)) Vsin & X sin c In like manner, if the three angles, A, B, C are given to find c, the side between A and B. SPHERICAL TRIGONOMETRY. 277 Let A + B 4- C = S, sin J c=-^ ^ -1 = ; or, ^sin B X sin C cos } ,^ V<^o«(iS-Brx"cos(iS-C) ^ V'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 was explained. Plane Trig. Schol. BND OP SPHERICAL TRIGONOMETRY ^Ti fM NOTES ON THE FIRST BOOK OF THE ELEMENTS. DEFINITIONS. - I. In the definitions a few changes have been made, of which it is neces- sary to give some account. One of these changes respects the first 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, as every good definition ought certainly to be. That it is not comrertible 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 aline. 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 of 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, because it is in fact an inference deduced from comparing the defi- nitions 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, better 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 length, 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 boundary which is common to two solids which are contiguous, or which divides one solid into two contiguous parts, is called a superfi- cies ; Thus, if 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, this thick- ness must either be a part of the thickness of the solid AG, or the solid BM, or a part of the thickness of each of them. It cannot be a part of the thicfe^ 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 still thei 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 BCGF, the boundary of the solid BM, does nevertheless remain; therefore the super- ficies BCGF has no thickness, but only length and breadth. " The boundary ©f a superficies is called a line ; or a line is the common boundary of two superficies that are contiguous, or it is that which dividos 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 boun- 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 have any, this must be part either of the breadth of the superficies ABCD or 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 breadthofthe superficies ABCD; be- cause, if this be removed from the su- perficies KBCL, the line BC, which is the boundary of the superficies G M F B K 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 na thickness, as was shown ; therefore a line has neither breadth nor thick- ness, but only length. " The boundary of a line is called a point, or a point is a common boun- dary or extremity of two lines that are contiguous : Thus, if B be the ex- NOTES 28P tremity of the line AB, or tlie common extremity of the two lines AB, KB, this extremity is called a point,- and has no length : For if it have any, this length must either be part of the length of the 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 same as it was ; Nor is it part of the length of the line AB ; for if AB be removed from the line KB, the point B, which is the extremity of the line KB, does nevertheless remain : Therefore the point B has no length ; a^ And because a point is in a line, and H G r 'C B 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, and superficies are to be understood." III. Euclid has defined a straight line to be a line which (as we translate it) "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 our translation ; for the word which we render evenly is s^ias, equally, and is ac- cordingly translated ex isquo, and equaliter by Commandine and Gregory. The definition, therefore, is, that a straight line 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 rectae con- " gruere totam toti in infinitum productum si bina puncta unius binis al- " terius congruant, patet ex ipsa admodum clara rectitudinis idea quam 36 282 NOTES. "habemus." (Supplementum m lib. 3. § 550.) Now, that which Mr. Boscovich would consider as an inference from our idea of straightness, seems 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 penisal 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 Eudid'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 m which, if the position of two points he 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 given,, 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 wtiich Dr. Simson has given. Another definition different from both might have been adopted, viz. That those superficies are called plane, which are such, that 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 that of 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 " 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 circle, 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 10th 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 " 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 are 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- "not be both parallel to the same straight line." On this subject, how- ever, a fuller explanation is necessary, for which see the note on the 29th Prop. PROP. IV. and VIII. B. I. The IV. and VIII. propositions of the first book are tho 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 he constituted on the one, a figure every way equal to it may be con- stituted on the other. Thus if AB and 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 the Postulates or Axioms of geometry ; for the straight lines AB and DE being every 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 of the one are equal to the two ED, DF of the other, and the angle BAG, contained by the sides AB, AC of the one, equal to the angle EDF, con tained by the sides ED, DF of the other ; the triangles ABC and EDF are erery way equal. NOTES. 285 On AB let a triangle be constituted every way equal to the triangle DEF ; Chen 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 equal 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 ; therefore the angles ABC, ACB are 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 to 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 Imes 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 Alexandrinus 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 within 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 ED by construction, and BF and FD will be greater than BC and CA. NOTES. 287 It is evident, that if the angle BAG 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 domonstrated 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 Geometric 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 Avithout 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 demonstration 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 pre- sent instance. The defect may easily be suppHed, and Dr. Simson him- 288 NOTES. self 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 line 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. XXVn. and XXVHI. 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 Dug 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 lo 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. 289 monstrate 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 being *' 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 assumption 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 one of the lines from any point 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, OF 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 straight line DF, and therefore the definition ought to be, " that two straight " 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 prefer 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. Parallels.) Another definition that has been given of parallels is, that they 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, which, it must be acknowledged, is a perfectly good one, if it be understood 37 390 NOTES. by it, that the two lines called parallel, are such as make equal angles with A \g a a certain third line, but not with any line that falls upon them. It remams, 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 certainline 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 diiferent 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, Slc. 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. This 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 motion, but of time, seems to be involved in the notion oi 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- dently a circuitous method of coming at the truth. Thomas Simson, in the second edition of his Elements, has presented this Axiom in a simpler form. " If two points in a straight line are posited "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, CD 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 at 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 sides 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, "Thatjwo straight lines, which " cut one another, cannot be both parallel to the same straight line." This Axiom has been assumed by others, particularly by Ludlam, in his very useful little tract, entitled Rudiments of Mathematics. 292 NOTES. It is a proposition readily enough admitted as self-evident, and leads to 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 tv/o 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 ? 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 new Axiom has been made by a modern geometer, Franceschini, NOTES. 293 Professor of Mathematics in tlie University of Bologna, in an essay, which he entitles, La Teoria delle parallele rigorosamente dimonstrata, printed in his Opuscoli Mathematici, 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 to 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 difi^erent ways, one in the text, and the other in the notes. In the 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 line, as to magnitude, or cannot have an equa- 294 NOTES. tiori of any sort between them. This truth is involved in the distinction between homogeneous and heterogeneous quantities, (Euc. v. def. 4.), which has long been received in Geometry, but led only to negative con- sequences, 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 di. function of the quantities A, B, P ; that is to say, 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 the 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 Ihird 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 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 triangle 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, not liable to the same objection, which I know, from experience, to be of use in explain- NOTES. 295 ing the Elements. It proceeds, like that of the French Geometer, hy 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, which 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 different 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 BAG contained by the line AB and AG. GoR. If a line turn about a point from the position AG till it come into the position AG again, it describes angles which are together equal to four right angles. This is evident from the second Gor. to the 15th. 1. PROP. I. All the exterior angles of any rectilineal figure are together equal to four right angles. 1. Let the rectilineal figure be the triangle ABG, of which the exterior angles are DGA, FAB, GBG ; these angles are together equal to four right angles. Let the line GD, placed in the direction of BG produced, turn about the point G till it coincide with GE, a part of the side GA, and have described the exterior angle DGE or DGA. Let it then be carried along the line GA, till it be in the position AF, that is, in the direction of GA produced, and the point A remaining fixed, let it turn about A till it describe the angle FAB, and coincide with a part of the line AB. Let it next be car- ried along AB till it come into the position BG, and by turning about B, 296 NOTES. o I> let 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. n. 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 produced toward 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 AC is equal to AF, the angles AFC, ACF aie also equal (5. NOTES. 297 1.) ; but the exterior angle HAC is equal to the two interior and opposite mgles ACF, AFC, and therefore it is double of either of them, as of ACF. fow ACE is equal to HAC by construction, therefore ACE is double of .CF, and is bisected by the line CF. In the same manner, if FG be taken lual'to FC, and if CG be drawn, it may be shewn that CG bisects the mgle FCE, and so on continually. But if from a magnitude, as the an- jle ACE, there be taken its. half, and from the remainder FCE its lalf FCG, and from the remainder GCE its half, &c. a remainder will at length be found less than the given angle DCE.* K A Let GCE be the angle, whose half ECK is less 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 right angles. For if AGH be not equal to GHD, let it be greater, then add- ing BGH to both, the angles AGH, HGB are greater than the * Prop. 1. 1 Sup. The reference of this proposition involves nothing incons'stent with good reasoning, as the demonstration of it does not depend on any thing that has gone before, so that it may be introduced in any part of the Elements. 38 298 NOTES. angles DHG, 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 are 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 BAG. 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 ECB to both, and the two rg angles ECD, ECB' are jV 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 ; and the same an- NOTES. 299 gie DCA being tte exterior angle of the triangle ABC, is equal to the two angles ABC, CAB, wherefore the two angles ABC, CAB are equal 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 BAC. BOOK IL 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 a^e easily demonstrated by Algebraical formulas. BOOK III. DEFINITIONS. The definition which Euclid makes the first of this Book is that of equal eircles, 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, that if two magnitudes be double of two others, each of each, the sum or differ- ence of the first two is double of the sum or difiference of the other two, which are two cases of the 1st and 5th of the 5th Book. Th« justness of 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 -f B=2(C + D). For since A=C4-C, and B=D+D, adding equals to equals, A + B=(C-f 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, that " magnitudes are said to have a ratio to one another, when the less can be " multiplied so as to exceed the greater." NOTES. 301 DEF. 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 may 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, (fee. as before, and to divide the product by tjie 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, &lc. 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 maybe 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 been at first multiplied by 1000 ; and therefore, our standard of the pro- portionality of numbers may be expressed thus : If the first multiplied any number of times by 10, and then divided by the second, gives the same quo- 302 NOTES. tient 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 muhiplications 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 has 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 mK divided by B gives n for the quotient, wA=nB by the nature of division, so that when mA=;tB, mC=znt), 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 mky-nQ ; and, for the same reason, mC/'wDj which is another of the conditions of Euclid'^ definition. Lastly, when wiA is divided by B,Iet n be a whole number greater than the exact quotient, then mk/_nE, and because n is also greater than the quotient of mC divided by D, (which is the same with the other quotient), therefore mC /_nD. Therefore, uniting all these three conditions, we call A, B, C, D, propor- tionals, when they are such, that if mkynB, mC /wD ; if mA=znB, mC=: nD ; and if mA/wB, mC^wD, 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, whethet 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^bwr 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 than they would otherwise be, by avoiding all discus- sion about the magnitude of the remainder left, after B is taken out of mA as 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 it NOTES. 30a 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 mA into parts equal to B can be exactly performed, the other when it cannot be exactly performed whatever value be assigned 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 ad ah- surdum. M. P'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 maniere, (la reduction a absurde,) la " plupart des propositions qui regardent les incommensurables. L'idee de " i'inliui entre au moins implicitemens dans la notion de ces sortes de quan- " tites ; et comme nous n'avons qu'une idee negative de I'infini, on ne peut " demontrer directement, et a priori, tout ce qui concerne Tinfini mathema- " tique." (Encyclopidie, mot Geometrie.) 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 less 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 difierent 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 only 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 difiie- rent operations supposed to be perfo-rmed 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, as 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 merely 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, rattier ihaLU 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 all the three definitions is best seen when they are ranged in this order. It is then plain, that two equal ratios compound a ratio duplicate of either of them ; 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 cpquo, 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 ratio, 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 the 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 ttj T7j ~r\i "crjbe any series of ratios, such as described above, the ratio a Lf D hi A A B • -=-, or of A to E, is said to be compounded of the ratios -rr-, -^, &c. The ratio A A B •=^, is evidently determined by the ratios -^, -p^, &c. because if each of the hj a L> 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 of 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 anfong 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 Meyalr) Swta^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 bjAwhich 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 ne*cessary to observe, that some 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, tb 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 rectilineal figures 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 the circular spaces, to the ratios of the sums of those rectangles, or of the whole rectilineal figures. In doing this, he takes for granted a proposition, which, it is wonderful, that one who had studied 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, c, d, e,fj as many others ; and if A : B : : a : &, C : D : : c : (Z, 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 c, to the sum of h, d and/, or A+C+E : B+D 4-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 : c ; and consequently, B :D ::b: d, and B : F : : 5 : /; or, secondly, when all the ratios of A to B, C to D, E to F, 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 solid angle, the straight line, joining those points, falls wholly within the solid 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 figures 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 definitiouj^, 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 by Dr. Simson, as will presently appear. PROP. IV. This very elegant demonstration is from Legendre, and is much easier than that of Euclid. The demonstration given here of the 6th is also greatly simpler than 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 of 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. 311 BOOK III* DEF. 11, and PROP. I. These relate to similar and equal solids, a subject on wKich mistakes have 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 AE 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 ? 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 seems 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 conti- 312 NOTES. SUPPL. BOOK III. nually ; for, though two plane figures that are equal and similar can always be made to coincide, yet, with regard to solids that are equal and similar, if they have not a certain similarity in their position, there will be found just as many cases in which they cannot, as in which they can coincide. Even figures 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 being greater than the other. This Axiom may be rendered general by saying, That things of which the magnitude is determined by conditions that are exactly the same, are equal to oije 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 in 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 from one another. But, concerning those things that ^re 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, of two, or of three dimensions. NOTES. SUPPL. BOOK III. 313 Legendre in his Elements has made the same remark that na.? been just stated, 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. He 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 such 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 solids 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, which have co*e 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 speak of, and on looking slightly into Simson, having found a limitation of the usual description of equal solids, he had without much inquiry, set it down 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 really belonged to himself. This at least seems to be the most probable solution of the difficulty. I have entered into a fuller discussion of Legeiidre's mistake than I 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 V is evident would have been a pretension as ridiculous as ill-founded, if the r same observation had been made in a book like Simson's, which in this f country was in every body's hands, and which I had myself professedly f 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 I 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 think 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 he same number of equal and similar planes. But this is not true, as Dr. 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 obviously unequal, the one being contained within the 40 314 NOTES. SUPPL. BOOK III. other. Simson observes, that Euclid needed only to have added, that the equal and 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 n§w 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 himself 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 it 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 are 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 equal 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 siinilar to the triangle FBC, and the triangle ExiC 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 are 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,