IN MEMORIAM FLORIAN CAJORl E-UCLID'S ELEMENTS OF GEOMETKY: BOOKS I. 11. III. IV VI. AND PORTIONS OF BOOKS V. AND XI. WITH NOTES, EXAMPLES, EXERCISES, APPENDICES, AND A COLLECTION OF EXAMINATION PAPERS ARRANGED BY A. E. LAYNG, M.A., Head-master of Stafford Grammar School ; formerly Scholar of Sidney Sussex College, Cambridge. u . C. ^. lo, '■2, lU. It, 3. (J), U^y- LONDON. BLACKIE & SON, Limited, 50 OLD BAILEY, E.G. GLASGOW AND DUBLIN. PREFACE. In this edition of those parts of Euclid's Elements most com- monly read in schools and required for examinations, no fundamental changes have been made, the axioms and enunciations remaining the same as in Simson's edition. Minor changes have, however, been frequently introduced, when it appeared that, either by the insertion of an additional step or reference, the path of the begiimer could be made smoother, or conciseness and clearness gained by the removal of redundant and antiquated phrases. In some cases, moreover, Euclid's proofs have been displaced in favour of shorter ones based on his methods. Well-known abbreviations of words have been . freely employed throughout, but the symbols used in the text are on ly such as are allowed in all the principal public examinations, and these are employed from the very first in the belief that their use is easily acquired, and that, by enabling the proofs to be pre- sented in an attractive form, they help rather than hinder the beginner. In some of the examples a few other symbols in common use in modern geometry have, on account of their great convenience, Beeii employed. The text is so arranged that the enunciation, figure, and proof of each proposition are all in view together. Notes and easy exercises are also directly appended to the propositions to which they refer, and numerous examples, fully worked out, are inserted at intervals as models ; while at the end of the diff'erent books additional propo- sitions connected with those books are given, followed by collections of miscellaneous exercises. Great care has been taken to provide large and clear diagrams, and, by the use of varied lines, to mark the distinction between given lines and lines of construction, as well as between real and hypothetical constructions. Appendices containing some of the simpler theorems of modern geometry, and a few alternative proofs of propositions on methods somewhat different from Euclid's, and for this reason not inserted in the body of the work, have been added, together with a collection of the questions lately set in various public examinations. A. E. L. INTRODUCTORY NOTE. Geometry, as the name signifies, probably had its origin in land-surveying, of which the Egyptians had some knowledge as far back as 2000 B.C., and it was from Egypt that the Greeks, to whom the great development of the science which occurred between 600 B.C. and 400 a.D. is due, obtained their first knowledge of it. One of the most celebrated of the early Greek mathematicians was Euclid, who was born about 330 B. C. , and educated, probably, at Athens, but who afterwards settled at Alexandria, where he gained a great reputation as a teacher of geometry. Little is known of his life beyond a few sayings which tradition ascribes to him; among these is his reply to the Egyptian king that there was no royal road to geometry. Euclid was the author of several works on mathematics, of which the "Elements" is the most important. It at once became the standard text- book on elementary mathematics, a position which, so far as the geometrical part of it is concerned, it has now held for more than 2000 years. The first part of the work was compiled from the writings of the earlier Greek mathematicians, Pythagoras and Hippocrates. The Greek text, on which our English ones are based, is that of an edition prepared by Theon (the father of Hypatia) about 380 A.D. Books I., II., III., IV. and VI. of Euclid's Elements treat of Plane Geometry, Book V. of Proportion, Books VII., VIII. and IX. of rational numbers, Book X. of surd numbers, Books XI. and XII. of Solid Geometry, and Book XIII. contains additional propositions relating to the other books. Most modem Enghsh editions of Euclid follow a translation of the Greek text made by Robert Simson, who was born in 1687, and was professor of mathematics at Glasgow University, CONTENTS. Page Introductory Note, iv Abbreviations, vii Symbols, viii BOOK I. Definitions, with Notes and Exercises, 5 Postulates and Axioms, 16 Propositions 1-48, with Notes, Examples, and Exercises, - - - 18 Miscellaneous Examples, 80 Miscellaneous Exercises, - - 90 BOOK II. Definitions, with Notes and Exercises, 97 Propositions 1-14, with Notes, Examples, and Exercises, - - - 99 Miscellaneous Examples, - - - 120 Miscellaneous Exercises, 126 BOOK III. Definitions, . 129 Propositions 1-37, with Notes, Examples, and Exercises, - - 132 Miscellaneous Examples, - - - 176 Miscellaneous Exercises, 186 BOOK IV. Definitions, 193 Propositions 1-16, with Notes, Examples, and Exercises, - - 194 Miscellaneous Exercises, 214 BOOK V. Definitions, 217 Summary of Theorems, „ . 220 BOOK VI. Definitions, 222 Propositions 1-D, with Notes, Examples, and Exercises, - - - 223 Miscellaneous Examples, - 266 Miscellaneous Exercises, 278 VI CONTENTS. BOOK XI. Page Definitions, with Notes and Exercises, 281 Propositions 1-21, with Notes, Examples, and Exercises, - - - 288 Miscellaneous Examples, 308 Miscellaneous Exercises, 317 APPENDIX I. Transversals, 320 Harmonic Section, - - - - 322 Pole and Polar, 324 Miscellaneous Exercises, - - 326 APPENDIX 11. Alternative Proofs, 326 EXAMINATION PAPERS. College of Preceptors, 329 Civil Service, 333 South Kensington, 335 London University Matriculation, 340 Education Department, 342 Scotch Education Department, Leaving Certificate, - - . - 345 Oxford Local Examinations, -......- 347 Oxford Responsions, ,. . . . 349 Cambridge Local Examinations, 349 Cambridge Higher Local Examinations, 352 Cambridge Previous Examination, 353 Cambridge Mathematical Tripos, 353 Index, ~ . 356 ABBREVIATIONS OF WORDS. adj. for adjacent. invert. for invertendo. alt. n altitude. isos. II isosceles. alter n. n alternando. mag. II magnitude. ax. II axiom. max. II maximum. cent. n centre. mid. 1 1 middle. chd. chord. min. II minimum. circ<^ II circumscribed. mult. II multiple. com. II common. no. 1 1 number. compo. II componendo. opp. II opposite. comp* II complement. post. 1 1 postulate. const. II constant. prob. II problem. constr. II construction. prod. II produce. conv. 1 1 converse. prop. 1 1 proposition. cont^ II contained. prop"^ 1 1 proportion. cor. II corollary. pt. 1 1 point. def. definition. quad^ 1 1 quadrilateral desc. 1. describe. rad. II radius. diag. II diagonal. rect. It rectangle. diam. n diameter. recti II rectilineal. diflf. II difference. recf^ It rectangular. dist. II distance. reg. It regular. dup. 11 duplicate. reme It remaining. equiang. n equiangular. rem"^ II remainder. eqiiidist. m equidistant. req*^ It required. equilat. m equilateral. rt. II right. equirault. n equimultiple. segt It segment. ext>^ 1, exterior. sim'' II similar. ex. aeq. n ex sequali. sq. It square. fig. II figure. St.- ' It straight. harm. n harmonic. supp* It supplement. hyp. hypothesis. tang. It tangent. hypot. II hypotenuse. tet" It tetrahedron. insc*^ II inscribed. theor. II theorem. inf^ II interior. vert^ II vertical. join AB for "draw a straight line from A to B." rect. AB, CD ,, *' the rectangle contained by AB and CD.' SYMBOLS. The following symbols are used in the Text: — = for any one of the phrases "is equal to," "are equal to," " be equal to," or " equal to." > II " is greater than." < II "is less than." II "therefore." II "because." L II "angle." ^^ II "triangle." O " "parallelogram." A- II "perpendicular." II II "parallel." ::^ II "proportional." I. "circle." Oce II "circumference." The following symbols are used in Examples only : — AB^ for " the square described on AB." AB.CD II "the rectangle contained by AB and CD." " the sum of." " the difference between." " a is to 6 as c is to d." a:b: '.c:d A B " the ratio of A to B." A symhol, when used in the plural, is followed by the letter s. EUCLID'S ELEMENTS OF GEOMETRY. BOOK J. DEFINITIONS. 1. A point is that which has no parts, or which has no magnitude. 2. A line is length without breadth. 3. The extremities of a line are points. 4. A straight line is that which lies evenly between its extreme points. 5. A surface (or superficies) is that which has only length and breadth. 6. The extremities of a surface are lines. 7. A plane surface is that in which any two points being taken, the straight line between them lies wholly on that surface. NOTES. A point has position but no size, and so cannot be divided into parts, A straight line is sometimes called a right line. A plane surface is generally spoken of as "a plane." (e.g., By "a plane curve " is meant a curved line drawn upon a plane surface.) EXERCISES. 1. How many dimensions has (i.) a solid, (ii.) a surface, (iii.) a line, (iv.) a point? 2. How many dimensions has (i.) a sheet of paper, (ii.) a shadow, (iii.) a brick, (iv.) a cricket-ball, (v.) a bit of spider's web? 3. Are the surfaces of the walls of an ordinary room plane? 4. How would you show that the surface of a cricket-ball is not plane? 5. How many plane surfaces has a well-cut square of plate-glass? How many edge-lines? How many corner-points? 6. Is it possible for a straight line to lie wholly upon the polished surface of a cedar pencil ? Is this a plane surface ? 7. Is a point a magnitude? Is a line a magnitude? 8. If two points were taken at random on a plane, would a straight line through those points bridge over any part of the plane? ,>/(310) ^ EUCLID, BOOK I. A plane angle is the inclination of two lines to each other in a plane, which meet but are not in the same direction. 9. A plane rectilineal angle is the inclina- tion of two straight lines to each other in a plane, which meet but are not in the same straight line. 10. When a straight line standing on another straight line makes the adjacent angles equal, each is called a right angle. The line which stands upon the other is called a perpendicular to it. 11. An obtuse angle is greater than a right angle. 12. An acute angle is less than a right angle. 13. A term, or boundary, is the extremity of anything. ADDITIONAL DEFINITIONS. The vertex of an angle is the point at which the lines, which form the angle, meet. Adjacent angles are such as lie next each B other. Thus AOB and AOC are ad- jacent angles. The angles AOC and BOD are called opposite vertical angles. If a right angle is divided into any two parts, each part is called the complement of the other. Thus, if POQ is a right angle, the angle EOF is the comple- ment of the angle ROQ. S The angle EOS is called the supplement of EOQ. The bisector of an angle is a straight line which divides it into two equal angles. DEFINITIONS. NOTES. B Def. 8 is of no importance, for only angles contained by straight lines are dealt with in Elementary Geometry, and "angle" when it occurs always means plane rectilineal angle. An angle is generally denoted by three letters, the letter which stands at the vertex of the angle being placed between the other two. The same angle may be named in various ways ; for instance ABC, CBA, DBE, EBD, ABE, EBA, DBC, CBD are all names for the angle here represented. N.B. — The size of an angle does not depend on the length of the lines. If OA, OB, OC are three straight lines which meet at O, the angle AOC is the sum of the angles O AOB and BOC ; the angle BOC is the difference of the angles AOC and AOB. An angle equal to one-ninetieth part of a right angle is called a degree (written thus 1°). EXERCISES. 1. Write down names for all the angles in the accompanying figure which have (i.) vertex A, (ii.) vertex B, (iii.) vertex C, (iv.) vertex D, (v.) vertex O. 2. Name the angle which is equal to (i.) the sum of the angles ABO and CBO. /^V>t (ii.) the sum of the angles ABD and DBC. 4^C (iii.) the sum of the angles BCO and ACD. (^^P b' (iv.) the difference of the angles ADC and BI>G.^^$ (v.) the difference of the angles CD A and CDO.^^ (vi.) the difference of the angles BAD and BAC. i)AC 3. Mention a pair of adjacent angles with vertex O. 4. Mention a pair of opposite vertical angles with vertex O. 5. Mention an angle which is the supplement of the angle AOD. 6. Is an angle of 93° acute or obtuse? 7. What is the complement of an angle of 45°? 8. What is the supplement of an angle of 125°? 9. Are angles of 23° and 67° complementary? 10. Are angles of 68° and 102° supplementary? (810) - A 2 8 EUCLID, BOOK I. 14. A figure is that which is enclosed by one or more boundaries. 15. A circle is a plane figure contained by one line called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circum- ference are equal to one another. 16. This point is called the centre of the circle. 17. A diameter of a circle is a straight line drawn through the centre and terminated both ways by the circumference. 18. A semicircle is a figure contained by a dia- meter and the part of the circumference that it cuts ofi*. 19. A segment of a circle is a figure contained by a straight line and the part of the circumference that it cuts ofi*. ADDITIONAL DEFINITIONS. A radius of a circle is a straight line drawn from the centre to the circumference. (In the fig. of Def. 15, OP and OQ are radiL) An arc is part of a circumference. "^-v^^^ ^^^ Concentric circles are such as have the same point as centre. A chord of a circle is a straight line joining two points on its circum- ference. The perimeter of a figure {i.e. the measure round it) is the sum of the lengths of its sides. The area of a figure is the surface (or space) enclosed by its boundaries. DEFINITIONS. EXERCISES. 1. Is a circumference a figure ? 2. Is an angle a figure? 3. Is an arc a figure? 4. Is a semicircle a plane figure? 5. Can a figure be bounded by one straight line? 6. Mention a figure which is bounded by two lines. 7. If the diameter of a circle is 3 feet long, what is the length of its radius? 8. The radius of a circle is 7 inches long ; find the length of a diameter. 9. Draw a plane six-sided figure. 10. If two figures are equal in ares 11. In Figure 1 mention by name (i.) a radius, (ii.) a diameter. (iii.) an arc. (iv.) a chord. (v.) a segment. 12. In Figure 2 mention by name (i.) each circle, (ii.) their common radius, (iii.) their common chord, (iv.) their centres. 13. With a pair of compasses, describe (i.) three unequal circles which do not cut each other; (ii.) three equal circles which mutually cut each other; (iii.) three concentric circles. 14. Find the locus (or path) of a point which moves in a plane, so that its distance from a given fixed point is always the same. 15. If a line, one extremity of which is fixed, revolve in a plane, what is the locus of its other extremity? 10 EUCLID, BOOK 20. Rectilineal figures are such as are contained by straight lines. 21. A trilateral figure, or triangle, is contained by three straight lines. 22. A quadrilateral figure is contained by four straight lines. 23. A multilateral figure, or polygon, is con- tained by more than four straight lines. TRIANGLES. 24. An equilateral triangle has three equal sides. 25. An isosceles triangle has two of its sides equal. 26. A scalene triangle has three unequal sides. 27. A right-angled triangle has a right angle. 28. An obtuse-angled triangle has an obtuse angle. 29. An acute-angled triangle has three acute angles. ADDITIONAL DEFINITIONS. The side opposite to any angle of a triangle is said to subtend that angle. Ar In a right-angled triangle, the side which subtends the right angle is called the hypotenuse. (For example, if ABC is the right angle, AC B is the hypotenuse.) If a side BC of a triangle be produced to D, the angle ACD is called an exterior angle of the triangle. DEFINITIONS. 11 NOTES. In Defs. 24, 25, 26, triangles are classed according to sides; in Defs. 27, 28, 29, according to angles. It is often convenient to distinguish one particular side of a triangle by calling it the base. The angular point opposite to the base is called the vertex of the triangle. In the case of an isosceles triangle, the side which is not equal to either of the others is the base. EXERCISES. Is a semicircle a rectilineal figure? The side QE, of the triangle PQE, is produced to S ; mention (i.) an exterior angle of the triangle? (ii.) its adjacent interior angle? (iii.) the two opposite interior angles? Which side subtends the angle PRQ? Write down names for the five triangles in the accompanying figure. B ,D Draw two triangles standing on the same base and on the same side of it, with the vertex of each outside the other. Mention names for the eight triangles in the accompanying figure. Mention exterior angles of the triangle EDO. What angles does OC subtend? Mention a pair of vertically opposite angles with vertex O. If one side of an equilateral triangle is 3 feet long, find the perimeter of the triangle. One of the equal sides of an isosceles triangle is 3 feet long, and the base is 2 feet ; find its perimeter. What is the perimeter of an equilateral polygon of five sides, if the length of one side is 2 feet? 12 EUCLID, BOOK I. QUADKILATEKALS. 30. A square is a four -sided figure with all its sides equal, and its angles right angles. 31. An oblong is a four-sided figure with all its angles right angles, but only its op- posite sides equal. 32. A rhombus is a four-sided figure with all its sides equal, but its angles not right angles. 33. A rhomboid is a four -sided figure with its opposite sides only equal, and its angles not right angles. 34. All other four -sided figures are called trapeziums. ADDITIONAL DEFINITIONS. A diagonal of a quadrilateral is a straight line joining two opposite angles. (AC and BD are the diagonals of the quadrilateral ABCD.) To bisect is to divide into two equal parts. To trisect is to divide into three equal parts. DEFINITIONS. 13 NOTES. The names oblong, rhomboid, and trapezium are seldom used. Practic- ally, an oblong is spoken of as "a rectangle" or " a rectangular figure; " a rhomboid as " a parallelogram; " and a trapezium as " a quadrilateral." It will be shown hereafter (Prop. 34) that the class of figures defined as rect- angles includes the oblong, and the class defined as parallelograms includes the rhomboid. A four-sided figure is often denoted by two letters A which stand at opposite corners. Thus the rectangle ABCD may be referred to as "rectangle AC," or T) as "rectangle BD." EXERCISES. 1. What is meant by an equiangular figure? 2. What by a rectangular figure? 3. What by an equilateral figure? 4. What by a rectilineal figure? 5. Is a rhombus equilateral? 6. Is a square rectangular? 7. Show how to draw a line which shall divide a given square into (i.) two right-angled triangles; (ii.) two trapeziums. 8. If the length of one side of a rhombus is 2 feet, what is its peri- meter 1 9. Mention a dozen common objects which have a rectanglilar surface. 10. How many rectangular surfaces has a brick? 11. The diagonals of a quadrilateral ABCD cut at O. Write down names for all the different triangles formed. 12. If two right-angled triangles stand on opposite sides of a line which is their common hypotenuse, is the figure formed necessarily an oblong? 14 EUCLID, BOOK I. 35. Parallel straight lines are such as lie in the same plane, and which, being produced ever so far both ways, do not meet. ADDITIONAL DEFINITIONS. A parallelogram is a four-sided figure with its opposite sides parallel. / A rectangle is a right-angled parallelogram. A diameter of a parallelogram is a straight line joining two opposite angles {i.e. a diagonal). If a straight line EFG meets two parallel straight lines AB and CD, such a pair of angles as AFG and FGD are called alternate angles; such an angle as EFA is called an exterior angle, and FGC is the angle interior and opposite to EFA. If, through any point in a diameter of a parallelogram, straight lines are drawn parallel to the sides, the figure is divided into four par- allelograms. Of these four, the two through which the diameter passes, are called parallelograms about the diameter; and the other two, which complete the whole figure, are called the complements. (For example, AE and EC are complements.) A figure which has all its sides equal, and all its angles equal, is called a regular figure. A quadrilateral with two sides parallel, and two not parallel, is sometimes called a trapezoid. DEFINITIONS. 15 NOTE. Pentagon, hexagon, octagon, decagon, dodecagon, quindecagon are names used for polygons of 5, 6, 8, 10, 12, and 15 sides respeetively. EXEKCISES. 1. If two straight lines, in different planes, when produced ever so far both ways, did not meet, would they be necessarily parallel? 2. Show how to divide a quadrilateral into four triangles by lines drawn from a point within the figure. 3. Divide a pentagon into as many triangles as the figure has sides, by lines drawn from a point within it. 4. Define a regular hexagon. Make a sketch of the figure. 5. Define a regular decagon. Make a sketch of the figure. 6. If one side of a regular octagon is 3 inches long, what is its perimeter? 7. What are the marked angles called in the following diagrams? (i) (ii) 8. Give the meaning of the words, magnitude, rectilineal, acute, equilateral, isosceles, complement. 9. Draw two straight lines which cut. 10. Draw two straight lines which meet. 11. Class triangles according to their sides. 12. Make a sketch of a trapezoid. 13. Make a sketch of a right-angled isosceles triangle. 14. Has every triangle a hypotenuse? 15. Has every figure an area? Is an angle a figure? 16. Define a plane. 17. Does the magnitude of an angle depend on the length of its arms? 18. Define a superficies. 16 EUCLID, BOOK I. POSTULATES. Let it be granted : 1. That a straight line may be drawn from any one point to any other point. 2. That a terminated straight line may be produced to any length in a straight line. 3. That a circle may be described from any centre, at any distance from that centre. AXIOMS. 1. Things that 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 une(]|ual. 6. Things that are double of the same thing are equal. 7. Tilings that are halves of the same thing are equal. 8. Magnitudes which coincide with one another are equal. 9. The whole is greater than its part. 10. Two straight lines cannot enclose a space. 11. All right angles are equal to one another. 12. If a straight line meet two straight lines so as to make the two interior angles on the same side of it together less than two right angles, these two straight lines will, if produced, meet on that side on which the angles are less than two right angles. (310) POSTULATES, AXIOMS. 17 NOTES. The postulates are three problems which are assumed to be possible. They amount to a statement that certain instniments are necessary; and sufficient, for constructing the figures of the propositions. These instru- ments are: — (i) A ruler for drawing straight lines, but not graduated for measuring their lengths, (ii) Compasses for describing circles, but not to be used for carrying measurements from one place to another. The Axioms are theorems the truth of which is assumed to be evident without proof. Euclid calls them Common Notions. Axioms 1 to 7, and 9 are general, that is, they do not apply to geometrical magnitudes only. Axioms 8, 10, 11, and 12 are geometrical. The order of the earlier axioms is easily learned by noticing that it is the same as in the early rules of arithmetic: — Axioms 2 and 3 — Addition and Subtraction of equals. Axioms 4 and 5 — Addition and Subtraction of unequals. Axiom 6 — Multiplication. Axiom 7 — Division. Axiom 8 has been called Euclid's test of equality. It states that two magnitudes (whether lines, angles, or figures) are equal, if one can be so placed on the other that all their parts agree; if, in short, they coincide. Axiom 12 is not required before the 29th Proposition; and the beginner may postpone the consideration of it until that proposition is reached. EXERCISES. 1. How many straight lines can be drawn between two given points? 2. ABC, DEF are two equal straight lines ; the part AB is equal to the part DE; is BC equal to EF? 3. State the converse of Axiom 8. Is this always true? 4. Are the areas of two pages of the same book equal? 5. Is the area of the triangle ABC equal to the area of the triangle DBC in the accompanying figure? -p- 6. What is the least number of straight lines that can form a figure ? 7. Can two straight lines cut in more than one point? 8. AB and CD are equal straight lines; AB is bisected at E, CD is bisected at F ; is AE equal to CF? (310) B 18 EUCLID, BOOK 1. PEOPOSITION I. Problem. To describe an equilateral triangle on a given finite straight line. Let AB be the given st. line. It is re^ to desc. an equilat. ^^^ on AB. "With cent. A and rad. AB desc. BCD Post. 3. With cent. B and rad. AB desc. ACE Post. 3. From pt. C, wheVe the 08 cut, draw st. lines CA, CB Post.i. Then shall ^^ABC be equilat. For, since A is cent, of 0BCD /. AB=AC Def. 15. and, since B is cent, of 0ACE /. AB=BC Def. 15. Hence, since AC and BC are each of them equal to AB /. AC=BC Ax.i. and AB, AC, BC are all equal. Wherefore, the ^-^ ABC is equilat. and has been described on A B. Q.E.F. NOTES. The letters Q.E.F. which stand at the end of a problem, are the initials of the words Quod erat faciendiim {tvhich was to be made). In succeeding propositions when I. 1 occurs among the references printed on the right-hand side of the page, it stands for ** Book I., Pro- position 1 ". PROP. I. 19 The propositions are either Problems or Theorems. In a problem something has to be made, or done, with the ruler and compasses of the postulates. In a theorem some geometrical truth, or property of a figure, has to be proved, or deduced, from the axioms. The statement at the head of a proposition is called its enunciation. The condition, subject to which any proposed problem is to be solved, or theorem proved, is called the hypothesis* (For example, the hypothesis of Prop. I. is that the equilat. triangle must be constructed "on a given finite St. line" i.e. on a st. line whose position and length have both been fixed beforehand.) The beginner should carefully notice the three distinct parts of the pro- position; they are — 1st. The Particular Enunciation, or re-statement of the general enuncia- tion with reference to a figure. 2nd. The Construction, or directions for drawing such lines and circles as are needed to solve the problem. 3rd. The Proo/ that the figure made is really such as was required. He should also compare the statement "then shall ^A A"Rn be equilat." at the beginning of the proof with "wherefore .^ABC is equilat." at the end. And, in writing out a proposition, he should on no account attem{)t to draw the figure first, but should allow it to grow, step by step, as he writes down the construction. The letters Q.E.D., at the end of a theorem, stand for Quod erat demonstrandum (which was to he proved). EXERCISES. (Exs. 1 to 5 refer to the figure of Prop I.) 1. If F be the other point of intersection of the circles, prove that FAB is an equilateral triangle. 2. Prove that FBCA is an equilateral figure. 3. Find a point P in AB produced such that AP is double of AB. 4. On AB as base describe an isosceles triangle with its sides each double of AB. 5. If with centre A and radius less than AB a circle be described cutting AB in P and AC in Q, prove that PB is equal to QC. 6. P and Q are fixed points ; it is required to find a point which shall be equidistant from P and Q. 7. Show how to obtain a straight line which shall be treble of a given finite straight line. 8. Two concentric circles have centre O; radii OA, OB of the smaller circle are produced to meet the circumference of the larger circle in C and D respectively; prove that AC is equal to BD. * In a problem the given conditions are also called the data. EUCLID, BOOK PROPOSITION II. Problem. From a given point to draw a straight line equal to a given straight line. Let A be the given pt., and BC the given st. line. It is req^ to draw from A a st. line=BC. Join AB Post. 1, On AB desc. an equilat. --^DAB i.L Produce DA, DB to E, F Post. 2. With cent. B, rad. BC desc. 0CGH cutting DF at G...Post. 3. With cent. D, rad. DG desc. 0GKL cutting DE at L... Post. 3, Then shall AL=BC. For, since B is cent, of 0CGH .-. BC=BG Def. 15. And, since D is cent, of 0GKL /. DL=DG Def. 15. but part DA=part DB Constr. /. rem^ AL=rem'" BG Ax. 3. Hence, AL and BC are each of them equal to BG .-. AL=BC Ax. 1. Wherefore, from the given pt. A &c. Q.E.F. PROPOSITION III. Problem. From the greater of two given straight lines to cut off a fart equal to the less. Let AB and C be the two given st. lines, AB being the greater. It is req^ to cut off from AB a part^=C. From A draw AD=C L 2. X ^P With cent. A, rad. AD, desc. [ X ]E DEF cutting AB in E Post. 3. I ^ J ^ Then shaU AB=C. C— :— ^^ For, since A is cent, of 0DEF .*, AE=AD Def. 15. but AD=C Constr. /. AE=C Ax. 1. Wherefore, /ro»t the given st. line AB &c. Q.E.F. PROPS. II., III. 21 NOTES. The complex figure of Prop. 2. is caused by the restrictions in the postu- lates. (See note on Post. 3.) Prop. 2 should be practised with varied positions of the given line and point, and with different sets of letters for the figures. Since either end of the given line might be joined to the given point; the equilateral triangle described on either side of this line; and the sides of the triangle produced in either direction, it follows that, for any given Fig. 1. Fig. 2. positions of the line BC and the point A, eight lines each equal to BC can, by varying the figure, be obtained at A. Fig. 1, Fig. 2 are examples of such varieties. EXERCISES. 1. Draw the six other figures of the set to which Fig. 1 and Fig. 2 belong. 2. Make the figure and give the proof of Prop. 2 for the special case when the given point is in the given line. 3. In what special cases is it unnecessary to produce the sides of the equi- lateral triangle in Prop. 2? 4. In the figure of I. 2 AD, BD are produced to meet the circumference of the larger circle in P and Q respectively ; prove that AP is equal to BQ. 5. On a given straight line as base describe an isosceles triangle having each of its other sides equal to a given straight line. In what case is this impossible? 6. In the figure of I. 3, from AB, or AB produced, cut off a part equal to three times C. 7. On a given straight line as base construct a triangle having its other sides equal to two given straight lines. When is this impossible? 8. Equal circles, whose centres are O and Q, cut the line OQ at A and B respectively. Prove that OB is equal to AQ. 22 EUCLID, BOOK I. PKOPOSITION IV. Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles, contai?ied hy these sides, equal; they shall have their bases, or third sides, equal, and the two triangles shall he equal, and their other angles shall be equal, each to each, namely, those to which the equal sides are opposite. Let ABC, DEF be two -:^s, having AB=DE ^A jP AC=DF Z.BAC=z.EDF. Then shall BC=BP ^^ABC=^^DBP z.ABO=^DEF B C E F ;LA0B=Z.DPB. For, if ^^ABC be applied to .^DEF, with pt. A on D, and AB lying along DE, then pt. B must fall on E, V AB=DE Hyp. AC must lie along DF, •/ Z.BAC= ^EDF Hyp. pt. C must fall on F, V AC=DF Hyp. Now, since B falls on E, and C on F, BC must coincide with EF, or two st. lines would enclose a space which is impossible Ax. lo. Hence, since BC coincides with EF, /. BC=EF Ax.8 and, since the -ir:\s coincide, C^^ABC=.-:XDEF^ .-. \ ^ABC=2LDEF I Ax.8. i Z.ACB=^DFE J Wherefore, if two triangles &c. Q.E.D. NOTES. This proposition, the first of Euclid's theorems, is of great importance, and the beginner should not attempt to proceed further until he has thoroughly digested it, and can apply its results in examples. A triangle has seven parts: 3 sides, 3 angles, and an area. When all the parts of one triangle are respectively equal to all the part of another triangle, the triangles are said to be "equal in all respects." PROP. IV. 23 The 4th Prop, may be enunciated shortly thus : If two sides and the included angle of one triangle are known to be equal to two sides and the included angle of another, these triangles must be equal in all respects. Prop. 4 is proved by "the method of superposition," i.e. it is shown that one of the triangles could be so placed on the other as exactly to cover it without overlapping, and then, since all their parts would coincide, the truth of the proposition follows from Axiom 8. In this, and all succeeding theorems, the particular enunciation is ar- ranged thus: the known facts, which form the hypothesis, stand first; then follow, printed in dark type, those which must not be taken for granted, but are to be proved. This part is called the conclusion. The proof of the following theorem is given as an example of the way in which the results of Prop. 4 may be used : Theorem. The diagonals of an oblong are equal. A^ ^D Let ABCD be an oblong, AC and BD its diagonals; then shall AC=BD. For in ^^s ABC, DBC, 'AB=DC Del 31. BC is common to both ^^s ,rt. Z_ABC=rt.Z.DCB Ax.ii. B~ /. AC=BD by Prop. 4. O E D EXERCISES. 1. If two triangles have two sides of one equal to two sides of the other, must the triangles be equal in all respects ? 2. AB is a straight line; D is its middle point; DC is a line at right angles to AB ; prove that CA is equal to CB. 3. The diagonals of a square are equal. 4. ABC is an isosceles triangle having AB equal to AC ; AD bisects the vertical angle BAC and meets the base BC at D; prove that BC is bisected at D. 5. The bisector of the vertical angle of any isosceles triangle is at right angles to the base. 6. D and E are the middle points of the equal sides AB, AC of an isosceles triangle ; prove that the triangles ABE and ACD are equal in all respects. 7. In the figure of I, 2, if AG and BL be joined, AG- is equal to BL, 8. In the figure of I. 2, if AG and BL be joined, prove that the angle AGD is equal to the angle BLD. 9. In the figure of I. 3, if with centre A and radius AB a circle be de- scribed, and AD be produced to meet this circle in G, the triangles AGE and ABD are equal in all respects. 10. ABCDE is a regular pentagon; prove that BD is equal to CE. 24 EUCLID, BOOK I. PEOPOSITION V. Theorem. The angles at the base of an isosceles triangle are equal to one another ; and, if the equal sides he produced, the angles on the other side of the base are also equal. A Let ABC be an isos. -.^ having AB=AC, and let AB, AC be prod^ to D, E. Then shall (i) ^ABC=z.ACB, (ii) L DBC= L BOB. bZ ~,Ap In BD take any pt. F. /^-"^^^^ Ar From AE cut off AG=AF 1.3. ^f^ "^ Join BG, CF, Post. i. D Then, in the .^s AFC, AGB, ^^ ( AF=AG Constr. V ] AC=AB Hyp.; I L B AC is com. : FC=BG I Z.AFC=z.AGB*^ 1.4 ^ACF=^ABG J Again, the whole AF= whole AG Constr. and part AB=part AC Hyp. /. rem'^ BF=rem'^ CG Ax. 3. Hence, in ^^^^s FBC, GCB, BF=CG ^ FC=BG [ proved above. ^BFC=^CGB*J Z.FBC=^GCB\ ^FCB=^GBCJ '•'*• Now, the whole z.ABG=whole Z-ACF,) ^ ^ ' , ^^^ ^^^ )■ proved above. and part Z_GBC= part Z.FCB; j /. rem^ z.ABC=rem>- Z-ACB Ax. 3. which are the angles at the base, and, from above, Z-FBC=^GCB which are the angles on the other side the base. Wherefore, the angles at the base &c. Q.E.D. CoR. An equilat^ ^^^ must also be equiang'. { * N.B. — AGB and CGB are both names for the same angle. PROP. V. 25 NOTES. The 5th Prop, is proved by a double application of the 4th, and so offers little difficulty to one who has thoroughly understood the 4th. It may, however, help the beginner if he draws separate figures for the pairs of triangles dealt with. Thus, in the first pair of triangles : — AF is known to be equal to AG (constr.). A A AC is known to be equal to AB (given). A A FAC, BAG are both names of the same / \ / \ angle BAG, which is common to both tri- / \ / \ angles. / \ / \ Hence, since we find two sides, and the r - -;^f^ R^vl" \ angle contained by them, of .<^FAC equal A^""^*"^ ^"'^"'^A to the same three parts of ^^ BAG, we ^ know, from Prop. 4, that all the parts of these two triangles are equal. N.B. — In writing out a proof of the equality of two triangles, all parts of the left-hand figure should stand on the left, and those of the right- hand figure on the right, of the sign of equality. Also, in written work clearness is gained by writing the name of the angle partly within the angle symbol; e.g. /ABC. The first part of Prop. 5 may be enunciated thus — If tivo sides of a tri- angle are equal, tioo of its angles must also he equal. A corollary is the statement of some geometrical fact not mentioned in the enunciation, but the truth of which may be inferred from the proof of the proposition. EXERCISES. 1. In the figure of I. 3 if DE be joined Z_ ADE = ^AED. 2. In the figure of I. 5 prove that Z.rCG = Z.FBG. 3. If, in the figure of I. 5, FG be joined, prove that the .:^FBG is equal in area to the ^AFCG 4. If, in the figure of I. 5, FC and BG meet at H, prove that -^s BFH, CGH are equal in area. 5. In the figure of I. 1, if the circles cut at C and D, prove that the angle CAD is equal to the angle CBD. 6. If the middle point of the base of an isosceles triangle be joined to the vertex, the figure is divided into two right-angled triangles. 7. ABC is an isosceles triangle. On the side of AB remote from C, any point P is taken and joined to C and B. Prove that angle PBC is greater than angle PCB. 8. AOB is a diameter of a circle whose centre is 0. C is a point on the circumference. Prove that the angle ACB is equal to the sum of the angles CAB and CBA. 9. The opposite angles of a rhombus are equal. 26 EUCLID, BOOK I. PROPOSITION VI. Theorem. If two angles of a triangle are equal to one another, the sides also which subtend the equal angles, are equal. Let ABC be a .i^ A having Z.ABC=^ACB. ThenshaU AB=AC. D^ For, if AB is unequal to AC, one must be the greater. If possible, suppose AB > AC. From AB cut off BD=AC I. 3. Join DC Post. 1. B Then in .^s DBC, ABC, DB=AC Constr. BC is com. ^DBC=z.ACB Hyp. /. .^DBC=^ABC 1.4. i.e, the part=the whole, which is absurd Ax. 9. Hence, AB cannot be unequal to AC, i.e. AB=AC. Wherefore, if two angles &c. Q.E.D. CoR. An equiang'^ ^^ must also be equilat^ NOTES. Prop. 6 is the converse of the first part of Prop. 5; i.e. the hypothesis of Prop. 5 forms the conclusion of Prop, 6, and vice versd,. These two propositions may be stated thus: — Prop. 5 (i). If a triangle is isosceles, Hyp. two of its angles must be equal Conclusion. Prop. 6. If a triangle has two angles equal, Hyp. it must be isosceles Conclusion. PROP. VI. 27 Prop. 6 is proved by an indirect method. In order to show that AB and AC are equal, the question, Can they possibly be unequal? is first con- sidered ; and this it is shown can only be the case if the area of the whole triangle ABC is equal to the area of its part the triangle DBC, which is contrary to axiom. Thus the truth of the proposition is evident. This method of proof is sometimes called a reductio ad ahsurdum EXERCISES. 1. In the figure of I. 5, if FC and BG meet at 0, prove that OB is equal toOC. 2. In the figure of I. 6, prove that angle DCB is less than angle DBC. 3. ABC is an isosceles triangle, with AB equal to AC. The bisectors of the angles ABC and ACB meet at O. Prove that CO=BO. 4. ABC is a triangle; BA is produced to D; AD is cut off equal to AC, and DC joined. Prove that angle BCD is greater than angle BDC. 5. If, in the figure of I. 6, any point be taken within the triangle ABC, prove that angle OBC is less than angle ACB. 6. If, in the figure of I. 6, a point P be taken on the side of AC remote from B, the angle PCB is greater than the angle PBC. 7. ABC is an isosceles triangle, having AB equal to AC; BO, CO, the bisectors of the angles ABC and ACB, meet at O, and AO is joined. Prove that the triangles ABO and ACO are equal in all respects. 8. If, in the figure of I. 6, CD be produced to E, and EB joined, the angle EBC is greater than the angle ECB. 9. If, in the figure of Ex. 8, EB and EC are produced to G and H respectively, prove that the angle GBC is less than angle HCB. 10. If, in the figure of I. 1, the circles cut at C and D, prove that CD bisects the angle A^B. ~- — ^^^^ Prove also that CD bisects AB. 11. ABODE is a regular pentagon; prove that the triangle ACD is isosceles. 12. Explain the words Oeometry, Propoeition, Problem, Theorem, Corollary, Enunciation, Hypothesis, Postulate, Axiom, Finite, Hexagonal, Oc- tagonal. 28 EUCLID, BOOK PROPOSITION VII. Theorem. Oil the same base and on the same side of it there cannot be two triangles which have their sides, terminated in one extremity of the base, equal to one another, and likewise those terminated in the other extremity. For, if it be possible, let the .^^^ ABC, DBG, standing on the same side of BC, have AB=DB and also AC=DC. Case 1. When the vertex of each .^^ is outside the other. Join AD Post. 1. "V^^ n Then, since AB=DB Hyp. /. z.BAD=z_BDA i.5(i). but z.BAD>z_CAD Ax. 9. /. ^BDA>^CAD but z.CDA>z.BDA Ax. 9. much more /. L CDA > L CAD. B ' C Again, since DC=AC Hyp. .-. z.CDA=^CAD 1.5. i.e. the L CDA is both = and > L CAD which is impossible. Case 2. When the vertex D of one ^^ is within the other. Join AD Post. 1. /g Produce B A to E, and BD to F Post. 2. ^ Then, since AB=DB Hyp. /. ^EAD=^FDA i.5(ii). / 1^^^ but z.EAD>z.CAD Ax. 9. ' ^' .'. z.FDA>z.CAD but Z.CDA>z.FDA Ax. 9. much more .*. z.CDA> Z.CAD. B C Again, since DC=AC Hyp. .-. ^CDA=z.CAD 1.5. i.e. the L CDA is both = and > L CAD which is impossible. PROP. VII. Case 3. When the vertex of oiie .^^ A is on a side of the other. >/^\d This case needs no proof. B C Wherefore, on the same hose &c. Q.E.D. NOTE. This negative theorem is only required by Euclid for the 8th Proposition. Some writers have used another proof of Prop. 8, omitting the 7th Propo- sition altogether. See note on Prop. 8. EXERCISES. 1. Why does Case 3 "need no proof"? 2. Is it possible for two triangles, standing on the same side of the same base, to have their sides, which terminate in one extremity of the base, equal, and their sides which terminate in the other extremity, unequal? 3. On the same base, and on the same side of it, there can be but one equilateral triangle. 4. Two circles cannot cut in more than two points. 5. What axiom is implied in the proof of the 7th Proposition? 6. On the same base, but on opposite sides of it, are two acute-angled triangles, having their sides terminated in one extremity of the base equal, and, likewise those terminated in the other extremity; prove, by joining their vertices and using Proposition 5, that their vertical angles are equal. 7. ABC, DBC are two isosceles triangles standing on the same side of the same base BC; prove that the angle ABD is equal to the angle ACD. 8. ABC, DBC are two isosceles triangles standing on opposite sides of the same base BC; prove that the angle ABD is equal to the angle ACD. 9. AOB is an angle. OC is drawn within the angle AOB and OC is made equal to OB. If AB, AC, BC are joined prove that angle ACB is greater than angle ABC. 30 EUCLID, BOOK I. PKOPOSITION VIII. Theorem. If two triangles have two sides of one equal to two sides of the other, each to each, and have, likewise, their bases equal; the angle contained hy the two sides of the one is equal to the angle contained by the tivo sides, equal to them, of the other. a D ^^ Let ABC, DEF be two ^^s, having AB=DE, AC=DF, and BC=EF. Then shaU L BAC= L EDP. b C E F For, if ^^ABC be applied to ^^DEF with pt. B on E, and BC along EF, the pt. C must faU on F, V BC=EF Hyp. And if BA, AC did not fall on ED, DF, but had a diiferent position such as EG, GF, then DEF, GEF would be two ^^s standing on the same side of EF and having DE=GE, and DF=GF, which is impossible... J. 7. Hence BA, AC must fall on ED, DF, and Z.BAC must coincide with Z.EDF /.^BAC=iLEDF Ax. 8. Wherefore, if two triangles &c. Q.E.D. NOTES. Since the triangles have been shown to coincide it follows that they are equal in all respects. The 8th Proposition is a converse of the 4th, and, as is the case with most converse propositions, it is proved indirectly. It should be noticed that, when the hypothesis of a proposition contains more than one condition, there will be more than one converse. The following direct proof of Prop. 8 is independent of Prop. 7. Apply ^:^ABC to DEF so that their bases j^ coincide, but with their vertices on opposite sides of EF. Let GEF be the new position of .^ABC. Join DG. Then since ED=EG Hyp. /. Z.EDG=Z.EGD i.5. Again, since FD=rG Hyp. /. z.rDG=^rGD 1.5. Hence, the whole L EDF= whole L EGF ... Ax. 2. Q.E.D. i^F PROP. VIII. 31 The following theorem illustrates the application of Prop. 8 to examples. Theorem. The straight line which joins the vertex to the middle point of the base of an isosceles triangle, is at right angles to the base. Let ABC be an isos. ^r^ and D the mid. pt. of base BC; then shall AD be at rt, Z_ s to BC. A For, in ^^s ABD, ACD {BD=CD Hyp. AD is com. AB=AC Hyp /. Z_ADB=Z.ADC 1.8. g ^ ^ and these being equal adjacent Z-S, are rt. Z_s..., Def. lO. /. AD is at rt. Z_s to BC. Q.E.D. EXERCISES. 1. In the figure of 1. 1, if the circles cut at C and D, prove that angle ACB is equal to angle ADB. 2. In the figure of Euc. I. 5, if BG and CF cut at H, prove that — (i) Z.AHB=^AHC (ii) ^BHF=Z.CHG. 3. ABC is an isosceles triangle; D is the middle point of the base BC; prove that AD bisects the vertical angle BAC. 4. A diagonal of a rhombus bisects the angles through which it passes. 5. Two isosceles triangles stand on the same side of the same base ; prove that the line joining their vertices, when produced, meets the base at right angles. 6. Two circles, whose centres are O and Q, cut at P and E,, show that the angle OPQ is equal to the angle ORQ. 7. Show, by the method of superposition, that a square is divided by its diagonal into two triangles of equal area. 8. A rhomboid is bisected by its diagonal. 9. Two isosceles triangles stand on opposite sides of the same base ; show that the line joining their vertices bisects their vertical angles. 10. Every oblong is bisected by its diameter. 11. ABCD is an oblong whose diagonals cut at O; prove that triangle OAD is equal in area to triangle OBC. 32 EUCLID, BOOK I. PKOPOSITION IX. Problem. To bisect a given rectilineal migle. Let ABC be the given rect^ L . It is req'^ to bisect it. In AB take any pt. D. Aj From BC cut off BE=BD...I. 3. Join DE Post 1. j^/ On the side of DE remote from B desc. an equilat. .^DEF LL JoinBF Post 1 B E Then shall BF bisect lABO. In .-As DBF, EBF, DB=EB Constr. BF is com. .DF=EF (equilat'^::^) Constr. A z.DBF=Z.EBF is. Wherefore, the given rectilineal angle &c. Q.E.i^ EXERCISES. 1. Why is the equilateral triangle constructed on the side of DE remote from B? 2. Divide a given rectilineal angle into four equal parts. 3. In the figure of Prop. 9, prove that angle BDF is equal to BEF. 4. Prove the following construction for bisecting a given angle POQ. With centre O and any radius describe a circle cutting OP in R, and OQ in S. With centre R and any radius greater than half RS, and with centre S and the same radius, describe circles cutting at T. Join OT. 5. ABC is an isosceles triangle; the bisectors of the base angles meet at 0, and O is joined to the vertex A. Prove that OA bisects the vertical angle BAC. 6. The bisectors of the three angles of an equilateral triangle meet at a point. — [Note. Draw tioo bisectors, join their pt. of intersection to the remain- ing L , and then prove that this line bisects that Z .] PROPS. IX., X. 33 PROPOSITION X. Problem. To bisect a given finite straight liiie. Let AB be the given finite st. line. It IS req^ to bisect it. A D B On AB desc. an equilat..=^CAB I.i. Bisect Z.ACB by CD meeting A B at D 1.9. Then shall AB be bisected at D. In ^s CAD, CBD, CA=CB Constr. CD is com. iLACD=Z.BCD Constr. AD=BD 1.4. Wherefore, the given st. line &c. Q.E.F. EXERCISES. 1. What is meant by & finite straight line? 2. Prove the following construction for bisecting a given line PQ: With centre P, and any radius greater than half PQ; and with centre Q and the same radius; describe arcs cutting at R and S. Join SR meeting PQ at T. 3. Show how to divide a given line into eight equal parts. 4. Prove that the common chord of two equal circles which cut one another bisects the line which joins their centres. 5. If, in the figure of Proposition 10, CB be bisected at E, and AE be joined, meeting CD at O; prove that OD is equal to OE. 6. In the figure of I. 9, if BF and DE cut at H, prove that DE is bisected at H, and that FH is at right angles to DE. 7. Prove that the common chord of two circles which cut is at right angles to the line joining their centres. (310) C 34 EUCLID, BOOK I. PEOPOSITION XI. Problem. To draw a straight line at right angles to a given straight lirie^ from a given point in the same. Let AB be the st. line, and C the given pt. in it. It is req^ to draw from C a st. line at rt. Ls to AB. F B In AC take any pt. D. From CB cut off CE=CD L 3. On DE desc. an equilati ^^ DEF L i. Join CF Post. 1. Then CP shall be at rt. Lsto AB. In .^s FDC, FEC, DC=EC Constr. CF is com. DF=EF Constr. .-. z.DCF=z.ECF L8. and these, being equal adjacent L s, are rt. Z. s Def. lo. Wherefore, from the given point C &c. Q.E.F. EXERCISES. 1. Prove the following construction for this proposition: — With centre D and any radius greater than DC, and with centre E and the same radius, describe circles cutting at G. Join CG. 2. Find a point in a given straight line, of unlimited length, which shall be equidistant from two given external points. Draw figures for the various cases that may occur. For what special positions of the given points with respect to the given line is there — (i) no solution of the problem; (ii) an infinite number of solutions? 3. Find the locus of a point which moves in a plane so as always to be equidistant from two given fixed points in the plane. 4. Show, by a reductio ad absurdiim, that, on the same side of the line AB, but one straight line at right angles to AB can be drawn from the point C. PROPS. XI., XII. Of PKOPOSITION XII. Problem. To draw a straight line 'perpendicular to a given straight unlimited length, from a given point without it. Let AB be the given st. line of unlimited length, and C the given pt. without it. It is reef to draw from C a st. line ± to AB. Take any pt. D on the other side of AB. With cent. C and rad. CD desc. a EDF, cutting AB at E and F Post. 3. Bisect EF at G i. lo. Join CG Post. 1. Then CG shall be .l to AB. Join CE, CF, Post. 1. Then in -^s CEG, CFG |'EG=FG Constr. */ ^ CG is com. lCE=CF (radii) Constr. ,',LCGE==:LCGF 1.8. and these, being equal adj. Ls, are rt. Z.s...Def. lo. /. CG is i. to AB Del 10. Wherefore, /rowi the given point C &c. EXERCISES. 1. Why is the given line "of unlimited length"? 2. Why is the point D taken "on the other side " of AB? 3. What distinction does Euclid make between a line at right angles, and a perpendicular to another line ? 4. Construct an angle which shall be double of a given angle. 5. Prove that the perpendiculars, let fall from any point in the bisector of an angle upon the lines which contain the angle, are equal.* 6. The perpendiculars drawn from the ends of the base of an isosceles triangle to the opposite sides are equal.* 7. Find a point within an equilateral triangle which shall be equally distant from each of the angular points of the triangle. 8. P and Q are fixed points on opposite sides of a fixed straight line PS. Find a point T in RS such that the angle PTR may be equal to the angle QTR. In what case is this impossible? * Ifote.—Vae the method of Ex. 6, p. 32, i.e. draw one i.. 36 EUCLID, BOOK I. PEOPOSITION XIII. Theorem. The angles which one straight line makes with another straight line, on one side of it, a7'e either two right angles, or are together equal to tioo right angles. Let the st. Hue AB meet CD at B. Then shall either (i) z.s CBA, ABD be two rt. z.s, or (ii) L s CBA, ABD together=two rt. L s. (i) If z.CBA=^ABD, each is a rt. z_ Def lO. and /. 2_ s CBA, ABD are two rt. L s. (ii) If L CBA is not= z. ABD, from B draw BE at rt. z.s to CD....I. ii. Then, since L CBE= L s CBA, ABE, add L EBD to each, C B D /. Z-sCBE, EBD = z.sCBA, ABE, EBD ax. 2. Again, since z. ABD= ^s ABE, EBD, add L CBA to each, /. Z-sCBA, ABD=z.sCBA, ABE, EBD Ax. 2. Hence, z-s CBA, ABD= AS CBE, EBD Ax. 1. But Z.S CBE, EBD are two rt. z.s Constr. /. Z.S CBA, ABD are together=two rt. z.s. Wherefore, the angles which &c. Q.E.D EXERCISES. 1. State the axiom assumed in the statement "Z_CBE=Z_s CBA, ABE," 2. AOB and COD are right angles, C lying within the angle AOB ; prove that the angle AOC is equal to the angle BOD. 3. If, in the figure of Prop. 1, AB be produced both ways and meet the circle BCD in P, and the circle ACE in Q, prove that the angle CAP is equal to the angle CBQ. 4. In an isosceles triangle, the angles at the base together with the angles on the other side the base, are equal to four right angles. 5. OA, OB, OC, OD are radii of a circle ABCD. OB is at right angles to OA, and OC to OD. Prove that AC is equal to BD. 6. The bisectors of adjacent angles are at right angles. PROPS. XIII., XIV. 37 PEOPOSITION XIV. Theorem. If^ at a point in a straight line, two other straight lines, on opposite sides of it, make the adjacent angles together equal to two right angles; these two straight lines must he in one and the same straight line. At the pt. B, iu the st. line A) AB, let CB, BD, on opp. sides of AB, make z.s CBA, ABD together=two rt. Z.s. / E. Then OB shall be in the ^ same st. line with BD. c B D For, if not, if possible produce CB in some other direction BE. Then, since AB meets st. line CBE at B /. Z.S CBA, ABE=two rt. z.s 1.13. But ilsCBA, ABD=two rt. z.s Hyp. /. Z.S CBA, ABE= ^s CBA, ABD Ax. 1. Take away the com. L CBA, .*. rems l ABE=remg ^ ABD Ax.3. or, the part = the whole, which is absurd Ax. 9. Hence, CB is in the same st. line with BD. Wherefore, if at a point &c. Q.E.D. EXERCISES. 1. Of what proposition is the 14th the converse? - 2. What method of proof is adopted in Proposition 14? 3. In Prop. 14 Euclid first makes use of Axiom 11. Point out where this occurs. 4. If, in the figure of Prop. 11, a second line, CG, be drawn from C at right angles to AB, but on the other side of it; then will CF and CG be in one and the same straight line. 5. If the angles on the other side of the base of a triangle, whose sides have been produced, are equal, prove that the triangle is isosceles. Of what theorem is this the converse? 6. Two squares have a corner-point common; if two of their sides are in one straight line, then two other sides will also be in one straight line. 38 EUCLID, BOOK I. PEOPOSITION XV. Theorem. If two straight lines cut one another, the opposite vertical angles are equal. Let the st. lines AB and CD cut at E. ^^-^ ^^^ Then shaU ^AED= ^CBB, " ^' and ^ABC=-lDEB. ^y^ ^ Since AE meets CD at E, /. Z.S AEC, AED=two rt. Z-s Lis. Since CE meets AB at E, /. ^s AEC, CEB=two rt. Z-s 1. 13. Hence, Z.s AEC, AED=z.s AEC, CEB Ax. i. and Ai. ii. Take away the com. L AEC, /. rems z. AED=rem8 Z.CEB Ax.3. In the same way it may be shown that L AEC= L DEB. Wherefore, if two straight lines &c. Q.E.D. Cor. 1. If two straight lines cut, the four angles formed are together equal to four right angles. CoR. 2. If any number of straight lines meet at a point, the sum of all the angles formed is equal to four right angles. EXERCISES. 1. Write out in full the proof of the second part of this proposition, that angle AEC is equal to angle DEB. 2. AB and CD bisect each other at 0; prove that the triangles BOC and AOD are equal in all respects. 3. AOB, COD are diameters of a circle; prove that AC is equal to BD. 4. Could a perfect pavement be formed entirely of equal regular hexagons whose angles are each of 120°? 5. Any point is taken within a rectangle, and is joined to the four angular points. Prove that all the angles of all the triangles so formed are together equal to eight right angles. 6. The bisectors of opposite vertical angles are in the same straight line. PROPS. XV., XVI. 39 PROPOSITION XVI. Theorem. If one side of a triangle be produced, the exterior angle is greater than either of the opposite interior angles. Let ABC be a ^^ having one of its sides, BC, produced to D. Then shall (i) L ACD > L BAG, and (ii) z. ACD > ^ ABC. Bisect AC at E i. lo. Join BE Post. 1. Produce BE to F Post. 2. CutoffEF=BE 1.3. ^ '^ \ ^ Join CF Post. 1. \ Then, in ^s ABE, ECF ^"^ AE=EC Constr. BE=EF Constr. Z.AEB=z.CEF 1.15. /. z.BAE=z.ECF 1.4. But z_ECD>z.ECF Ax.a .*. ^ECD>z.BAE i.e. Z_ACD>Z.BAC. In the same way, if BC be bisected, and AC produced to G, it may be shown that Z.BCG {i.e. Z. ACD) > ^ ABC. Wherefore, if one side of a triangle &c. Q.E.D. EXERCISES. 1. Write out in full the proof of the second part of this proposition. 2. If, in the figure of Prop. 16, AF be joined, then AF is equal to BC. 3. In the figure of I. 7, case 2, prove that angle ADB is greater than ACB. 4. Prove that the angles at the base of an isosceles triangle are together less than two right angles. 5. A triangle can have but one right angle. 6. In the figure of Prop. 5, if FC and BG meet in H, prove that angle FHG is greater than BAG. 7. In the figure of I. 16 prove that triangles ABC FBC are equal in area. 40 EUCLID, BOOK I. PROPOSITION XVII. Theorem. Any two angles of a triangle are together less than two right angles. Let ABC be a .:^. Then shall any two of its z. s be together less than two rt. Z.S. A B C D Produce BC to D Post. 2. Thenjexf^ ^ACD>opp. inf" Z.ABC 1. 16. Add z_ ACB to each /. ^s ACD, ACB> ^s ABC, ACB Ax.4. but z_s ACD,ACB=two rt. Ls 1. 13. ,-. Z-sABC, ACB side AC. Then shaU ^ACB>aABC. From AB cut off AD = AC 1.3. Join DC Post. 1. Then, since AD=AC Constr. /. z_ADC=z.ACD T.5. Butexf z.ADC>opp. inf^ Z.DBCof .^CBD.... I.i6. /. also Z-ACD>^DBC but z.ACB>Z.ACD Ax. 9. much more /. ^ ACB> ^DBC (or ABC). Wherefore, the greater side &c. Q.E.D. NOTE. In this proposition the base and vertical angle are not considered; the base may, of course, be the (jreatest of the three sides, as it is in the above figure. If the two sides are equal, we know from Prop. 5 that they subtend equal angles. This proposition then is an extension of the 5th, and shows that, if the two sides are not equal, the greater of the two subtends a greater angle than the other subtends. EXERCISES. 1. What is the hypothesis in this proposition? 2. In the figure of Prop. 5, show that the angle ABG is greater than the angle AGE. 3. In a scalene triangle the greatest side subtends the greatest of the three angles. 4. Enunciate the converse of the following proposition: — " If a triangle have two of its sides unequal, the greater of these two sides subtends an angle which is greater than that subtended by the other side." 5. ABCD is a quadrilateral. AB is equal to AD, but BC is less than DC. Prove that angle ABC is greater than angle ADC. 42 EUCLID, BOOK I. PEOPOS'ITION XIX. Theorem. J The greater angle of every triangle is opposite to the greater side. Let ABC be a ^^ having 2_ ABC> Z. ACB. Then shaU AOAB. For, if AC be not>AB, then must either (i) AC=AB, or (ii) AC the third side. Produce BA to D Post. 2. CutoffAD=AC 1.3. Join DC Post.i. A) Then, since AC = AD .Constr. Ay^ j ,\ z.ACD=z.ADC 1.5. y/\ j But ^BCD>z.ACD Ax.9. y/^ \j :. also z.BCD>2LADC ^ ^^ B C Hence BD>BC 1.19. i.e. BA, AC>BC Constr. In the same way it may be shown that AC, CB>BA, and CB, BA>AC. Wherefore, any two sides of a triangle &c. Q.E.D. EXERCISES. 1. Write out in full the proof that AC, CB are greater than BA. 2. Prove, by a similar construction to that of the proposition, that the difference of any two sides of a triangle is less than the third side. 3. Any three sides of a quadrilateral are greater than the fourth side. 4. The sum of the four sides of a quadrilateral is greater than the sum of its two diagonals. 5. The sum of the lines joining any point within a triangle to its angles is greater than half the sum of its sides. 6. In the figure of Prop. 6, prove that the perimeter of the triangle ABC / is greater than the perimeter of triangle DBC. 7. In the figure of I. 16, prove that BC and CF are greater than twice BE. 8. Points are taken on the circumference of a circle and joined in order, forming a polygon. Show that the perimeter of the polygon increases as the number of sides is increased. 9. Two sides of any triangle are together greater than twice the straight line joining the vertex to the middle point of the base. 44 EUCLID, BOOK I. PKOPOSITION XXI. Theorem. //, from the ends of a side of a triangle, two straight lines he drawn to a point within the triangle; these two straight lines are together less than the other tvjo sides of the triangle, but (mitain a greater angle. Let ABC be a ^^, and, from the ends of the base BC, let BD, CD be drawn to a pt. D within the ^::\. Then shaU (i) BD, DCBE 1.20. ^^^ Add EC to each, /. BA, AOBE, EC. ^ And, in .^EDC, DE, EODC 1.20 Add BD to each, /. BE, EOBD, DC. Much more ,'. BA, AC>BD, DC. (ii) Again, in .::^EDC, exf Z-BDC>opp. int'* /.DEC. I 16 And, in .^ABE, exf LDEOopp. inf ABAC 1. 16 Much more /. z.BDC> .iBAC. Wherefore, if from the ends of a side &c. Q.E.D. EXERCISES. 1. State an axiom which is assumed in this proposition. 2. If any point be taken within a quadrilateral and joined to two opposite angular points, the new quadrilateral thus formed will have a less perimeter than the other. 3. If a point O within a triangle ABC be joined to the angles, then OA, OB, OC are together less than the perimeter of the triangle. PROPS. XXL, XXII. 45 PROPOSITION XXII. Problem. To make a triangle whose sides shall he equal to three given straight lines, any two of which are greater than the third. Let A, Bj and C be the given st. lines, any two being > the third. It IS req^ to make a ^^ with sides respectively =: A, B, and C. M D\ F G H E Take a st. line DE terminated at D, but unlimited towards E, From DE cut ofF DF=A, rG=B, GH=C i. 's. With cent. F and rad. FD desc. a DLK Post. 3. With C3nt. G and rad. GH desc. a HMK, Post. 3. cutting the other in K. Join KF, KG Post, l Then shaU ^^ KFG have its sides=A, B, and O. Since F is cent, of DLK /. FK=FD Def.l5. but FD=A Constr. .'. FK=A Ax.L Since G is cent, of HMK .'. GK=GH Def.l5. but GH=C Constr. /. GK=C Ax.L Also FG=B Constr. Wherefore, a .^:::\KFG has been made &c. Q.E.F. NOTE. In Practical Geometry, when unrestricted use of compasses can be made, a triangle with sides of given lengths would be constructed thus : — If AD, B, and C "" '^ are the given lengths, extend the compasses till the distance between their feet is the length B, and then with cent. A describe an ^ ^ arc. In the same way with cent. D, but radius C, describe another arc, cutting the former at E. Rule lines EA and ED, 46 EUCLID, BOOK I. PEOPOSITION XXIII. Problem. At a given poiiit in a given straight line to make an angle equal to a given angle. Let AB be the given st. line, A the given pt. in it, and CDE the given L. It is re^ to make at A an L=^ L CDE. In DC and DE take any pts. C and E. Join CE Post. 1. On AB make a ^^A AFG with its side rA=CD AG=DE1- I. GF=EC Then shall lFAG = z.CDE. In ^s FAG, CDE rA=CD^ AG=DE[ Consti. GF=ECJ /. ^FAG=Z.CDE 1.8. Wherefore, at the given pt. A in the given st. line &c. Q.E.F. EXERCISES. 1. Prove that two triangles, satisfying the conditions of the problem, can be obtained with the construction of Prop. 22. 2. What previous problem is a special case of Prop. 22? 3. Show, by diagrams, that in Prop. 22 the restriction " any two of which are greater than the third " is necessary. 4. If, in the figure of Prop. 22, KD and KH be joined, the perimeter of KDH is more than double the sum of the lines A, B and C. 5. Construct a rhombus having given the length of a diagonal, and one of the angles through which it passes. 6. If one angle of a triangle is equal to the sum of the other two angles, the triangle can be divided into two isosceles triangles. 7. Construct a quadrilateral whose sides shall be respectively equal to those of a given quadrilateral. 8. Construct a triangle, having given the base, one of the angles at the base, and a line equal to the sum of its two sides. PROP. XXIII. 47 EXAMPLES. I. To construct a rectilineal figure whose sides and angles sliall he equal respectively to those of a given rectilineal figure. Let ABODE be the given figure. Join AC, AD, dividing it into -^As. Make a ..^YGrB. with its sides equal to those of .^ ABC (FG=AB &c.)...i. 22. On FH make a ^r^FKH with its sides equal to those of ^^ADC (FK=AD and HK=CD) 1.22. On FK make a ^AFMK with its sides equal to those of ^^AED (FM=AE, &c.) 1.22. Then fig. FGHKMhas (by constr.) its sides^to those of the given figure. Also, by Euc. I. 8, the three pairs of ^A s have their Z.s equal, each to each, since their sides are equal. ,*, the sum of the three Z.s at F=the sum of the three Z.s at A, and so on. Hence the figure is also equiangular to ABODE. Q.E.F. II. To construct a triangle, having given the base, one of the angles at the base, and the difference of the sides. Let A be the given L, BO the given base, A E^ and D the given difference of sides. ^ D G. At B, in BO, make L OBE=A. . .1. 23. From BE cut off BF=D...I.3. Join FO. F^ AtOmake Z.FOG=Z_OFE 123. OG meeting BE in G. ^ q Then GBC shaU be the ^^ reqd. For, since ^GFO=Z_GOF Constr. /. GF=GO 1.6. /, the difference of GB and GO is BF, and BF=D. Also L GBO=A Constr. Hence, ^^^GBO has one of its base Z.s=A, stands upon the given base BO, and has the difference of its sides=D. Q.E.F. 48 EUCLID, BOOK I. PEOPOSITION XXIV. Theorem. If tivo triaTigles have two sides of the one equal to two sides of the other J each to each, hut the angle contained hy the two sides of the one greater than the angle contained hy the two sides of the other; the hase of that which has the greater angle is greater than the hase of the other. Let ABC, DEF be two .^s, having AB=DE, a D AC=DF, but Z.BAOZ.EDF. Then shaU BOEP. Of the two sides DE, DF, let DE be the one which is not > the other. At pt. D in DE, and on the same side of it as F, make the z_EDG= Z.BAC i. 28. Cut off DG=DF or AC is. Join EG, GF Post. i. Then, in ^^s ABC, DEG i AB=DE Hyp. V -j AC=DG Constr. [z.BAC=Z-EDG Constr. .-. BC=EG 1.4. Again, since DG=DF Constr. /. z.DGF=^DFG 1.5. But ^DGF>^EGF Ax. 9. .-.also ^DFO^EGF. But 2LEFG>^DFG Ax. 9. much more /. z.EFG> Z.EGF. Hence EG>EF 1. 19. But BC=EG Proved above. /. BOEF. Wherefore, if two triangles &c. Q.E.D. NOTE. The shorter of the two sides DE, DF is selected in the construction (if the sides are not equal) in order to ensure that F shall lie on the side of EG remote from D. Otherwise three cases might occur:— Case i. as above; Case ii. when F fell on the same side as D (in which case produce DF and DG in order to prove EG > EF by the method of Prop. 7, Case 2) ; and Case iii. when F fell on EG. PROPS. XXIV., XXV. 49 PEOPOSITION XXV. Theorem. If two trioAigles have two sides of one equal to two sides of the othei\ 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 is greater than the angle contained by the sides of the other. Let ABC, DEF be two .^s, having AB=DE, AC=DF, but BOEF. Then shaU ^BAC>z.BDP. For, if z_BAC is not> L EDF, then either (i) 2LBAC= Z.EDF, or (ii)z.BACDE. From AB cut off BG=DE 1.3. Join GC Post. 1. Then in .^a GBC, DEF, GB=DE Constr. BC=EF Hyp. lGBC= lBEF Hm .-. z.GCB=2LDFE 1.4. But LACB=LT)FE Hyp. /. ^GCB=z_ACB Ax.i. or, the part = the whole, which is absurd Ax. 9. ,*, AB cannot be unequal to DE, i.e. AB=DE. Hence, in .^s ABC, DEF, AB=DE proved above. BC=EF Hyp. AABC=Z_DEF Hyp. AC=DF ) ^bac=^edfJ '■*■ PROP. XXVI. 51 Case II. Let ^^s ABC, DEF have z.ABC=Z.DEF, Z.ACB=^DFE, AB=:DE. Then sMll BC=BF, AC=DF, ^BAC=iLBDP. For, if BC is not=EF, one must be the greater. If possible, suppose BC > EF. From BC cut off BG=EF 1.3. Join AG Post. 1. Then in .^s ABG, DEF, AB: BG: ^ABG: /. z.AGB= But z_ACB=^DFE Hyp. /. ^AGB=z.ACB Ax.i. or, ext^ L of .-rrX AGC=an opp. int'^ L , which is impossible I. ic. /. BC cannot be unequal to EF, i.e. BC=EF. Hence, in .^s, ABC, DEF, AB=DE Hyp. BC = EF proved above. Z.ABC=Z_DEF Hyp. ( AC=DF ) **• (z.bac=z.edfI Wherefore, if two triangles &c. Q.E.D. \ :DE Hyp. :EF Constr. :Z.DEF Hyp. :Z.DFE 1.4. .1.4. NOTE. It follows from I. 4 that the areas of the triangles are equal. Hence Prop. 26 may be stated shortly thus: — If tivo angles and a side of one triangle are equal to two angles and the corresponding side of another, the triangles are equal in all respects. 52 EUCLID, BOOK I. PEOPOSITION XXVI. Theorem. NOTES. Prop. 26 completes one section of Book I. Among the most important results of this section are those of Props. 4, 8, and 26, which deal with four of the six cases which occur in the general question: — If two triangles have three parts of the one known to he equal to three -parts of the other, each to each, must the triangles be equal in all respects? Case i. ^Two sides and the in- Prop. IV. J eluded angle. ( -:;:::\s equal in all respects. Case ii. r Three sides. Prop. VIII. s -=^s equal in all respects i. (See note to Prop. 8.) Case hi. /Two angles and the Prop. XXVI. (i) J adjacent side. (. ^^ s equal in all respects. Case iv. I^Two angles and an Prop. XXVI. (ii) -^ opposite side. ( -^:::\ s equal in all respects. Case v. Exception. Three angles. - opp. inf^ L EFG. 1. 16. but z.AEF=^EFG Hyp. i.e. Z. AEF is both>, and=, Z.EFG, which is absurd. Hence AB and CD cannot meet towards B and D. In the same way it may be shown that they cannot meet when prod*^ towards A and C. /. AB is II to CD Def. 35. Wherefore, if a straight line &c. Q.E.D. EXERCISES. 1. Show that AB and CD cannot meet towards A and C. 2. What is the hypothesis of this proposition? 3. Enunciate the converse of Prop. 27. 4. In the figure of I. 16, if AF be joined, prove that AF is parallel to BC. Prove also that AB is parallel to FC. 5. If, in the figure of Prop. 1, the circles cut at C and F ; prove that AC is parallel to BF. 6. Opposite sides of an oblong are parallel. 7. Opposite sides of a rhombus are parallel. 8. If two straight lines bisect one another, the figure formed by joining their extremities is a parallelogram. 9. Every rhomboid is a parallelogram. PROPS. XXVII., XXVIII. 55 PEOPOSITION XXVIII. Theorem. If a straight liiie, falling on two other straight lines, make the ex- terior angle equal to the interior and opposite angle upon the same side of the line; or, make the two interior angles on the same side together equal to two right angles; the two straight lines shall be parallel. Part I. Let st. line EF, falling on AB and CD, make exf^ L EGB=iirt^ opp. L GHD. Then shaU AB be ll to CD. /^ For, since ^EGB=^GHD Hyp. A ^ B and z.EGB=^AGH 1. 15. / /. ^AGH=z.GHD Ax.i. c -fr D and these being alt. z. s, / /.AB is II to CD I. 27. f/ Part II. Let EF, falling on AB and CD, make inf^ Z-S BGH, GHD togetlier=two rt. /.s. Then shall AB be ll to CD. For, since z. s BGH, GHD=two rt. z.s Hyp. and Z-s BGH, AGH=two rt. z_s 1.13. /. ^sBGH, AGH=z.sBGH, GHD Ax. 1. Take away the com. L BGH, /. rems z.AGH=rem« Z_GHD Ax. 3. and these being alt. L s, /. ABis II to CD 1.27. Wherefore, if a straight line &.C. Q.E.D. EXERCISES. 1. What is meant by the words " on the same side " in the second part of the enunciation? 2. Prove Part I., assuming ext^ angle EGA equal to int^ angle GHC. 3. Prove Part II., assuming the two interior angles AGH and GHC together equal to two right angles. 4. Mention two other pairs of exterior and interior angles to be found in the figure, which have not been mentioned before. 5. Enunciate the converse of each of the two theorems contained in the enunciation of Prop. 28. 6. Prove that the opposite sides of a square are parallel. 7. Prove that straight lines at right angles to the same straight line are parallel to one another. 56 EUCLID, BOOK I. PROPOSITION XXIX. Theorem. If a straight line fall on two 'parallel straight lines, it makes the alternate angles equal; an exterior angle equal to the interior and opposite angle upon the same side; and two interior aiigles on the same side together equal to two right angles. Let AB and CD be || s, and let st. line EGHF fall on them. Then shall (i) l AaH=alt. L GHD (ii) extr L BGB=inti- opp. L GHD (iii) two intr Z-S BGH, GHD=two rt. z_s. C Tfl D F (i) For, if L AGH is not= L GHD, one must be the greater. If possible, suppose L AGH > L GHD Add Z-BGHtoeach, /. Z.S AGH, BGH>^sBGH, GHD Ax. 4. But Z.S AGH, BGH=two rt. ^s 1.13. /. L s BGH, GHD < two rt. L s. /. AB, CD will meet, if prod** towards B and D Ax. 12. which is impossible, for they are || , Hyp. Hence, z_ AGH cannot be unequal to Z.GHD i.e. iLAGH=z.GHD. (ii) Again, since L AGH= L GHD proved above. and z.AGH=z.EGB 1.15. /. z.EGB=^GHD Ax.i. (iii) Also, since Z_ EGB= Z. GHD provedabove. . Add L BGH to each /. ^sEGB, BGH=z.sBGH,GHD ax. 2. but z_s EGB, BGH=two rt. Z.S 1.13. /. Z-sBGH, GHD=twort. z.a Ax. 1. Wherefore, if a straight line &c. Q.E.D. PROP. XXIX. 57 NOTES. Prop. 29 contains three converse propositions — those of 27, 28 part i, and 28 part ii. It is in this proposition that Euclid first makes use of Axiom 12. Euclid's definition of parallels being a negative one, it became necessary to assume some positive fact relating to them before any of the properties of parallels could be proved. But the nature of the subject is such that there are no ^^ common notions" respecting parallels. Euclid, therefore, requested that the theorem, known as Axiom 12 (and which he probably placed among the postulates) should be taken for granted. He proved, however, its converse. (See note on Prop. 17.) Many attempts have been made to improve on Euclid's treatment of this part of Elementary Geometry. But, as the fundamental difficulty can never be entirely avoided, it may be doubted whether a little gain in clearness is of such value as to justify any tampering with Euclid's text ; especially as a change in either definition or axiom necessitates changes, in some cases considerable, in the propositions. The following is generally regarded as the best of the various substitutes for Euclid's axiom which have been suggested. "Two straight lines which cut cannot both be parallel to the same straight tine." EXERCISES. 1. Prove, by a reductio ad absurdum that, with the hypothesis and figure of Prop. 29, angle BGH is equal to angle GHC. 2. If two straight lines which meet are both parallel to the same straight line, these two straight lines are in one and the same straight line. 3. A straight line drawn parallel to the base of an isosceles triangle makes equal angles with the sides. 4. The angle between two straight lines is equal to the angle between two others which are parallel to them respectively. 5. If a perpendicular be drawn to each of two parallels, these perpendiculars are themselves parallel. 6. A is a point equidistant from two parallel straight lines; show that the portion of any line through A, which is intercepted by the parallels, is bisected at A. Also, if two lines be drawn through A, show that the triangles formed by them with the parallels are equal in all respects. 7. The diagonals of a parallelogram bisect each other. 8. If a straight line DE be drawn parallel to the base BC of an isosceles triangle ABC, the trapezoid DBCE will have each pair of its opposite angles together equal to two right angles. 58 EUCLID, BOOK I PEOPOSITION XXX. Theorem. Straight lines that are parallel to the same straight line are parallel to one another. Let St. lines AB, CD be each || to EF. Then shaU AB be || to CD. Draw GHK cutting the lines in G, H, A j B and K. Then,since ABis |1 to EF Hyp. ^AGH=alt.Z.GHF 1.29. ^ /H ^ and, since EF is || to CD Hyp. /. ext'^z.GHF=int'^Z.HKD 1.29. c / D Hence ^AGH=^HKD Ax. 1. '^ and these being alt. L s, .-. ABis II to CD 1.27. Wherefore, straight liJies that are &c. Q.E.D. PEOPOSITION XXXI. Problem. 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 st. line. It is req^ to draw, through A, a st. line || to BC. In BC take any pt. D. Join AD Post. 1. A At A, in AD, make L DAE ^ = L ADC, but on the opp. side of AD L23. Produce EA to F Post. 2. B ^ C Then shall BF be H to BC. For, since AD, meeting two st. lines, makes L EAD=alt, L ADC.Constr. /. EFis II toBC L2r. Wherefore, through the given point &c. Q.E.F. EXERCISES. 1. Through a given point draw a straight line which shall make a given angle with a given straight line. 2. Through three given points draw three straight lines so as to form a triangle two of whose angles shall be equal to given angles. PROP. XXX., XXXI., XXXII. 59 PEOPOSITION XXXII. Theorem. If a side of a triangle he produced, the exterior angle is equal to the tivo opposite interior angles; and the three angles of every triangle are together equal to two right angles. Let ABC be a ^^ with BC produced to D. Then shall (i) extr z.ACD=^s CAB, ABO. (ii) L^ CAB, ABC, BCA=two rt. ^Ls. B C D Through C draw CE II to AB I. 31. (i) Then since CE is || to AB :, Z-ECA=alt. /.CAB i. 29. (i) andext^ ^ECD=int^ ^ABC i. 29. (ii) .-. whole 2LACD=z.sCAB, ABC Ax. 2. (ii) Again, since z. ACD= z.s CAB, ABC... provedabove. Add Z.BCA to each, /. z-s BCA, ACD=z.s CAB, ABC, BCA Ax. 2. But Z.S BCA, ACD=two rt. Z.s 1. 13. /. Z.S CAB, ABC, BCA=two rt. z.s. Wherefore, if a side of a triangle &c. Q.E.D. CoR. 1. All the interior angles of a rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. For, by joining any pt. O inside the figure to each of the angles, the figure is divided into as many ..^s as it has sides. Now, the three Z_sof a ^^=two rt. Z.S...I.32. Hence, twice as many rt. Ls as the fig. has =all the Z. s of all the ^i:As, =all the inf" Z_s of fig. with z.s at O, ^=all the inV L s of fig. with 4 rt. Ls 1. 15. Cor. 2. 60 EUCLID, BOOK I. PEOPOSITION XXXII. Theorem. Cor. 2. All the exterior angles of a rectilineal figure are together equal to four right angles. D B For, any iiit^Z. ABC with its adj. exf^Z. ABD=two rt. Z.s 1. 13. Hence, all the ext^ L s with all the inf^ L s =twice as many rt. Z.s as the fig. has sides =all the inf^Z-S with 4 rt. Z.s 1.32. Cor.i. Take away the com. int^Z-s, /. all the e.WLs=Ji rt. Ls. NOTE. The exterior angles must be formed by producing the sides so that no two produced lines can cut ; and the rectilineal figure must have no re-entrant angles, that is, all its angles must have their vertices pointing outwards. EXAMPLES. I. Find the magnitude of an interior angle of a regular decagon. The 10 intr /.s with 4 rt. Z_s=20 rt. Z.s 1.32. Cor.i. /, the 10 inf Z_s=16 rt. Z.s... Ax. 3. ,', one int"^ L =tt ^^ ^ '^^ ^ ^'S- i» regular. =1 of art. L. Hence, the no. of degrees in an angle of a reg. decagon is f x 90=144°. II, Construct a right-angled triangle having given the hypotenuse and the sum of the other two sides. Let A be the given hypot. and BC the given sum of sides. At C in BC make an Z.BCD=to half a rt. L. A From BC cut off BE=A 1.3. ^^F With cent. B and rad. BE desc. an arc ^-^"^^\\ cutting CD at F. Join BF. ^^-^^^ \\^ Draw FG ± to BC I.i2. ^^-^^ | \ \^ Then FBG shall be the ^^ reqd. B G E C For, since L GCF is half a rt. Z- ) ^ , , _^^ \ Constr. and Z_ FGC is a rt. Z. ) /. Z_ GFC is half a rt. L 1. 32. Hence, ^GFC=Z.GCF Ax. i. and .*. GF=GC 1.6. Wherefore, ^^Y^Qi has the sum of its sides=BC, has a right angle FGB, and its hypot. FB=A. Q.E.F. PROP. XXXII. 61 EXERCISES. 1. What had been proved about the exterior angle of a triangle previous to Prop. 32? 2. If two triangles have two angles of one equal to two angles of the other, the third angle of the one is equal to the third angle of the other. 3. ABC, DEF are equilateral triangles, prove that angle ABC is equal to angle DEF. 4. Prove that all the interior angles of any quadrilateral are together equal to four right angles. 5. Show that each angle of an equilateral triangle is equal to two-thirds ^^ of a right angle. 6. Trisect a right angle. 7. One angle of a triangle is the complement of another ; prove that the triangle is right-angled. 8. Through three given points draw three straight lines so as to form an equilateral triangle. 9. If two triangles have two angles of the one equal to two angles of the other, then must the third angle of the one be equal to the third angle of the other. 10. ABC is an isosceles triangle having AB equal to AC ; BA is produced to D; prove that angle DAC is double of angle ABC. 11. If an isosceles triangle has each base angle double of the vertical, find their magnitude. 12. Find the size of an interior angle of a regular (i) pentagon, (ii) hexagon, (iii) octagon, (iv) quindecagon, taking the right angle as unit of measurement. Find also the magnitude of each angle in degrees. 13. One angle of a regular polygon contains 135°; find the number of sides. 14. Could a pavement be formed of tiles which were all equal regular (i) hexagons? (ii) pentagons? 15. Find the magnitude of each angle of a right-angled isosceles triangle. 16. Construct an isosceles triangle having the angles at the base each equal to one-sixth of the vertical angle. 17. Construct a right-angled triangle having given the hypotenuse and the difference of the other two sides. Draw both the figures which the construction yields. 18. From a point O within a triangle ABC perpendiculars OM, ON are drawn to the sides AB and AC ; prove that the angles MON and MAN are together equal to two right angles. 19. ABC is a triangle and the exterior angles at B and C are bisected by BD and CD meeting at D ; show that angle BDC, with half angle BAC, make up a right angle. 20. Trisect an angle equal to one -fourth of a right angle. {The 07ily angles tohich can be trisected with ruler and compasses alone are the right angle, or its half, fourth, eighth, i t- -c A ^ D E F Then, since ABCD is a O? Hyp. /, AD=BC I. 34(i). and, since EBCF is a O, Hyp. /. EF=BC i.34(i). B ' ^ Hence AD=EF Ax. i. ^ig. i. To each add DE (in Fig. i.), or take away DE (in Fig. -i.). ,', the whole, or rem'^, AE=whole, or rem^, DF Ax. 2, or Ax. 3. Hence, in ..r^s ABE, DCF, A E D F r AB=DC 1.34. \ / \ j */ j AE = DF Piovedabove. \ / \ / l.^BAE=z.CDF i.29(ii). \i jj /. ^ABE=^DCF 1.4. ^j,jg 2 ^ Now, if from the whole trapezium ABCF the ^:^ ABE be taken away, the OEBCF is left. Or, if from the same trapezium the ^^rrXDCF be taken away, the OABCD is left. Hence these renin's are equal Ax. 3. i.e. OABCD= OEBCF. Wherefore, parallelograms on the same base kc. Q.E.D. NOTES. In the enunciations of this and succeeding propositions, the word "equaV must be understood to mean ^^ equal in area,'''' not "equal in all respects." The altitude of a parallelogram is the perpendicular distance between the parallels between which it stands, and the altitude of a triangle the length of the perpendicular from its vertex on its base, or base produced. PROPS. XXXV., XXXV r. 65 PROPOSITION XXXVI. Theorem. Parallelograms on equal bases, and betiveen the same parallels, are equal to one another. Let £7s ABCD, EFGH be on = bases BC, FG, and between the same || s AH, BG. Then shaU OABCD=BFGH. Join BE, CH Post. 1. Then, since BC=FG Hyp. andEH=FG 1.34. /. BC=EH Ar.l. Hence, BC and EH being = and || , .*. BE is II to CH , 1.33. andEBCHisa O- Now, Os ABCD, EBCH are on the same base BC, and between the same || s AH, BC, .-. OABCD=OEBCH 1.35. Also Os EBCH, EFGH are on the same base EH, and between the same || s BG, EH, .-. OEBCH=OEFGH 1.35. Hence OABCD=ZZ7EFGH Ax. 1. Wherefore, parallelograms on equal bases &c. Q.E.D. EXERCISES. 1. What special form of Axiom 3 is used in Prop. 35? 2. The perimeter of a rectangle is less than that of any other parallelogram standing on the same base and having the same area. 3. AFCD, EBGF are equal parallelograms standing on equal bases FC, EF, in the same straight line, but not towards the same parts. If AB, DG be joined, prove that ABGD is a parallelogram. 4. Construct a rhombus equal to a given parallelogram. 5. ABCD, EBCF are parallelograms on the same base but on opposite sides of it; prove that, if AE and DF be joined, AEFD is a parallelo- gram; and that its area is equal to the sum of the areas of ABCD and EBCF. 6. If two parallelograms stand on the same base, but the altitude of one is double that of the other, the area of the former is double that of the latter. 7. If, in Fig. 1 of Case 2, Prop. 35, BE and CD meet at O, prove that trapezium ABOD is equal to trapezium EOCF. (310) B 66 EUCLID, BOOK I. PROPOSITION XXXVII. Theorem. Triangles on the same base, and between the same parallels, are equal to one another. Let .^^ s ABC, DBG be on the same base BC, and between the same 1| s AD, BC. Then shaU ^ABC=^DBC. Produce AD both ways Post. 2. Through B, draw BE || to AC, i. 31. and meeting AD prod*^ in E. Through C, draw CF || to BD, 1.31. and meeting AD prod*^ in F. Then, since EBCA, DBCF are Os on the same base BC, and between the same || s, /. OEBCA=ODBCF 1.35. But .^ABC is half OEBCA) ..... and ^DBC is half ODBCF j ^ /. .=^ABC=.^DBC Ax. 7. Wherefore, triangles on the same base &c. Q.E.D. PROPOSITION XXXVIII. Theorem. Triangles on equal bases, and between the same parallels, are equal to one another. Let .^^ ABC, DEF be on the equal bases BC, EF, and between the same || s AD, BF. Then shall .^ABC=.::ADEF. g A D h Produce AD both ways Post. 2. Through B, draw BG |i to AC,....i. 31. and meeting AD prod*^ in G. Through F draw FH || to ED, i. 31. b" and meeting AD prod*^ in H. Then, since GBCA, DEFH are Os on equal bases BC, EF, and between the same || s, .-. ZI7GBCA=/I7DEFH 1.36. But .-AABC is half OGBCA ] and ^DEF is half ODEFHj ^'"^" /. .-^ABC=.^DEF...., Ax. 7. Wherefore, triangles on equal bases &c. Q.E.D. NOTE. It follows from Prop. 88 that triangles on equal bases and with a common vertex are equal in area. Also, that if two triangles of the same altitude are on unequal bases that which has the greater base has the greater area. PROPS. XXXVIL, XXXVIII. 67 EXAMPLES. I. To construct a triangle equal in area to a given quadrilateral. Let ABCD be the given quadl Join AC. Through B draw BE || to AC and meeting DC prod^ in E 1. 31. Join AE. Then AED shall be the reqd ^^. For, .^AEC=.:^ABC 1. 37. Add ^^A(JD to each, .•. the whole -^AED=fig. ABCD. Q.E.F. II. The straight line joining the middle points of two sides of any triangle cuts off a triangle whose area is one-fourth of that of the xohole triangle. Let ABC be a ^^, and let D and E, the mid. pts. of its sides AB, AC, be joined. Then .:-^ADE shall be one-fourth of ..^ABC. Join BE. Then, <.^BAE=..^BEC i. 38, note. /. .-rXBAE is half .^ ABC, also ^^E AD= ^^EDB i. 38, note. .*. .i^EAD is half ^^BAE. Hence ^^EAD is one-fourth of -.r^XABC. Q.E.D. EXERCISES. 1. Given a triangle, construct (i) a right-angled triangle of the same area. (ii) a triangle whose area shall be double that of the given triangle, (iii) a triangle whose area shall be half that of the given triangle, (iv) a triangle equal to the given triangle in area and having one of its sides equal to a given straight line. 2. The four triangles into which a parallelogram is divided by its diagonals are equal in area. 3. In the figure of Prop. 37, prove that triangle ABD is equal to triangle ACD. And if AC, BD cut in 0, prove that (i) triangle AOE is equal to triangle BOA ; (ii) triangle OAB is equal to triangle OCD. 4. Triangles of equal altitude, on opposite sides of the same base, are equal. 5. If D, E, are the middle points of the sides AB, AC of a triangle ABC, and if CD and BE cut at O, prove that ^^AOE=^:ABOD. 6. In the figure of I. 38, prove that trapezium ABED is equal to ACFD. 7. Construct a triangle equal in area to a given rectilineal figure. 8. ABCD is a parallelogram. E is any point in the diagonal AC, or in AC produced. Prove that triangles EBC, EDC are equal in area. 9. Bisect a triangle by a straight line drawn through a given point in one of its sides. 10. Bisect a trapezium by a line drawn through one of its angles. 68 EUCLID, BOOK I. PROPOSITION XXXIX. Theorem. Equal triangles on the same hose, and on the same side of it, are between the same parallels. Let ABC, DBC be equal -.::As on the same base BC, and on the same side of it, and let AD be joined. Then shall AD be || to BC. ^ For, if not, if possible draw AE || to BC meeting BD, or BD produced, in E i. 31. \ ^^''^^s. \ Join EC Post.i. J^— — ^ Then, if AE is || to BC .:::AABC=-^EBC 1.37. but .:^ABC=.^DBC Hyp. and /. ^^EBC = .:::iDBC Ax. 1. or, the part = the whole, which is absurd Ax. 9. Hence, no other st. line but AD can. be |1 to BC, i.e. AD is || to BC. Wherefore, equal triangles on the same base &c. Q.E.D. PROPOSITION XL. Theorem. Equal triangles, on equal hoses in the same straight line, and towards the same parts, are between the same parallels. Let ABC, DEF, be equal ^^% on equal bases BC, EF in the same St. line, and towards the same parts, and let AD be joined. Then shall AD be || to BP. a d For, if not, if possible drawAGlltoBF 1.31. \ \ ^^Tx meeting DE, or DE produced, in G. Join GF Post. 1. g ^ — g -p Then, if AG is || to BF ^iABC=-^GEF 1.38. but ^^ABC=-^DEF Hyp. and /. ^^GEF=.^DEF Ax. 1. or, the part = the whole, which is absurd Ax. 9. Hence, no other st. line but AD can be || to BF, i.e. AD is || to BF. Wherefore, equal triangles on equal bases &c. Q.E.D. PROPS. XXXIX,, XL. 69 EXAMPLE. The straight line joining the middle points of the sides of a triangle is parallel to the base. Let E, F be the middle points of the sides AB, AC of -:^ ABC, Join EF, BF, CE, Then shall EF be || to BC. A For, since AE=EB Hyp. /. .^FAE=.:^FEB I. 38 note g and since AF=FC Hyp. .*. ^^FAE=^AFEC 1.38. Hence, -^FEB=^^FEC Ax.i. B "C and they are on the same base EF, /, BC is II to EF 1.39. Q.E.D. EXERCISES. * 1. If, in the above figure, D be the middle point of BC, prove that DE is parallel to AC. 2. Of what propositions are the 39th and 40th converse respectively? 3. Prove Prop. 39, supposing AE to meet CD produced in E. 4. Prove Prop. 40, supposing AG to meet DF in G. 5. In the figure of I. 6, if FG be joined, prove that FG is parallel to BC. 6. The figure formed by joining the middle points of the sides of any quadrilateral is a parallelogram, and its area is half that of the quadrilateral. 7. If the middle points of the sides of a triangle be joined, the triangle is divided into four equal triangles. 8. In the figure of the above example, prove that (i) EF is equal to half BC. (ii) If BF, CE cut at O, triangles OBE, OCF are equal in area, (iii) Triangles OEA, OFA are equal in area, (iv) Triangle OBE is one-third of triangle CBE. 9. If two equal triangles have a common vertex, and their bases in a straight line, their bases are equal. Hence, show that if OC in the above figure is bisected at Q, EC will be trisected in and Q. 10. If, of the four triangles into which the diagonals divide a quadrilateral, any two opposite ones are equal, the quadrilateral has two of its sides parallel. 11. In a right-angled triangle, the straight line joining the right angle to the middle point of the hypotenuse is half the hypotenuse. 12. What is the locus of the vertex of a triangle whose base is fixed, and whose area remains constant? 70 EUCLID, BOOK I. PKOPOSITION XLI. Theorem. If a 'parallelogram and a triangle be on the same base, and between the same parallels, the parallelogram shall be double of the triangle. Let OABCD and ^2kEBC be on the same base BC, and between the same || s AE, BC. Then shall OABCD be double of ^^BBC. A D F. B V, Join AC Post. 1. Then, since AE is || to BC Hyp. .-. .^ABC=.^EBC 1.37. But OABCD is double of .^ABC ..i. 34(iii). /. OABCD is double of -:^EBC Ax. 6. Wherefore, if a parallelogram &c. Q.E.D. EXERCISES. 1. In what special form is Axiom 6 used in this proposition? 2. Prove that if a parallelogram and a triangle stand on equal bases and between the same parallels the parallelogram is double of the tri- angle. 3. Construct a rectangle which shall be double of a given triangle. 4. If any point be taken in a side of a parallelogram and joined to the opposite angles, one of the triangles so formed is equal to the sum of the other two. 5. Any point O is taken within a parallelogram ABCD; prove that the sum of the triangles OAD, OBC is half the parallelogram, 6. Any point O is taken without a parallelogram ABCD ; prove that the difference of the triangles OAD, OBC is half the parallelogram. 7. Construct a triangle equal in area to a given parallelogram. 8. Construct a right-angled triangle equal in area to a given rectangle. 9. If in the figure of Prop. 41, BE and DC cut at O, and AO be joined; then triangles AOD, OCE are equal in area. PROPS. XLI., XLII. 71 PROPOSITION XLII. Problem. To make a parallelogram equal to a given triangle, and having an angle equal to a given rectilineal angle. Let ABC be the given ^^, and D the given L. It is req^ to make a O = -^^-4^(7, and having an L=^D. B^ E ^c L Bisect BC in E i.io. At E, in CE, make z.CEF=D 1.23. Through A draw AFG II to BC, I. 31. and meeting EF at F. Through C draw CG || to EF, 1. 31. and meeting AFG at G. Then shall FBCG be the O reqd. Join AE. Then, since BE=EC Constr. /. ^:iABE=.:^AEC i.ss. Hence -=:^ABC is double of .-rxAEC, but OFECG is double of ^^AEC i.4i. /. OFECG=^ABC Ax. 6. and it has an l.FEC=D Constr. Wherefore, a parallelogram has been made &c. Q.E.F. EXERCISES. 1. Construct a right-angled triangle equal to a given square. 2. Construct a rectangle equal to a given triangle. 3. Make a triangle equal to a given parallelogram and having an angle equal to a given angle. 4. Construct an isosceles triangle equal to a given square. 5. Construct a triangle equal to a given rectangle and having one of its angles equal to half a right angle. 6. Make a rectangle equal in area to a given parallelogram, having one of its angles at a given point, and one of its sides in a given direction, 72 EUCLID, BOOK I. PKOPOSITION XLIII. Theorem. The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another. Let ABCD be a O, AC its diam., EH, FG Os about AC, and BK, KD the complements. Then shall comp* BK=compt KD. B G C For, since EH is a O, .-. ^::::^AEK=.-:AAHK I. 34(iii). And, since GF is a O, ,\ ^^KGC=.^KFC i.34(ui). /. ^sAEK, KGC=^sAHK,KFC Ax. 2. But the whole .^A ABC = whole .^^ADO i. 34(iii). /, the rem'^ BK=rem'^ KD.... ax. 3. Wherefore, the complements &c. Q.E.D. EXERCISES. 1. Define "Parallelograms about the diameter," and "Complements," 2. If a point not on a diameter were taken, and through it parallels drawn to the sides, would any of the four parallelograms formed be necessarily equal? 3. In the figure of I. 43 prove that, (i) OHB=OED. (ii) Z.AEK=^KGC. (iii) EH is II to GF. 4. Prove that, if K is the middle point of AC, the complements are equal in all respects. 5. Prove that in the case of a rectangle the complements are rectangles. 6. Parallelograms about the diameter of a square are squares. PROPS. XLIII., XLIV. 73 PROPOSITION XLIV. Problem. To a given straight line to apply a parallelogram equal to a given triangle and having an angle equal to a given angle. Let AB be the given st. line, C the given .^^^ and D the given L . It is req^ to apply to AB a CJ = C, and having an L-= lD. Make OEBFG=.-::iC, having A.EBF= Z.D, and side BF in the direction of AB prod^ 1.42.* Through A draw AH || to BE i. 31. and meeting GE prod*^ in H, Join HB Post. 1. Now, since AH is || to FG Constr. .'. Z-sAHG, HGF=two rt. z.s...i.29(iii). and .-. Z.S BHG, HGF by the previous corollary, BA, AC Join OA. ^y^ Then shall OA bisect L BAG, / Draw the I.s OL, OM, ON from />""^^^ ^^\\ O to the sides L 12. Y. ^^^^ ^^ 1 Then, in ^^s OBN, OBL, i Z.OBN=Z.OBL Constr. V irt. Z.ONB=rt. Z.OLB Ax. n. I^OB is com. /. ON=OL L2c{ii). Similarly it may be proved that OM=OL /. ON=OM Ax. 1. Now, since L ONA and L OMA are rt. L s ,', sq, on OA=sqs. on ON, NA and sq. on OA=sqs. on OM, MA /, sqs. on ON, NA=sqs. on OM, MA. But since ON=OM, sq. on ON=sq. on OM ,*, sq. on NA=sq. on MA .*. NA=MA. Hence in .:^s NAO, MAO {NA=MA Proved above. AO is com. ON=OM Proved above. /. Z.NAO=^MAO L8. i.e. the bisector of ^ A passes through O. Q.E.D. Hence the following constructions: — To find points equidistant from three given straight lines. (i) Inside the triangle formed by the lines one point can be found. Bisect two of the angles of this triangle; the point is where these bisectors cut. (ii) Outside this triangle three points can be found. Bisect the exterior angles formed by producing two of the sides; the points of intersection of these bisectors can be proved to be equidistant from the sides by the method used in XI. 90 EUCLID, BOOK I. MISCELLANEOUS EXEECISES. 1. Are any two angles of a triangle greater than the third? Give a reason for your answer. 2. BAG, EAD are two equal angles; if the points B, A, D be in the same straight line, the points G, A, E can also be in the same straight line. 3. If two equilateral triangles be described on opposite sides of a straight line, the line joining their vertices will bisect the line. 4. If from the vertex of a triangle a perpendicular be drawn to the base, the difference of the squares on the sides is equal to the difference of the squares on the segments of the base. 5. ABG is a right-angled triangle, G being the right angle. Prove that if P be a point on BC between G and B, AP is less than AB, but if P be in GB produced, AP is greater than AB. 6. D is any point on the side BG of the triangle ABG. If the angle ABG be an obtuse angle, prove that AD is greater than AB but less than AG. 7. If, upon the same base AB, two triangles BAG, ABD be constructed, having the angle BAG equal to ABD and ABG equal to BAD, prove that the triangles BDC, ADG are equal in all respects. 8. BD, CE bisect the equal angles of the isosceles triangle ABG, and meet the opposite sides in D and E; prove that GD, DE, and BE are all equal. 9. A point A is taken in the circumference of a circle whose centre is and a circle is described having A as centre and meeting the first circle in B and G; prove that AG bisects the angle BAG. 10. In any right-angled triangle, the distance of the middle point of the hypotenuse from the right angle is half the length of the hypotenuse. 11. If AB, AG be equal sides of an isosceles triangle, and a circle with centre B and distance BA cut AG (or AG produced) in E, and BE be taken in AB (or AB produced, if E lies in AG produced) equal to GE, prove that the angle GEA is equal to the angle EAG. 12. Eind a point D in the side BG of a triangle ABG such that AD may be half the sum of AB and AC. 13. One interior angle of a regular polygon contains 165°; find the number of sides. 14. How many acres, roods, and poles are there in a triangular field whose longest side is 27 poles, and the distance of the opposite corner from this side 18 poles? 15. ABDE, BEGC are squares on the sides AB, BC of a triangle ABC. AE, CD are joined ; show that AE and CD are equal. MISCELLANEOUS EXERCISES. 91 16. The base of a triangle, whose sides are unequal, is divided into two segments by the line bisecting the vertical angle; prove that the greater segment is adjacent to the greater side. 17. If a point P be taken inside a quadrilateral ABCD, prove that the sum of the distances of P from the angular points is the least possible when P is situated at the intersection of the diagonals. 18. ABCD is a parallelogram and CE drawn parallel to BD meets AD produced in E; prove that AD is equal to DE, 19. AOB, COD are two straight lines intersecting 0; if the triangles AOC, BOD be equal in area, BC shall be parallel to AD. 20. ABC is a given triangle; construct a triangle of equal area having its base in AB and its vertex in a given straight line parallel to AB. 21. Find a square which shall be equal to three given squares. 22. Describe an isosceles right-angled triangle equal to a given square. 23. ABC is a triangle, D, E the middle points of AB, AC respectively; prove that the triangle BED is equal to the triangle CEF, where F is the point of intersection of BE, CD, 24. The sum of the four sides of any quadrilateral is less than twice the sum of its two diagonals. 25. Find the magnitude of one angle of a regular figure of 100 sides, 26. From a point P outside an angle draw a straight line cutting the lines containing the angle in B and C such tha,t PB shall be equal to BC. 27. Find a point in a given straight line, the difference of the distances of which from two given points on the same side of the line, shall be the greatest possible. 28. An obtuse -angled triangle ABC, having the angle ABC obtuse, is turned over about its side BC; prove that the line joining the two positions of A is perpendicular to BC produced. 29. In the base BC of a triangle ABC any point D is taken; draw a straight line such that, if the triangle ABC be folded along this line, the point A shall fall on the point D. 30. In a right-angled triangle the square whose diagonal is the side sub- tending the right angle is equal to the squares whose diagonals are the sides containing the right angle. 31. On a given straight line describe a triangle which shall be equal to a given parallelogram, and have one of its angles equal to a given rectilineal angle. 32. If the opposite angles of a quadrilateral are equal, it is a parallelogram. 33. If a quadrilateral be bisected by each of its diagonals, it is a parallel- ogram. 34. Through a given point draw a straight line which shall be equidistant from two other given points. 92 EUCLID, BOOK I. 35. If two adjacent corners of a rhombus be fixed, the other corners lie on fixed circles; but if two opposite corners be fixed the other corners lie on a fixed straight line. 36. ABCD is a rectangle; O is any point in the diagonal BD; show that the sum of the squares on OA and OC is equal to the sum of the squares on OB and OD. 37. Any point P is taken in the line joining an angular point A of a tri- angle to the middle point of the opposite side BC; prove that the triangles APB, APC are equal. 38. In any triangle straight lines are drawn from each angle to any point in the opposite side; prove that the sum of the lengths of these lines is greater than the semi-perimeter but less than three times the semi- perimeter of the triangle. 39. If ABC be a triangle, in which the angle A is a right angle, and BIC, CF be drawn bisecting the opposite sides respectively, show that four times the sum of the squares on BE and CF is equal to five times the square on BC. 40. Perpendiculars AD, BE, CF are drawn from the angular points A, B, C of a triangle upon the sides respectively opposite to them; prove that the sum of the squares upon BD, CE, AF is equal to the sum of the squares upon CD, AE, BF. 41. The four straight lines which bisect the successive angles of a parallel- ogram include a rectangle, and if the larger side of the parallelogram be twice the shorter, the diagonal of the rectangle is equal to the shorter side. 42. ABCD is an oblique parallelogram; from A a perpendicular is drawn to the side AB, from B a perpendicular is drawn to the side BC, from C a perpendicular is drawn to the side CD, from D a perpen- dicular is drawn to the side DA ; prove that these perpendiculars will by their intersections form a parallelogram equiangular to ABCD. 43. If two sides of a triangle be given, the area will be greatest when they contain a right angle. 44. Given three sides of a quadrilateral, and the angles adjacent to the fourth side, construct the figure. 45. Construct a triangle, two of whose angles are given, which shall have its vertex at a given point and its base in a given straight line. 46. AB, AC are two straight lines ; in AB three points P, Q, R are taken such that Q is equidistant from P and R, show that the perpen- dicular from Q on AC is equal to half the sum of the perpendiculars from P and R on AC. 47. ABC is a right-angled isosceles triangle, A being the right angle. Any line is drawn through A, and perpendiculars BM, CN from B and MISCELLANEOUS EXERCISES. 93 C are drawn to this line. Prove that MN is equal either to the sum or difference of BM and CN. 48. The point of intersection of the diagonals of the square described on the hypotenuse of a right-angled triangle is equidistant from the two sides which contain the right angle. 49. If, in the figure of I. 35, the diagonals of the parallelograms be drawn, prove that the two triangles which have each of them two of these diagonals for two of their sides will be equal. 50. Two sides AB, AD of a quadrilateral ABCD are given in position and magnitude, and the area of the quadrilateral is given; find the locus of the middle point of AC. 51. Prove that two of the angles of an equilateral triangle are together equal to one of the angles of an equiangular six-sided polygon. 52. Construct a triangle having given the perimeter and two of its angles. 53. A chord of a circle, whose radius is 10 feet, is 16 feet long; find its distance from the centre. 54. Given a pair of compasses; find a point C to which a right line AB must be produced so that AC is equal to three times AB. 55. If two triangles have the sides of one parallel to the sides of the other, the angles of the one are equal to the angles of the other. 56. The sides of a rectangular floor are 16 feet and 12 feet long respectively, find the distance between its opposite corners. 57. Find the magnitude of one interior angle of a regular polygon of n sides. 58. If lines be drawn from an angular point of a regular hexagon, to all the other angles, prove that the angles between these lines are equal. 59. The leaf of a book is turned down, so that the corner always lies on the same line of printing ; find the locus of the foot of the perpendicular from the corner on the crease. 60. Find a point at given distances from the arms of a given angle. 61. If K be the common angular point of the parallelograms about the diameter of a parallelogram, and BD the other diameter, the differ- ence of the parallelograms is equal to twice the triangle BKD. 62. Of all parallelograms which can be formed with diameters of given lengths, the rhombus is the greatest. 63. Trisect a parallelogram by lines drawn through an angular point. 64. If AA', BB', CC, DD' be equal lengths cut off from the sides of the parallelogram ABCD taken in order, then will A'B'C'D' be also a parallelogram. 65. Through the angular points A, B, C of a triangle are drawn three parallel straight lines meeting the opposite sides, or those sides pro- duced, in A', B', C respectively; prove that the triangles AB'C, BC'A', CA'B' are all equal. 94 EUCLID, BOOK I. 66. If through D, the middle point of the hypotenuse BC of a right-angled triangle ABC, DE be drawn at right angles to BC and meeting AC at E, prove that the square on EC is equal to the squares on EA, AB. 67. Between two given straight lines place a straight line equal to one, and parallel to another, given straight line. 68. Find the locus of the middle point of the hypotenuse of all right-angled triangles having a hypotenuse of given length and a common rt. angle. 69. Divide a given triangle into five equal parts. 70. Find a point within a triangle which shall be equidistant from the sides of the triangle. 71. Find a point equidistant from the angles of a triangle. 72. If L, M, N be the feet of the perpendiculars, drawn to the sides of a triangle ABC, from any point P within the triangle, then the sum of the squares on AM, BN and CL is equal to the sum of the squares on AN, BL and CM. 73. In the figure of I. 47, prove that the area of the hexagon formed by a side of each square and the three straight lines which join the adja- cent corners of the squares is equal to four times the area of the original triangle together with twice the square on the hypotenuse. 74. The area of a quadrilateral is equal to that of the triangle, two of whose sides are respectively equal to the diagonals of the quadri- lateral, and the angle included by these sides equal to the angle between the diagonals. 75. Construct a square equal to the difference of two given squares. 76. ABC is a triangle, and from A a line AD is drawn to the base making the angle BAD equal to ACB, a second line AE is drawn to meet the base in E so that AE is equal to AD. Show that the angle CAE is equal to the angle ABC. 77. If, in a right-angled triangle, the square on one of the sides containing the right angle be three times that on the other, the angle subtended by the first is double of that subtended by the second. 78. Construct a right-angled triangle having given the hypotenuse and one side. 79. ABC is a triangle in which ABC is a right angle, AB is 3 inches long, and BC is 4 inches long. CE is drawn at right angles to AC and equal to three times BC. Find the length of AE. 80. Given one side of a triangle, and the segments into which one of the other sides is divided by a perpendicular drawn to it from the ex- tremity of the first side, construct the triangle. 81. If the diagonals of a parallelogram are equal it is rectangular. 82. If the bisectors of two angles adjacent to one side of a parallelogram meet on the opposite side, what will be the ratio of the unequal sides? MISCELLANEOUS EXERCISES. 95 83. If the equal sides AB, AC of an isosceles triangle be produced to D and E so that AD=2AB, and AE=2AC, and if CD, BE meet in F, show that F is one of the points of trisection of CD and BE. 84. If the interior angle at one angular point of a triangle, and the exterior angle at another be bisected by straight lines, the angle contained by these two bisectors is equal to half the third angle of the tri- angle. 85. Describe a parallelogram equal and equiangular to a given parallel- ogram, and such that two of its opposite sides shall be at a given distance from each other. 86. E, F, G, H are points in the sides AB, BC, CD, DA respectively of a parallelogram ABCD such that AH=FC and AE=CG; show that EFG-H is a parallelogram. 87. The sum of the perpendiculars, from any point within an equilateral triangle, upon the sides, is constant. 88. Describe a circle which shall pass through two given points and have its centre in a given straight line. Is this always possible? 89. Describe a parallelogram equal to a given square, having an angle equal to half a right angle, and one side equal to a given straight line longer than a side of the square. 90. The area of a square is to the area of the equilateral triangle described on one of its sides in the ratio of 4 to V3. 91. ABCD is a quadrilateral. DA is produced to E, AB to F. BC to G, and CD to H. The straight lines which bisect the angles EAB, FBC, GCD, HDA are drawn. Show that any two opposite angles of the quadrilateral formed by these lines are together equal to two right angles. 92. The bisectors of the angles of a parallelogram form a rectangle whose diagonals are parallel to the sides of the parallelogram and equal to the difference between them. 93. If through the angular points of a triangle ABC there be drawn three parallel straight lines AD, BE, CF meeting the opposite sides, or those sides produced, in D, E, F, then will the area of the triangle DEF be double that of ABC. 94. A', B', C are the middle points of the sides of a triangle ABC, and through A, B, C are drawn three parallel straight lines meeting B'C, C'A', A'B' in a, h, c respectively; prove that triangle abc is half triangle ABC, and that be passes through A, ca through B, ab through C. 95. In a given triangle inscribe a parallelogram equal to half the triangle so that one side is in the same straight line with one side of the triangle and one extremity at a given point in that side. 96 EUCLID, BOOK I. 96. Given a line 1 inch long, show how to construct a triangle whose base shall be 2 inches, one of its sides 3 inches, and area 1^ square inches. 97. Prove that, if be any point in the plane of a parallelogram ABG13, and the parallelograms OAEB, OBFC, OCGD, ODHA be completed, then EFGH will be a parallelogram whose area is double that of the parallelogram ABCD. 98. In a right-angled triangle ABC the sides AB, AC which contain the right angle are equal to one another. A second right-angled triangle is described, having the sides containing the right angle together equal to AB and AC but not equal to one another. Prove that this triangle is less than the triangle ABC. 99. The angular points of one triangle lie on the sides of another ; if the latter triangle be thus divided into four equal parts, prove that the lines joining its angular points with the corresponding angular points of the former triangle will be bisected by the sides of the former. 100. Trisect a triangle by lines drawn from a point in one of its sides. BOOK II. DEFINITIONS. 1. A rectangle is said to be contained by any two of its sides which meet. 2. In any parallelogram, one of the paral- 'F ,■ ,^ lelograms about the diameter together h/ . J^ """p^ with the two complements, is called / ^-''' / / a gnomon. ^--- - ' 'q NOTES. In Book II. rectangles are often mentioned which are not actually con- structed, and such an expression as "rect. AB, CD" must often be understood to mean "the area which would be enclosed if a rectangle were constructed having two of its adjacent sides equal to AB and CD respectively." The rectangle contained by two lines is sometimes spoken of as the rectangle under those lines. A gnomon is named by three letters standing at opposite corners of the parallelograms; thus AKG, or HFC, is a name for the gnomon drawn to illustrate def. 2, Two magnitudes are said to be commensurable when some unit of mea- surement can be found which is contained in both of them an exact number of times. If the sides of a rectangle are commensurable lines its area can be exactly expressed arithmetically, but if the sides are incommensurable lengths (such, for instance, as the length of the side and the length of the diagonal of a square) its area can only be approximately expressed arithmetically. Book II. treats, mainly, of the properties of rectangles. EXERCISES. 1. Prove that, if two adjacent sides of one rectangle are equal, respectively, to two adjacent sides of another, the rectangles are equal in area. 2. Why cannot every parallelogram be correctly said to be contained by two of its adjacent sides? (310) ® 98 EUCLID, BOOK II. ADDITIONAL DEFINITIONS. If any point be taken in a line, it is called a point of section. The parts into which a line is divided at any point are called segments of the line. If any point be taken in a line produced, the line is said to be divided externally. (For example, the line AB is divided ^ C B D internally at C, and externally at D; AC " and CB are internal segments; AD and BD are external segments.) The projection of a point on a line is the foot of the perpendicular on the line from the point. The projection of a line on another line is ^B that part of the second line which is intercepted between the feet of the per- pendiculars to the second line from the extremities of the first. (For example, EF is the projection of ABonCD.) C E F D EXERCISES. 1. What is the area of a rectangle, one of whose sides is a feet long and another h yards? 2. Two adjacent sides of an oblique parallelogram are 3 feet and 4 feet long respectively: is its area greater than, equal to, or less than 12 square feet? 3. What is the area of a triangle standing on the same base and having the same altitude as the rectangle in Ex. 1 ? 4. Find approximately, to within one square inch, the area of a rectangle two of whose sides are each 3 feet long, and the other two each equal in length to the diagonal of a square whose side is 3 feet. 5. A line PQ is divided internally at O and externally at E,, mention its internal and external segments. 6. Show that the length of the line PQ is equal to half the sum of its internal segments together with half the difference of its external segments, 7. ABCD is a rectangle; in AD any points E and F are taken, and through E, F are drawn EG, FH parallel to AB and meeting BC in G and H ; prove that EGHF is a rectangle. 8. What is the altitude of a rectangle whose area is ah and base c? 9. A line AB is 3 feet long ; find the projection of AB on CD when AB is inclined to CD at an angle of (i) 60°; (ii) 45°; (iii) 30°. PROP. I. 99 PEOPOSITION I. Theorem. 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, and the several parts of the divided line. Let AB and CD be two st. lines, and let CD be divided into any number of parts in E, F. Then shall rect. AB, CD=rect. AB, CB; rect. AB, EP; and rect. AB, FD. Q E F D A B From C draw CG at rt. Z-S to CD i. ii. Cutoff CG = AB 1.3. Through G draw GH II to CD I. 31. Through E, F, D draw EK, FL, DH II to CG, 1.31. and meeting GH in K, L, H. Then, fig. CH=figs. CK, EL and FH,^ and all these tigs, are rectangles Constr. &L 29 But, since CG=AB Constr. /. fig. CH is rect. AB, CD. And, since Ek=CG 1.84. =AB .-. fig. CK is rect. AB, CE. Similarly, fig. EL is rect. AB, EF, and fig. FH is rect. AB, FD. .-. rect. AB, CD = rect. AB, CE, rect. AB, EF and rect. AB, FD. Wherefore, if there be two straight lines &c. Q.E.D. EXEECISE. Prove that if two straight lines be each divided into any number of parts (say three), the rectangle contained by the two lines is equal to the sum of all the rectangles contained by all the parts of one taken separately with all the parts of the other. * The whole is equal to the sum of its parts. 100 '. EUCLID, BOOK II. PEOPOSITION II. Theorem. If a straight line he divided into any two parts, the rectangles con- tained hy the whole and each of the parts are together equal to the square on the whole line. Let the st. line AB be divided into any two parts at C. Then shall rect. AB, AC with rect. AB, BC = sq. on AB. On ABdesc. sq. ADEB 1.46. Through C draw OF || to AD or BE, and meeting DE at F i. 3i. Then, figs. AF and CE=fig. AE. But, since AD=AB Constr. /. fig. AF is rect. AB, AC. And, since BE=AB Comtr. .-. fig. CE is rect. AB, BC. Also, fig. AE is sq. on AB Constr. .'. rect. AB, AC with rect. AB, BC=sq. on AB. Wherefore, if a straight line &c. Q.E.D. A C R t) F E NOTES. Prop. 2 is the particular case of prop. 1 in which the two given lines are equal. If, in prop. 1, AB contained a units of length; if CE, EF, FD contained 6, c, d units respectively, and, consequently, CD contained h + c + d units; the area of the rectangle CH would be a(6 + c + d) square units, the areas of the rectangles CK, EL, FH would be ah, ac, ad square units respectively; and the enunciation of the theorem becomes the statement of the algebraical identity a{b + c + d...) = ab + ac + ad... . EXEKCISES. 1, The square on any straight line is equal to the rectangle contained by its double and its half. 2, Construct a rectangle equal to a given rectangle in area, but having one of its sides three times the length of one of the sides of the given rectangle. 3, If the sides of the given rectangle in Ex. 2 are a and 6, the difference of the perimeters of the rectangles is 4(a- -). 3 ^i4pia^ (^h^ij-^. PROP. III. PKOPOSITION III. Theorem. If a straight line be divided into any two parts, the rectangle con- tained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square on the aforesaid part. Let the st. line AB be divided into any two parts at C. Then shall rect. AB, AC=rect. AC, OB with sq. on AC. C B On AC desc. the sq. ADEC 1.46. Through B draw BF |I to AD or CE, and meeting DE prod* in F 1. 31. Then, fig. AF=figs. AE and CF. ^ ^ But, since AD=AC Constr. /. fig. AF is rect. AB, AC. And since CE=AC Constr. /. fig. CF is rect. AC, CB. Also fig. AE is sq. on AC Constr. .*. rect. AB, AC=rect. AC, CB with sq. on AC. Wherefore, if a straight line &c. Q.E.D. F NOTE. This proposition is the particular case of Prop. 1, in which one part of the divided line is equal to the undivided line. EXERCISES. 1. Which is "the aforesaid part " in the proof as it is given above? 2. Write out the proof of the proposition, making the square upon EC instead of on AC. 3. State an algebraical identity which corresponds to this proposition. 4. Enunciate a geometrical theorem corresponding to {a + h)a -ab = a^. 6. Prove the following theorem: — If a straight line be produced to any point, the rectangle contained by the whole line so produced and the given line is equal to the rec- tangle contained by the given line and the part produced, together with the square on the former. 102 EUCLID, BOOK II. PEOPOSITION IV. Theorem. If a straight liiie he divided iiito any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained hy the parts. Let the st. line AB be divided into any two parts at C. Then shall sq. on AB=sqs. on AC and OB, with twice rect. AC. CB. On AB desc. sq. ADEB i. 46. Join BD. Through C draw CFG II to AD or BE, and meeting BD in F, and DE in G 1.31. Through F draw HFK II to AB or DE, and meeting AD in H, and BE in K i.3i. H B K Then, Z_CFB=z.ADB, 1.29. but Z.ADB=z.ABD, 1.6. /. ^CFB=aABD /. CF=CB L6. But CF=BK and CB=FK 1.34. /. OCFKB is equilat. And, since L CBK is a rt. L Constr. ,*. all its L s are rt. Lb i. 46 Cor. Hence, fig. CFKB is a sq Def. 30. Similarly it may be shown that HDGF is a sq. Now, fig. AE=figs. HG, CK, AF, and FE; =sqs. on HF and CB, with figs. AF and FE... Above. =sqs. on AC and CB, with figs. AF and FE...I. 34. =sqs. on AC and CB, with twice comp* AF....I. 43. =sqs. on AC and CB, with twice rect. AC, CF; i.e. sq. on AB=sqs. on AC and CB, with twice rect. AC, CB..Above. Wherefore, if a straight line &c. Q.E.D. CoR. — From this prop, it is manifest that Os about the diam. of a sq. are likewise sqs. PROP. V, 103 PEOPOSITION V. Theorem. If a straight line he divided into two equals and also into tivo unequal parts, the rectangle contained by the unequal parts together with the square on the line between the points of section, is equal to the square on half the line. Let the st. line AB be bisected at C, and divided unequally at D. Then shall rect. AD, DB with sq. on CD=sq. on CB. it. C T) B E Hi? On CB desc. sq. CEFB 1.46. Join BE. Through D draw DGH 1| to CE or BF, and meeting BEinG, EFinH 1.31. Through G draw KGL || to CB or EF, and meeting CE in K, BF in L 1.31. Through A draw AM || to CE or BF, meeting LK prod<^ in M. ..i. 3i. Then, rect. AD,DB with sq. on CD = rect. AD,DG with sq. on CD. . 11. 4 Cor, =rect. AD,DGwithsq.onKG..i. 34. =fig. AG with fig. KH 11.4 Cor. =figs.AK,CG,andKH, =figs. AK, GF, and KH i. 48. =figs.CL,GF,andKH 1.36. =fig.CF, =Sq. on CB Constr. Wherefore, if a straight line &c. Q.E.D. CoR. — From this prop, it is manifest that the difference of the squares on two unequal st. lines is equal to the rectangle contained by their sum and difference. (See page 106.) 104 EUCLID, BOOK II. PEOPOSITION VI. Theorem. If a straight line he bisected and produced to any point, the rectangle contained hy the whole line thus produced and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part pro- duced. Let the st. line AB be bisected at C and prod*^ to D. Then shall rect. AD, DB with sq. on CB=sq. on CD. ^, C V, y p G / L M K E HE OnCDdesc. sq. CEFD 1.46. Join DE. Through B draw BGH || to CE or DF, and meeting DE in G, EF in H 1.31. Through G draw KGL |! to CD or EF, and meeting CE in K, DF in L 1.31. Through A draw AM || to CE or DF, meeting LK prod^ in M...I. 31. Then, rect. AD, DBwith sq. onCB=rect. AD,DL withsq. on CB..ii.4Cor. =rect.AD,DLwithsq.onKG..i. 34. =fig. AL with fig. KH 11.4 Cor. =figs. AK and CL with fig. KH, =figs.CGandCLwithfig.KH..i. 36. =figs.GFandCLwithfig.KH..i. 43. =fig. CF, =sq. on CD .Constr. Wherefore, if a straight line &c. Q.E.D. PROP. VII. 105 PROPOSITION VII. Theorem. If a straight line he divided into any two parts, the squares on the whole line and on one of the parts are equal to twice the rectangle con- tained hy the whole and that part, together with the square on the other part. Let the st. line AB be divided into any two parts at C, Then shaU sqs. on AB and BC = twice rect. AB, BC with sq. on AC. ^ c i^ On AB desc. sq. ADEB, and complete the figure as in Prop 4. H D E Then, sqs.onABandBC=figs. AEandCK ii. 4Cor. =tigs. AK, FE, HG and CK =figs. AK, AF, CK and HG .1. 43. =figs. AK, AK and HG, =twice rect. AB, BK and sq. on AC.li. 4 Cor. = twice rect. AB, BC and sq. on AC.li. 4 Cor. Wherefore, if a straight line &c. Q.E.D. NOTES. From Prop 4 we know that — I. The square on the sum of two lines is equal to the sum of the squares on those lines, together with twice the rectangle contained by them. By help of Prop. 7 it can be shown that — II. The square on the difference of two lines is less than the sum of the squares on those lines by twice the rectangle contained by them. For sqs. on AB and BC=twice rect. AB, BC with sq. on AC... Prop. 7. ,*. sq. on AC alone < sqs. on AB and BC by twice rect. AB, BC. And AC is the difference of AB and BC, ,*, sq. on diflf. of AB and BC < sum of sqs. on AB and BC by twice rect. AB, BC. EXERCISES. 1. Interpret the theorems I. and II. algebraically. 2. Prove that, if AB is divided at C, the sqs. on AB and AC are equal to twice the rectangle AB, AC together with the square on BC. (310) a2 106 EUCLID, BOOK II. NOTES. The following is an independent proof of the important corollary to Prop, 5: — The difference of the squares of two unequal straight lines is equal to the rectangle contained by their sum and difference. Let CB be the greater of the two lines, and from it cut off CD equal to the less. {Construction as in Prop. 5.) A. c r» -E G / k K / E H I Then, diff. of sqs. on CB and CD=diff. of figs. CF and KH II.4. =gnomon CLH, =figs. CL and GF, =figs. AK andGF 1.36. =figs. AK and CG 1.43. =fig. AG, =rect. AJD, DG, =rect. AD, DB II. 4. Now, AD is sum of AC and CD, =sum of CB and CD. And DB is diff. of CB and CD. /, diff. of sqs. on CB and CD=rect. contained by their sum and diff. Q.E.D. D B Suppose AC to contain a units of length, and CD to contain 6 units; then DB con- ^' ■— tains a-h units. Then, rect. AD, DB becomes (a + 6) (a - 6), sq. on CD 6^, sq. on CB a^, and the corollary of Prop. 5, expressed in algebraical form, becomes PROPS. IV., v., VI., VII. 107 NOTES. Props. 5 and 6 may be included in one enunciation, thus: — If a straight line be divided into two equal segments and (either inter- nally or externally) into two unequal segments, the rectangle contained by the unequal segments is equal to the difference of the squares on the half line and on the line between the points of section. Prop. 6 may also be deduced from Prop. 5, thus: — Produce BA to E, making AE = BD. tt- a c T^ D Then ED is bisected at C, and divided *^ ' ' « unequally at D. ,*, rect. EB, BD with sq. on CB=sq. on CD II. 6. ButEA=BD Capstr. /, EB=AD. /. rect. AD, DB with sq. on CB=sq. on CD. Q.E.D. EXERCISES. 1. The square on any line is four times the square on its half. 2. Express Prop, 4 in algebraical form. 3. In the figure of Prop. 5, mention names for — (i) the line between the points of section; (ii) the two unequal segments; (iii) the figure which is equal to the difference between the squares on half the line and on the smaller segment. 4. In the figure of Prop. 6, prove that — (i) CD=half the difference of AD and DB; (ii) perimeter of gnomon CLH=perimeter of rect. AD, DB. (iii) AC=half the sum of AD and DB; (iv) AK=Dr. 5. Prove that the sum of the sum and difference of two lines is double of the greater. 6. Prove that the difference of the sum and difference of two lines is double of the less. 7. Express Prop. 5 in algebraic form. 8. Construct a rectangle equal to the difference of two given squares. 9. If a straight line be divided into two equal and into two unequal segments, the square on the whole line is equal to four times the rectangle contained by the unequal segments, together with four times the square on the line between the points of section. 10. Divide a given straight line into two parts such that the rectangle con- tained by those parts shall be a maximum. 11. By help of the theorems I. and II., prove that the square on the sum of two straight lines together with the square on their difference is double of the sum of the squares on the two lines. (Prop. 7, note.) 108 EUCLID, BOOK II. PEOPOSITION VITI. Theorem. If a straight line he divided into any two parts, four times the rec- tangle contained hy the whole line and one part, together with the square on the other part, is equal to the square on the straight line which is made up of the whole and that part. Let the st. line AB be divided into any two parts at C. Then shall four times rect. AB, BC with sq. on AC = sq. on the sum of AB and BC. Produce AB to D. CutoffBD = BC 1.3. On A© desc. sq. AEFD, 1.46. and complete the double fig. as in Prop. 4. ThenGP=GK ii.4. =CB 1.34. =BD CoDBtr, =BK II. 4. =GC 1.34. Hence, four times rect. AB, BC with sq. on AC =four times rect. AB, BD with sq. on AC Constr. =four times fig. AK with fig. XH II. 4. =twice figs. AK, AG, CK with fig. XH, =twice figs. AK, MP, GR with fig. XH i. 86. =figs. AK, KF; MP, PL; twice fig. GR with fig. XH. . .1. 48. =figs. AK, KF, MP, PL, GR, BN, XH i.46. Ex. i. =fig. AF, =sq. on AD Constr. =sq. on sum of AB and BC. Wherefore, if a straight line &c. Q.E.D. I ^ C R 1 p ■ftf G K / N X P / R / E H J L t NOTES. No use is made of this proposition by Euclid, and it is of very little importance. The following is another form of the enunciation: — The square on the sum of two lines exceeds the square on their difference by four times the rectangle contained by the lines. EXERCISES. 1. In the figure of Prop 8, prove (i) BN=KO; (ii) AN=CL; (iii) C0= four times BN; (iv) "PL=Rr. 2. Express the enunciation of Prop. 8, as given in the note, algebraically. PROPS. VIII., IX. 109 PKOPOSITION IX. Theorem. If a straight line he divided into two equal and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. Let the st. line AB be bisected at C, and divided unequally at D. Then shall sqs. on AD, DB=twice sqs. on AC, CD. From C draw CE at rt. Z.S to AB I. n. Cut off CE=AC I. 3. Join AE, EB. Through D draw DF || to CE, meeting EB in F 1.31. Through F draw FG || to AB, meeting CE in G I. 31. Join AF. Then, since AC=CE Constr. /. ^CAE=^CEA 1.5. but Z. ACE is a rt. L Constr. /. Z-sCAE, CEA together=a rt. L 1.32. Hence Z_CAE and Z.CEA each = half a rt. L. Similarly it may be shown that L CBE and L CEB each =half a rt. Z. . /. whole L AEB is a rt. Z. . Again, since GF is |I to AB Constr. /. z_EGF=z_ECB 1.29. but Z. ECB is a rt. Z. Constr. /. z_EGFisart. Z-. But L GEF is half a rt. L Above. .*. z. GFE is half a rt. L 1.32. Hence z.GFE=^GEF, /. GF=GE 1.6. Similarly it may be shown that z. FDB is a rt. Z. , and that DF=DB. Now, sqs. on AD and DB=sqs. on AD and DF Above. =sq. on AF 1.47. =sqs. on AE and EF i. 47. =sqs. on AC, CE and on EG, GF....I. 47. =twice sqs. on AC and GF Above. =twice sqs. on AC and CD 1. 34. Wherefore, if a straight line &c. Q.E.D. 110 EUCLID, BOOK II. PROPOSITION X. Theorem. If a straight line he bisected and produced to any point, the square on the whole liiie thus produced and the square on the part of it pro- duced are together double of the square on half the line bisected, and of the square on the line made up of the half and the part produced. Let the st. line AB be bisected at C and prod*^ to D. Then shaU sqs. on AD, DB=twice the sqs. on AC, CD. From C draw CE at rt. Z-s to AB i. ii. CutoffCE=AC 1.3. Join AE, EB. Through E draw EF II to AD 1.31. Through D draw DF II to CE and meeting EF at F 1.31. Since CE is || to DF Constr. /. Z-sCEF, EFD=twort. z-s 1.29. /. Z.S BEF, EFD sqs. on BC and AC by twice the rect. BC, CD. A. For, sq. on BD = sqs. on BC and CD with twice rect. BC, CD il. 4. Add sq. on AD to -each, .*. sqs. on AD and BD = sqs. on BC, CD and AD with twice rect. BC, CD. S C" But, sqs. on AD and BD=sq. on AB 1.47. And, sqs. on AD and CD=sq. on AC 1.47. Hence, sq. on AB=sqs. on BC and AC with twice rect. BC, CD. i.e. sq. on AB >sqs. on BC and AC by twice rect. BC, CD. Wherefore, in obtuse-angled triangles &c. Q.E.D. NOTE. Since CD is the projection of AC on BC, Prop. 12 may be stated thus : In an obtuse-angled triangle, the square on the side opposite to the obtuse angle is greater than the squares on the sides which contain it, by twice the rectangle contained by one of these two sides and the projection of the other upon it. EXERCISES. 1. If BE be perpendicular to AC produced, prove that the square on AB is equal to the sqs. on BC and AC with twice rect. AC, CE. 2. Two right-angled triangles ACB, ADB stand on the same side of the same hypotenuse AB ; if AD and BC cut at 0, prove that rect. AO, OD=rect. BO, OC. 3. The sides of a triangle are 4, 5, and 7; is it obtuse-angled? 4. The sides of a triangle are 4, 6, and 7, find the length of the projection of the side 4 upon the side 5. 5. Find the ratio of CD to AC when Z. ACB is— (i) 120°; (ii) 135°; (iii) 150°. 6. If AC and BC are each 3 feet long, and the angle ACB is 120°, find the length of AB to within an eighth of an inch. 116 EUCLID, BOOK II. PEOPOSITION XIII. Theorem. In every triangle the square on the side subtending an acute angle is less than the squares on the sides containing that angle by twice the rectangle contained by either of these sides, and the straight line inter- cepted between the perpendicular let fall upon it from the opposite angle and the acute angle. Let ABC be any ^^, having an acute L at C, and let AD be drawn ± to BC, or BC prod**. Then shall sq. on AB < sqs. on BC and AC by twice the rect. BC, CD. A A A (i) A (") IK (i") BD C B D C D B" C Case 1. — If the Z. at B is a rt. z., D coincides with B, and the rect. BC, CD becomes the sq. on BC, and the truth of the prop, is evident I. 4'?-, Case 2. — If the L at B is not a rt. Z. . Since, in fig. ii., BC is divided at D, and, in fig. iii., DC is divided at B, /„ in both figs., sqs. on BC and CD = twice rect. BC, CD with sq. on BD...II. T. Add sq. on AD to each, /. sqs. on BC, CD, AD = twice rect. BC, CD with sqs. on BD, AD. ,*. sqs. on BC and AC = twice rect. BC, CD with sq. on AB....I. 47. i.e. sq. on AB < sqs. on BC and AC by twice rect. BC, CD. "Wherefore, in every triangle &c. Q.E.D. NOTE. Since BD is the projection of AB on BC, the prop, may be stated thus: — In any triangle the square on a side opposite to an acute angle is less than the sum of the squares on the sides which contain the acute angle, by twice the rectangle contained by one of these sides and the projection of the other upon it. PROP. XIII. 117 EXAMPLE. // DE he drawn parallel to the base BC of an isosceles triangle ABC, the square on BE is equal to the square on CE with the rectangle contained by BC and DE. (In this Example the notation AB^ for sq. onAB, BC. CD for rect. BC, CD, andthe algebraical. symbols -+ and - have been used, as they, are often very convenient in dealing with deductions on Book 11. , and are generally allowed for such purposes, but it must be remembered that they should not be used in writing out Book-work.) Join BE. Draw DM, EN ±s to BC 1.12. Then in .^s DBM, ECN, DM=EN 1.34. Z.DBM=Z_ECN 1.6. .Z.DMB=Z.ENC Constr .-. BM=CN. 1.26. Now, BE2=EC2 + BC2 - 2 BC . CN., =EC2 + BC.(BC-2CN) =EC2 + BC.(BC-CN-BM) Above. =EC2 + BC.MN. II. 1. =EC2 + BC.DE 1.34. Q.E.D. EXERCISES. 1. What line, in fig. iii., is the intercept between the foot of the perpen- dicular and the acute angle? 2. If BE be drawn perpendicular to AC, in fig. of Prop. 13, prove that the rect. AC, CE=rect. BC, CD. 3. The sides of a triangle are 4, 5, 6 ; is it acute- angled? 4. The sides of a triangle are 4, 5, 6 ; find the length of the projection of the side 4 upon the side 5. 5. Find the ratio of CD to AC when the angle ACB is (i) 30°, (ii) 46°, (iii) 60°. 6. State and prove the converse of Prop. 13. 7. Express the results of Props. 12 and 13 algebraically. 118 EUCLID, BOOK II. PEOPOSITION XIV. Problem. To describe a square that shall he equal to a figure. Let A be the given rect^ fig. It is re^ to desc. a sq.=A. rectilineal C D Make the rectangle BCDE=A I, 45. Then, if BE=ED, the fig. BCDE is the req'^ sq. But if not, produce BE to F. Cut off Er=ED ' 1.3. Bisect BF at G L lo. With cent. G and rad. GF desc. a semicircle BHF. Produce DE to meet the circumference in H. Then shall sq. on EH=A. Join GH. Then, since BF is bisected at G, and divided unequally at E, .*, rect. BE, EF with sq. on GE=sq. on GF n. 5. =sq. on GH Constr. =sqs. on GE, EH L 47. Take away the com. sq. on GE, /. sq. on EH=rect. BE, EF, =rect. BE, ED Constr. =fig. BCDE, = A .Constr. Wherefore, has been described &e. Q.E.F. PROP. XIV. 119 EXEKCISES. 1. Construct a square equal in area to a given parallelogram. 2. Construct a square equal in area to a given triangle. 3. Construct a rectangle equal in area to a given square, and having the sum of two of its adjacent sides equal to a given straight line. 4. Construct a parallelogram equal in area to a given square, having the sum of its altitude and base equal to a given straight line and having an angle equal to a given angle. 5. In what cases are Exs. 3 ajid 4 impossible? 6. Construct a rectangle having given its area and perimeter. 7. Produce a given straight line so that the rectangle contained by the given line and the part produced may be equal to the square on an- other given line. 8. What is the length of the side of a square equal in area to a rhombus whose side is 3 inches and one of its angles 30°? 9. If in II. 14 the given rectilineal figure be that of I. 47, show how to determine the required square graphically. 10. The side of an equilateral triangle is a; find the side of a square equal in area to the triangle. 11. Write down a quadratic equation of which Prop. 14 is a geometrical solution. 12. What is meant by the quadrature of a given rectilineal figure, and how would you proceed practically in the case of a pentagon? 120 EUCLID, BOOK II. MISCELLANEOUS EXAMPLES. I. The sum of the squares on the sides of any triangle is equal to twice the square on half the base, together with twice the square on the line joining the vertex to the middle point of the base. Let ABC be any .t:^, and D the mid. pt. of BC. Then shall sqs. on AB and AC = twice sqs. on BD and AD. B Tr"E t Draw AE ± to BC, or BC prod^ 1.12. Then, if ADB is the obtuse L , AB2=AD2 + BD2 + 2BD. DE 11. 12. and AC2=:AD2 + CD^ _ 2CD. DE 11. 13. =AD2 + BD2-2BD. DE Hyp. /. AB2 + AC2=2AD2 + 2BD2. Q.E.D. N.B. — This result is very important. MISCELLANEOUS EXAMPLES. 121 II. The sum of the squares on the four sides of any quadrilateral is equal to the sum of the squares on its diagonals, together with four times the square on the line joining the middle points of its diagonals. Let ABCD be a quad^, AC, BD its diags., and E, F the mid. pts. of AC andBD. Then shall sqs. on AB, BC, CD, DA = sqs. on AC and BD with four times sq. on BF. Join BE, ED. Then, since E is mid. pt. of AC, ,*, sqs. on AB, BC=twice sqs. on BE, AE Ex. L and sqs. on CD, DA=twice sqs. on DE, AE Ex. L /. sqs. on AB, BC, CD, DA =twice sqs. on BE, DE and four times sq. on AE. =four times sqs. on BF, EF and AE Ex. L =four times sq. on EF, with sqs. on AC, BD IL 4Ex.i. Q.E.D. 122 EUCLID, BOOK II. III. The difference of the squares on the sides of any triangle is equal to twice the rectangle contained by the base and the line intercepted between the middle point of the base and the foot of the perpendicular from the vertex to the base, or the base produced. Let ABC be any .i^, D the mid. pt. of the base, E the foot of the 4. from A on BC, or BC prod<^. Then shall diflf. of sqs. on AB and AC = twice rect. BC, DB. D E Join AD. Then, if lADC is acute, AC2=AD2 + DC2-2CD.de ..ii.is. =AD2 + BD2-2CD.de Hyp. And AB2= AD2 + BD^ + 2 BD . DE . AB2 - AC2=2 BD . DE + 2 CD . DE. =2 (BD + CD). DE =2 BC . DE. 11.12. .Ax. 3. .II. 1. Q.E.D. EXERCISES. 1. "Write out the proof when the angle ADB is acute. 2. Prove the theorem without using either Props. 12 or 18. 3. Examine the case when ADC is a right angle. MISCELLANEOUS EXAMPLES. 123 IV. To divide a st line into two parts so that the rectangle contained by the parts may be a maximum. Let AB be the given st. line. A C 15 H Bisect AB at C. Then shall rect. AC, CB be a maximum. For, take any other pt. D in AB. Then, since AB is bisected at C, and divided unequally at D, .*. rect. AD, DB with sq. on CD=sq. on CB II. 6. =rect. AC, CB, i.e. rect. AD, DB alone < rect. AC, CB. Q.E.F. V. In a right-angled triangle, the square on the perpendicular drawn from the right angle to the hypotenuse is equal to the recta/ngle contained by the segments into which it divides the hypotenuse. Let ABC be a rt. -angled .^^, having the rt. Z. at A, and let AD be the ± toBC. Then shall sq. on AD = rect. BD, DC. For, since BC is divided at D, .'. sq. on BC=sqs. on BD, DC with twice rect. BD, DC n. 4. But sq. on BC=sqs. on BA, AC L4r. =sqs. on AD, BD and on AD, DC HT. =sqs. on BD, DC and twice sq. on AD. /, sqs. on BD, DO and twice sq. on AD =sqs. on BD, DC with twice rect. BD, DC Ax.i. Take away the com. sqs. on BD, DC, ,*, twice sq. on AD=twice rect. BD, DC, .*. sq. on AD=rect. BD, DC. Q.E.D. 124 EUCLID, BOOK II. VI. To divide a straight line into two parts so that their rectangle may he equal to the square on a given straight line which is not greater than half the former line. Let AB and C be the given st. lines. It is reqd to divide AB into two parts so that the rect. contained by the parts may be equal to the sq. on C. Bisect AB at D i.io. With cent. D and rad. DA desc. a semicircle. From A draw AE at rt. Z.s to AB I. u. Cutoff AE=C 1.3. Through E draw EF || to AB and cutting the semicircle at F i. 31. Draw FG || to AE and meeting AB at G- l. si. Then shall rect. AG, GB = sq. on C. {The proof is similar to that of Prop. I4.) VII. To divide a given straight line into two parts so that the difference of the squares on the parts may he equal to a given area. Let AB be the given st. line, and C the given area. To AB apply a rect. ADEB=C i. 46. Bisect AB at F, and BE at G 1. 10. From FB cut off FH=BG 1.3. F H Then H shall be the reqd pt. D E For, AH2 - HB2=( AH + HB) . (AH - HB) 11. 6 Cor. =AB.(AF + rH-HB) =AB,(BF + FH-HB) =AB.2FH Constr. =AB.BE Constr. =C Constr. Q.E.F. MISCELLANEOUS EXAMPLES. 125 VIII. If ABC and AD are two straight lines, and P is the middle point of BC, then the projection of A P on AD is half the sum of the projections of AB and AC on AD. i ^ 1 >^ ^ X ^^^ J5 ^^ s ^>^ R .^^^ n M N From B, P, and C draw ±s BM, PL and CN to AD..... L12. Through B draw BRS || to AD, meeting PL in R, and CN in S L 3i. Through P draw PT 1| to AD, meeting CN in T 1.31. Then, since PT and BS are || to AD Constr. /. extr/_CPT=Z-CAD L29. = Z.PBR L29. Also extr Z-CTP=Z.CNA L 29. =a rt, Z- Constr. andextr Z.PIIB=Z_PLA L 29. =a rt. Z- Constr. Hence, in ^^As CPT, PBR, {Z_CPT=Z.PBR Above, rt. Z.CTP=rt. Z_PRB Above. CP=PB Hyp. /. Pr=BR L26. ButPT=LN, andBR=ML..... 1.34. /. LN=ML. Now, projection of AP on AD=AL, =AM and ML, =half of 2 AM and 2 ML, =half the sum of 2 AM and MN, =half the sum of AM and AN, =half sum of projections of AB and AC on AD. Q.E.D. EXERCISES. 1. Prove that PL is half the sum of BM and CN in the above figure. 2. If A were taken between B and C, prove that PL would be equal to half the difference of BM and CN. 126 EUCLID, BOOK II. MISCELLANEOUS EXEECISES. 1. Divide a straight line into two parts so that the rectangle contained by the whole line and one part may be double of the square on the other part. 2. Enunciate a geometrical theorem of which the algebraical expression is 3. If a straight line be trisected, the square on the whole line is equal to nine times the square on one of the parts. 4. If in the figure of II. 6 the produced part be equal to the original line, then the square on half the line bisected is one-eleventh of the whole figure. 5. Produce a given straight line to a point, so that the rectangle contained by the whole line thus produced and the part produced shall be equal to the square on the original line. 6. Any rectangle is half the rectangle contained by the diameters of the squares on two of its adjacent sides. 7. ABODE is a straight line, C being the middle point of BD, Prove that the square on AC together with the rectangle BE, DE is equal to the square on EC together with the rectangle AB, AD. 8. On AB a square ABOD is described, and the angles ACE, ACE are made each equal to half the angle of an equilateral triangle, thus inscribing an equilateral triangle OEF in the square. Prove that AB is divided at E so that the square on one part is double the rectangle contained by the whole and the other part. 9. If a straight line is divided so that the sum of the squares on the whole and on one part is equal to three times the square on the other part, the line is divided in medial section. 10. AB is a diameter of a circle and AC is a chord; if CD be drawn per- pendicular to AB, then the square on AC is equal to the rectangle contained by AB and AD. 11. A straight line AB is bisected in 0, and frdm A a straight line AD is drawn at right angles to AB and equal to AC. A point E is taken in AB produced such that CE is equal to BD. Prove that the rect- angle contained by AE and EB is equal to the square on AB. 12. The sum of the squares on the sides of any parallelogram is equal to the sum of the squares on its diameters. 13. If a straight line AD be divided at B and 0, the rectangle contained by AC and BD is equal to the sum of the rectangles contained by AB and CD, and by BO and AD. 14. Divide a straight line into two parts so that the sum of the squares on the parts may be a minimum. 15. The rectangle contained by the sum and difference of the sides of a triangle is equal to twice the rectangle contained by the base and the line intercepted between the middle point of the base and the foot of the perpendicular on the base from the vertex. MISCELLANEOUS EXERCISES. 127 16. If the square on the perpendicular from the vertex to the base of a triangle be equal to the rectangle contained by the segments of the base, the vertical angle is a right angle. 17. ACDB is a straight line and D bisects CB; prove that the square on AC is less than the sum of the squares on AD and DB by twice the rectangle AD, DB. 18. ABC is an equilateral triangle and D is any point in BC. Prove that the square on BC is equal to the rectangle contained by BD and DC together with the square on AD. 19. AB is divided into any two parts in C, and AC and BC are bisected at D and E ; show that the square on AE with three times the square on BE is equal to the square on BD with three times the square on AD. 20. If EF is parallel to BC the base of an isosceles triangle ABC, prove that the difference of the squares on BF and BE is equal to the 'rectangle contained by EF and BC. 21. In II. 11, show that the rectangle contained by the sum and difference of the parts is equal to the rectangle contained by the parts. 22. If a straight line AB be bisected at C and produced to D so that the square on AD is twice the square on CD, prove that the square on AB will be twice the square on BD. 23. If a straight line AB be bisected at C and produced to D so that the square on AD is three times the square on CD, and if CB be bisected at E, show that the square on ED is three times the square on EB. 24. ABC is a triangle, and on the side of BC remote from A a square BDEC is described; prove that the difference of the squares on AB and AC is equal to the difference of the squares on AD and AE. 25. Find the locus of a point which moves so that the sum of the squares of its distances from two fixed points is constant. 26. The square on a straight line drawn from the vertex of an isosceles triangle to any point in the base, is less than the square on a side by the rectangle contained by the segments of the base. 27. Construct a rectangle which shall be equal in area to a given square, and the sum of whose sides shall be of given length. 28. The sum of the squares on the diagonals of any quadrilateral is equal to twice the sum of the squares on the lines joining the middle points of opposite sides. 29. The squares on the diagonals of a four-sided figure having two parallel sides are equal to the squares on its two non-parallel sides together with twice the rectangle contained by the parallel sides. 30. In any triangle, three times the sum of the squares on the sides is equal to four times the sum of the squares on the lines drawn from the angles to the middle points of the opposite sides. 31. If a straight line be divided into two pairs of unequal parts, the squares on the greatest and least of the four parts are together greater than the squares on the other two. 128 EUCLID, BOOK II. 32. The base AB of triangle ABC is bisected in D, and in DB a point E is taken such that DC is equal to DE ; show that the squares on AC and CB are together equal to the squares on AE and EB. 33. Given a square and one side of a rectangle equal in area to the square, find the other side. 34. Construct a rectangle equal to a given square and having the difference of its sides equal to a given line. 35. AB is bisected at C and produced to D ; prove that the rectangle AC, AD is equal to the rectangle BC, BD together with twice the square onBC. 36. ABCD is a rectangle, and O is any point; prove that the sum of the squares on OA and OC is equal to the suna of the squares on OB and OD. 37. The diagonals of a quadrilateral meet in E, and F is the middle point of the straight line joining the middle points of the diagonals ; prove that the sum of the squares on the straight lines joining E to the angular points of the quadrilateral is greater than the sum of the squares on the straight lines joining F to the same points by four times the square on EF. 38. Produce a given straight line so that the square on the whole line thus produced may be double the square on the part produced. 39. Show that the locus of the vertex of a triangle whose base is fixed, and the difference erf the squares on whose sides is constant, is one or other of two straight lines. 40. If with the middle point of the line joining the points of bisection of the diagonals of any quadrilateral as centre and any radius a circle be described, prove that the sum of the squares of the distances of any point on the circumference from the angles of the quadrilateral, is constant. BOOK 111. DEFINITIONS. Equal circles are those of which the diameters (or the radii) are equal. [This is not a definition but a theorem : for, if the circles be applied to one another so that their centres coincide, their circum- ferences must also coincide, since their radii are equal.] 2. A straight line is said to touch a circle when it meets the circle but, being produced, does not cut it. 3. Circles are said to touch one another which meet but do not cat one an- other. ADDITIONAL DEFINITIONS. A straight line which cuts a circle is Called a secant. A straight line which meets but, being produced, does not cut the circle is called a tangent. When two circles touch, and one is entirely within the other, they are said to have internal contact. When two circles touch, and one is entirely without the other, they are said to have external contact. ( 310 ) I 130 EUCLID, BOOK III. 4. 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. 5. The chord on which the greater perpen- dicular falls is said to be further from the centre. 6. (See Book L, def. 19.) 7. The angle of a segment is contained by the straight line and the circumference. 8. The angle in a segment is the angle contained by two straight lines drawn from any point in the circumference to the extremities of the straight line which is the base of the segment. 9. An angle is said to stand upon that part of the circumference intercepted between the straight lines which contain the angle. (For example, angle BAG stands upon the -^ arc BBC.) ADDITIONAL DEFINITIONS. (For defs. of arc, chord, &c., see Book I., page 8.) If the circumference of a circle be divided into two unequal parts the greater part is called the major arc, and the less the minor arc. An arc is said to be convex or concave with respect to a point, according as the lines drawn from the point to the ends of the arc lie altogether without the circle or not. (For example, arc AD is convex, and arc BC concave with respect to the point P.) DEFINITIONS. 131 10. A sector of a circle is the figure con- tained by two straight lines drawn from the centre and the part of the circum- ference between them. 11. Similar segments of circles are ^^^=<^^ those which contain equal angles. ^^^^^^ ADDITIONAL DEFINITIONS. A sector, the bounding radii of which are at right angles to one another, is called a A figure is said to be inscribed in a circle, when each of its angular points is on the circumference. A figure is said to be described about (or to circumscribe) a circle, when each of its sides touches the circumference. If be the centre of a circle ABC, the angle BOG which stands upon the arc BC is called the angle at the centre; and the angle BAG is called the angle at the cir- cumference. A line drawn at right angles to a tangent from its point of contact is called a normal. Gircles are said to cut at right angles when the tangents at the point where they cut are at right angles. NOTE. In Book III. Euclid deals with the simpler properties of circles. 132 EUCLID, BOOK III. PROPOSITION I. Problem. To find the centre of a given circle. Let ABC be the given 0. It is req^ to find its cent. Draw any chd. AB. Bisect AB at D I. lO. Through D draw CDE at rt. z.s to AB meeting the Qce at C and E I- ii- Bisect CE at F i. lo. Then shall F be the cent. C For, if not, if possible, left some pt. G outside the line CE be the cent. Join GA, GB, GD. Then, in .^^ GAD, GBD, rAD=BD Constr. *.• \ GD is com. IGA=GB Radii. .-. Z.GDA=^GDB 1.8. .*. L GD A is a rt. Z. . But Z.FDA is art./. Constr. .-. Z.GDA=z.FDA. i.e. the whole = its part, which is absurd. Hence the cent, cannot be outside the line CE. .*, F, the mid. pt. of CE, is the cent. Wherefore, has been found &c. Q.E.F. Cor. — If, in a circle, a straight line bisect a chord and be also at right angles to it, the centre of the circle is in that straiglit line. PROP. I. 133 NOTE. The following extension of Prop. 1 is important: — To find the centre of a given arc, or segment. Let A, B, C be three pts. on the Qcq. Join AB, BC. Bisect AB, BC at D, E 1. 10. Through D, E draw lines at rt. Z_s to AB, BC Lil. Then these lines will, if prodz_DAE...., 1. 16. /. z_DEB>z.DBE, .-. DB>DE 1.19. But DB=DF Radii. /. DF>DE, i.e. the part > the whole, which is absurd. Hence, the pt. E cannot fall without the 0ABC. Similarly, it may be shown that E cannot fall on the Oce, /. the pt. E falls within the 0. But E is any pt. in AB, ,*. AB lies entirely within the 0. Wherefore, if any two points &c. Q.E.D. PROPS. II., III. 135 PEOPOSITION III. Theorem. If a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it shall cut it at right angles; and, conversely/, if it cut it at right angles it shall bisect it. Part I. — In the ©ABC let the diam. CD bisect the chd. AB at E. Then shall CD be at rt. z.s to AB. Find the cent I.io. Join OA, OB. Then in .^s OAE, OBE, rAE=EB Hyp. *.* -| OE is com. [0A=0B Radii. .'. lOEA=lO^B 1.8. /. CE is at rt. Z. s to AB. Part II.— In the ©ABC, let the diam. CD cut the chord AB at rt. z. s at E. Then shall CD bisect AB at E. Find O the cent 1. 10. Join OA, OB. Then, since OA=OB Radii. /. z.OAE=z.OBE 1.5. Hence, in ^^s OAE, OBE, L0AE= LOBE Above. rt. z.OEA=rt lOEB Hyp. OE is com. .-. AE=BE i.26(i.) i.e. AB is bisected at E. Wherefore, if a straight line &c. Q.E.F, 136 EUCLID, BOOK III. NOTES. The 2^^ part of Prop. 3 is the converse of the 1st part, and they are both converse of the corollary to Prop. 1. These three results are important. Prop. I. cor. — A st. line drawn at rt. L.S to a chord of a Q through its mid. pt. passes through the cent. Prop. III. part i. — A st. line from the cent, to the mid. pt. of a chord of a is at rt. Ls to the chord. Prop. III. part ii. — A ± from the cent, of a Q on any chord bisects that chord. EXAMPLES. I. If two circles cut one another, and through one of their points of inter- section straight lines he drawn equally inclined to the line joining their centres and terminated by the circumferences, these straight lines will be equal. Let A be a pt. of intersection of the 0s ABD, AEC, and let BAG DAE be equally inclined to OQ the line joining their cents. Then shaU BC=DE. DrawOM, QN xs to BC 1.12. Draw QR || to BC, meeting OM at R 1. 31. Then, since Z_ s at M and N are rt. L s, /. OM is II to QN I. 29 (iii.) Hence RMNQ is a 0> .•.QR=MN 1.34. But chd. AB is bisected at M, and AC at N III. 3 (ii.) Hence MN is half BC. /. QR is half BC. And L at R = Z. at M = a rt. L I. 29(ii.) Similarly, if ±s OK, QL and || OS be drawn to DE, it may be shown that OS is half DE, and that Z. at S is a rt. Z_. Also, extrZ_RQO=int'-z.BrQ i. 29 (ii.) = Z-EGO Hyp. =extrz_SOQ I.29(ii.) PROP. III. 137 Hence, in ..^^s, ROQ, SOQ, ^ORQ=Z.OSQ, ^RQO=^SOQ, OQ is com: /. RQ=OS I.2C. Hence BC=DE Ax. 6. Q.E.D. II. Through a given point within a circle to draw the shortest 'possible chord. Let ABC be the 0, and P the given pt. Find O the cent III. i. Join OP. Through P draw chd. BC at rt. Z.S to OP I. ii. Then BC shall he < any other chd. through P, such as DE. Draw 0M± to DE 1. 12. Join 00, OD. Then, since Z_OMP is a rt. L Constr. /.Z- 0PM is < art. Z. i. ir. /. OM < OP 1.19. Now, sq. on OC=sqs. on OP, PC I. 47. and sq. on OD=sqs. on OM, MD 1.47. Hence sqs. on OP, PC=sqs. on OM, MD. But sq. on OP > sq. on OM, .*. sq. on PC < sq. on MD. .*. PC < MD. But PC is half BC, and MD is half DE 111.3. (ii.) BC < DE. Q.E.D. EXERCISES. 1. Prove part ii. of Prop. 3, using I. 47 instead of I. 26. 2. If from the centre of a circle a perpendicular be drawn to the nearer of two parallel chords, it will, if produced, bisect the other. 3. Find the locus of the middle points of a system of parallel chords of a circle. 4. The portions of a common chord of two concentric circles, which are intercepted between the circumferences, are eqvial. 5. If two circles cut, any two parallel straight lines drawn through the points of intersection and terminated by the circumferences, are equal. 6. Through a point of intersection of two circles which cut one another, draw the greatest possible straight line terminated by their circum- ferences. 138 EUCLID, BOOK III. PEOPOSITION IV. Theorem. If, in a circle, two straight lines cut one another, which do not both pass through the centre, they do not bisect one another. Let ABCD be a ; AC and BD two chords which cut at E and which do not both pass through the cent. Then AC and BD shall not bisect each other. Case 1. — If one pass through the cent, it is a diam. and is bisected by the cent. /, it cannot be bisected by the other which does not. . . Case 2. — If neither pass through the cent, if possible, suppose ^^""^'^^ ^ AE=EC, and BE=ED. ^\.^_^^^ Find O the cent ill.i. Join OE. Then, if E is the mid. pt. of AC, ^OEA is a rt.z. iii.3(i.) And, if E is the mid. pt. of BD, Z_OEB is a rt.Z- llL3(i.) /. z_OEA=z.OEB. i.e. the part = the whole, which is absurd. Hence, AC and BD cannot bisect each other. Wherefore, if in a circle &c. Q.E.D. EXERCISES. 1. Draw two straight lines in a circle which shall bisect each other. 2. Draw two chords in a circle, neither of which pass through the centre, and one of which is bisected by the other. 3. Prove that, in a circle, if a chord which does not pass through the centre bisect another chord it cannot cut it at right angles. 4. Prove the converse of Ex. 3. 5. If a parallelogram be inscribed in a circle it must be rectangular. PROPS. IV., v., VI. 139 PEOPOSITION V. Theorem. If two circles cut one another they cannot have the same centre. Let the 0s ABC, ADE cut at A. Then they shall not have a com. cent. For, if possible, let O be their cent. Join OA. Draw OCE meeting the Qces at C and E. Then, since O is cent, of ©ABC, /. OA=OC. And, since O is cent, of ©ADE, /. OA=OE. Hence, OC=OE, i.e. the part=the whole, which is absurd. Wherefore, if two circles cut &c. Q.E.D. PKOPOSITION VI. Theorem. If two circles touch one another internally they cannot have the same centre. Let the ©s ABC, ADE touch internally at A. Then they shall not have a com. cent. For, if possible, let O be their cent. Join OA. -^ Draw OEC meeting the Qces at E and C. Then, since O is cent, of ©ABC, /. OA=OC. And, since O is cent, of ©ADE, /. OA=OE. Hence, OC=OE, i.e. the whole =its part, which is absurd. Wherefore, if two circles touch &c. Q.E.D. EXERCISES. 1. Prove that two concentric circles cannot cut. 2. Prove that two concentric circles cannot touch. 3. Of what propositions are Exs. 1 and 2 respectively converse? 140 EUCLID, BOOK III. PEOPOSITION YII. Theorem. If any point he taken in the diameter of a circle which is not the centre, of all the straight lines which can he drawn from this point to the circumference the greatest is that in which the centre is, and the other part of the diameter is the least; and, of any others, that which is nearer to the straight line which passes through the centre is always greater than one more remote; also, from the same point there can he drawn to- the circumference only two straight lines equal to one another, one on each side of the shortest line. Let ABCD be a 0, and O its cent.; let F be any other pt. within it ; let AOFD be the diam. through F; and let FB, FC, FE be any other st. lines from F to the Qce. Then (i) FA shall be tne greatest line; (ii) FB shaU be >FC, FOFE; (iii) FD shaU be the least line; and (iv) from F but one st. line = FB can be dra-s^m to the Oce. Join OB, OC, OE. At O in OF make Z.FOG=Z.FOE 1.23. Join FG. Then (i), in .r^BOF, FO, OB>FB 1.20. ButOB=OA Rad. /. FA>FB. i.e. FA is the greatest line. PROP. VII. 141 And (ii), in -^s EOF, COF, OB=OC Rad. OF is com. Z.FOB>Z.FOC Ax. 9. /. FB>FC 1.24. Similarly it may be shown that FOFE. Also (iii), in ^iOFE, OF, FE>OE 1.20. ButOD=OE Rad. /. OF, FE>OD. Take away the com. part OF, /. rem'^FE>rem'^FD. i.e. FD is the least line. Again (iv), in <^s OFG, OFE, OG=OE Rad. OF is com. Z.FOG=Z.FOE Constr. .-. FG=FE 1.4. And, from F no other line can be drawn to the Oce=FE. For, if possible, suppose FH=FE. Then, since FG=FE Above. /. FG=FH. i.e. a line nearer to FA=one more remote, which is impossible Part (iii). /. but one st. line can be drawn to the Qce from F=FE. Wherefore, if any point &c. Q.E.D. EXEECISES. 1. If from the end of a diameter any number of chords be drawn in a circle — (i) the diameter itself is the greatest of these chords ; (ii) a chord nearer to the diameter is greater than one more remote; (iii) there can be drawn but two equal chords from the point, one on each side of the diameter. 2. Write out in full a proof that FC is greater than FE. 3. From F draw a straight line to the circumference which shall be equal toFB. 142 EUCLID, BOOK III. PEOPOSITION VIII. Theorem. If any point he taken without a circle and straight lines he drawn from it to the circumference, of those which fall on the concave part of the circumference, the greatest is that which parses through the centre, and one nearer to that which passes through the centre is always greater than one more remote; hut of those which fall on the convex part of the circumference, the lea^t is that which, when produced, passes through the centre, and one nearer to it is always less than one more remote; also from the same point there can he drawn to the circumference only two straight lines equal to one another, 07ie on each side of the shortest line. Let ABCD be a 0, and its cent.; let F be any pt. without it ; let FDOA be drawn from A through the cent.; and let FGB, FHC, FKE be any other lines from F to the Qce. Then (i) FA shall be the greatest line; (ii) FB shall be >FC, FOFB; (iii) FD shall be the least line; (iv) FG shaU be FB 1. 20. But OB=OA Had. /. FA>FB. i.e. FA is the greatest line. PROP. VIII. 143 And (ii) in .^s EOF, COF, OB=OC Rad. OF is com. Z.FOB>z.FOC Ax. 9. /. FB>FC 1.24. Similarly it may be shown that FC>FE. Also (iii) in ^OFG, FG, OG>FO 1.20. But OG=OD Rad. /. rem'^FG>rem^FD. i.e. FD is the least line. And (iv) in .-rlOFH, OG, FGDB, DB>DA ill. 7(u). But this is contrary to hypothesis. Hence, D must be the cent. Wherefore, if a point &c. Q.E.D. NOTES. Euclid gave also a direct proof of this prop, depending on the cor. to Prop. 1. It should be noticed that the point E might be supposed to fall within the angle CDB, or within BAD. EXERCISES. 1. Prove this prop, directly, by joining AB, BC, bisecting AB and BC at E and F, and proving that DE, DF are at right angles to AB and AC. 2. Write out an indirect proof of the prop., supposing E, the centre, to fall within the angle CDB. PROPS. IX., X. 145 PROPOSITION X. Theorem. One circumference of a circle cannot cut another at more than two points. For, if it be possible, let 0ACDE cut ©BCDE at the pts. C, D, E. Find O the cent, of ©ACDE in. i. Join 00, OD, OE. Then, since O is cent, of 0ACDE, /. OC=OD=OE Radii. And, since from a i)t. O within 0BCDE three=st. lines are drawn to the Qce, /. O is cent, of 0BCDE iii. 9. i.e., pt. O is the com. cent, of two 0s which cut, which is impossible in. 5. "Wherefore, one circumference &c. Q.E.D. NOTE. Euclid gives two demonstrations of this prop, also, the other being similar to that indicated in the note to Prop. 9. It should be noticed in the above proof that the centre of one circle might be supposed to fall on or outside the circumference of the other circle. EXERCISES. 1. Write out another proof of Prop. 10, depending on III. 1, cor. 2. Complete the proof given above, by considering the cases mentioned in the note. 3. Two circumferences cannot have more than two points in common. (310) K 146 EUCLID, BOOK III. PEOPOSITION XI. Theorem. If two circles touch 07ie another internally, the straight line which joins their centres, being produced, shall pass through the point of contact. Let 0ABC touch 0ADE internally at A; and let O and Q be their cents. Then shall OQ prodd pass through A. A C For if not, let it, if possible, pass otherwise as OQDB. Join AO, AQ. Then, AQ, QO>AO 1.20. But AO=OB Radu. .-. AQ, QO>OB. Take away the com. part OQ, .-, rem"^ AQ>QB. But AQ=QD Radii. .*. QB>QB, i.e. the less > the greater, which is absurd. Hence, OQ when prod*^ must pass through A. Wherefore, if two circles &c. Q.E.D. PROPS. XL, XII. 147 PEOPOSITION XII. Theorem. If two circles toxich one another externally^ the straight line which joins their centres shall pass through the 'point of contact. Let 0ABC touch 0ADE externally at A, and let and Q be their cents. Then shall st. line OQ pass through A. =0C1 ^ -=QDJ For if not, let it, if possible, pass otherwise as OCDQ. Join AO AQ. Then, in .^AOQ, OA, QA>OCDQ I. But 0A=< and QA: /. OC,QD>OCDQ. i.e. part > the whole, which is absurd. Hence, st. line OQ must pass through A. Wherefore, if two circles &c. Q.E.D. NOTE. The words " the point of contact" in Props. 9 and 10 appear to assume that the circles can touch in but one point, although this is not proved until the succeeding proposition is reached. But, if the words are taken in the sense "the point of contact under consideration," it follows from the proofs that, if there were a second point of contact, the line joining the centres would, if produced, pass through it also, EXERCISE. State and prove the converse of (i) Prop. 11; (ii) Prop. 12. 148 EUCLID, BOOK III. PKOPOSITION XIII. Theorem. One circle cannot touch another at more than one point, whether on the inside or outside. For, if it be possible, let 0ABO touch 0DBC at the pts. B and C (i) Internally. Join BC. Bisect BC at E i. lo. Through E draw ADEF at rt.^s to BC i. ii. Then, since B and C are pts. in the Qces of both 0s, .*. BC falls within both 0s III. 2. And, since AF bisects BC at rt. L s, .*. the cents, of both 0s are in AF ill, i cor. .*. AF passes through a pt. of contact III. ii. i.e. AF passes through the extremity of a line which it bisects at rt. L s, which is absurd. (ii) Externally. A Join BC. Then, since B and C are pts. in the ;' '; Oce of 0ABC, \ / /. BC falls within ABC in 2 And, since B and C are pts. in the Oce of 0DBC, .-. BC falls within 0DBC ill. 2. i.e. the chd. BC is within each of two 0s which are without each other, which is absurd. "Wherefore, one circle cannot &c. Q.E.D. PROPS. XIII., XIV. 149 PROPOSITION XIV. Theorem. Equal straight lines in a circle are equally distant from the centre; and, conversely, those which are equally distant from the centre are equal to one another. Part I.— Let AB, CD be equal chds. of 0ABCD. Then shall AB, CD be equidist. from the cent. Find O the cent in. i. Draw OE, OF ±s to AB, CD 1. 12. Join OA, OD. Then, since OE, OF are ± to AB, CD, /. AB, CD are bisected at E, F<....iii. 3. But AB=CD Hyp. /. AE=DF, .', sq. on AE=sq. on DF. Also, since OA=OD Radii. .*, sq. on OA=sq. on OD. Hence, sqs. on AE, OE=sqs. on DF, OF i. 47. But sq. on AE=sq. on DF Above. /, rem^^ sq. on 0E=rem8^ sq. on OF. /. OE=OF, i.e. AB and CD are equidist. from O. Part II.— Let the ±s OE, OF be Then shall chd. AB=chd. CD. For, since OE=OF Hyp. /. sq. on OE=sq. on OF. And, since OA=OD Radii. /. sq. on OA=sq. on OD. Hence, sqs. on OE, AE=sqs. on OF, DF 1.47. But sq. on OE=sq. on OF ...Above, .*, rem^ sq. on AE=rems sq. on DF. /. AE==DF. But, since OE, OF are x to AB, CD, /. E, F are mid. pts. of AB, CD iii. 3. Hence, AB=CD Ax. 6. Wherefore, equal straight lines &c. Q.E.D. 150 EUCLID, BOOK III. PEOPOSITION XV. Theorem. 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, conversely, the greater is nearer to the centre than the less. A Part I. — Let ABCD be a 0, AD a diam., O the cent, and let the chd. BC be nearer to the cent, than EF, Then shall (i) AD be the greatest line; (ii) BOBF. From O draw ±s OM, ON to BC, EF 1.12. Join OB, OC, OE. Then (i) OB, OC>BC 1.20. But OB=OA, and OC=OD Radii. /. OA,OD>BC, i.e. AD is the greatest line. And (ii) since OM rem^ sq. on EN, /. BM>EN. But, since OM, ON are ± s to BC, EF Constr. /, M, N are mid. pts. of BC, EF ...iii. 3. Hence, BC>EF Ax. 6. PROP. XV. 151 Part II.— Let chd. BC >chd. EF, and let ±s OM, ON be drawn from the cent. O to these chds. Then shall 0M< ON. Join OB, OE. Then, since OM, ON are ± s to BC, EF Hyp. /. M, N are mid. pts. of BC, EF iii. 3. and, since BC>EF Hyp. /. BM>EN. /. sq. on BM>sq. on EN. But, since OB=OE Radii. /, sq. on OB=sq. on OE. Hence, sqs. on BM, OM=sqs. on EN, ON i. 47. But sq. on BM>sq. on EN Above. .*. rems sq. on OMOE. i.e. the part > the whole, which is absurd. Wherefore, if a straight line &c. Q.E.D. PROPOSITION XIX. Theorem. If a straight line touch a circle, and from the point of contact a straight line he drawn at right angles to the touching line, the centre of the circle shall be in that line. Let St. line DC touch ©ABC at C, and let C A be drawn at rt. Z. s to DC. Then shall CA pass through the cent. For, if not, if possible, let O the cent, be without AC. Join OC. Then, since DC is a tang, and OC the rad. to the pt. of contact, .', Z.DCO is a rt. L in. 18. But L DCA is a rt. z_ Hyp. /. ^DCA=z.DCO, i.e. the part=the whole, which is absurd. Wherefore, if a straight line &c. Q.E.D. PROPS. XVIII., XIX., XX. PROPOSITION XX. Theorem. 157 The angle at the centre of a circle is double of the angle at the cir- cumference standing on the same base, that is, on the saine arc. Let ABC be a 0, and let BOC, BAG be Z.s at the cent, and Oce, having the same arc BC for base. Then shall /.BOO be double of ^BAO. Case 1. — When the cent. O is on BA, or AC. In .^OAC, exf ^BOC=z.sOAC, OCA...I.32. But, since O A=OC Radii. /. Z.OAC=^OCA. Hence, Z.BOC is double of Z.OAC. Case 2. — Wlien the cent. O is within the z_BAC. Join AG, and prod. AG to meet the Oce in D. Then, as in Case 1, it may be shown that ^BGD is double of Z.GAB, and that ^ COD is double of ^GAC. /. whole L BGC is double of L BAG. Case 3. — When the cent. O is without the L BAG Join AG, and prod. AG to meet the Oce in D. Then, as in Case 1, it may be shown that L COD is double of L GAG, and that 2_B0D is double of Z.GAB. /. rem« A BGC is double of ABAC. Wherefore, the angle at the centre &c. Q.E.D. 158 EUCLID, BOOK III. PEOPOSITION XXI. Theorem. The angles in the same segment of a circle are equal to one another. Let ABCD be a 0, and let BAG, BDC be Z.s in the same segt BAG. Then shall z.BAC=aBDC. ^ ^D Gase 1, — When the seg* is > a semicircle. Find O the cent Ill.i. Join OB, OG. Then, since Z.BOG at the cent. 'S^ ^C and Z.BAG at the Qce are on the same base, /. z_ BOG is double of Z-BAG iii. 20. And, for the same reason, L BOG is double of L BDG. /. ^BAG=^BDG. Gase 2. — When the seg* is not > a semicircle. Find O the cent iii. 1. Join AO. Produce AO to meet the Qce in E. Join DE. Then, the seg* BAGE is > a semicircle, .*. Z-BAE=Z.BDE Case 1, above. Also, the seg* EBDG is > a semicircle, /. 2LEAG=/.EDG Casel. Hence, the whole Z,BAG=whole Z-BDG Ax. 2 Wherefore, the angles in the same segment &c. Q.E.D. EXERCISES. 1. ABC is a triangle inscribed in a circle whose centre is O. CD is per- pendicular to BC. Prove that angle BOD is equal to BAC. 2. If, in either figure of Prop. 21, AC and BD cut at F, then the triangle ABF is equiangular to DCF. 3. State and prove, by a reductio ad absurdum, the converse of Prop. 21. 4. In the figure of Ex. X. page 88, prove, by help of Ex. 3, that a circle will go round B, C, M, N. 6. Divide a given circle into two segments so that the angle in one segment may be double that in the other. PROPS. XXL, XXII. 159 PKOPOSITION XXII. Theorem. The opposite angles of any quadrilateral figure inscribed in a circle are together equal to two right angles. Let ABCD be a quad^ insc^ in 0ABCD. Then shall lb BAD, BOD together = two rt. lq. Join AC, BD. Then, since Z.s BAG, BDC are in the same seg*, /. ^BAC=^BDC III. 21. And, for a like reason, Z_DAC=^DBC III. 21. .'. L s BAG, DAG= Z.S BDC, DBC, i.e. z.BAD=^s BDC, DBG. To each add ^BCD, /. Z.S BAD, BCD= ^s BDC, DBC, BCD. =two rt. Z.S 1.32. Similarly, it may be shown that Z.S ADC, ABC=two rt. z.s. Wherefore, the opposite angles &c. Q.E.D. NOTE. Props. 21, 22 are very important, as are also the converse props. (See p. 177.) EXERCISES. 1. In the figure of Prop. 22, if BC be produced to E, angle DCE is equal to angle BAD. 2. If, in Prop. 22, the diagonals cut at F, the triangles ADF and BCF are equiangular. 3. If, in the figure of Prop. 22, AD and BC be produced to meet at G, the angle AGB is equal to the difference of the angles ACB and DBC. 4. In the figure of Ex. X., page 88, prove that (assuming the converse of III. 22), (i) a circle can be described about each of the quadrilaterals ANOM, BNOL, CLOM. (ii) a circle can be made to pass through the points P, B, O, C. (iii) if LN be joined, angle OBN is equal to angle OLN. (iv) if LM be joined, angle ACN is equal to angle ALM. 160 EUCLID, BOOK III. PROPOSITION XXIII. Theorem. On the same straight line, and on the same side of it, there cannot be two similar segments of circles not coinciding icith one another. For, if it be possible, on the same chd. EC, and on the same side of it, let there be two sim'' seg*^ BAG, BDC, not coinciding. Then, since the 0s cut at B and C, they cannot cut at any other pt III. lo. ,', one seg* falls within the other. D, _^ A " ■-^. B C Draw a chd. CA of the inner seg* and prod, it to meet the outer seg* in D. Join BA, BD. Then, since BAG, BDG are sim"" seg*% .•.2LBAC=z.BDC m.Def.ii. i.e. the exf^ Z. of a ^^ = iut'^ opp. L , which is impossible 1. 16. Wherefore, on the same base &c. Q.E.D. PROPOSITION XXIV. Theorem. Similar segments of circles on equal straight lines are equal. Let BAG, EDF be sim'" segt« of 0s on=st. lines BG, EF. Then shall segt BAC=segt EDF. A D- B C E F For, if segt BAG be applied to EDF so that pt. B fall on E and St. line BC lie along EF, the pt. C must fall on F, •.•BG=EF Hyp. And, since BC coincides with EF, .*. seg* BAG must coincide with EDF in. 23. /. segt BAG=segt EDF Ax. 8. Wherefore, similar segments &c. Q.E.D. PROPS. XXIII., XXIV., XXV. 161 PKOPOSITION XXV. Problem. A segment of a circle being given, to describe the circle of which it is a segment. Let ABC be the given seg*- It is req^ to complete the ©. Bisect BC at D i. lo. From Ddraw DA at rt. /_s to BC i. ii. Join AB. Then (i) if L DBA= L DAB, DB=DA 1.6. = DC CoDStr. i.e. DA, DB, DC are all equal, .'. D is the cent in. o. And, with D as cent, DB as rad., the can be completed. Again, (ii) if Z.DBA is not= Z.DAB, at pt. B in AB make L ABO= L DAB.. . .1. 23. Let BO meet AD, or AD prod^ at O. Join OC. Then in ..^b OBD, OCD, BD=CD Constr. OD is com. >t. z.BDO=rt. Z.GDO. /. OB=OC 1.4. And, since L ABO= L O AB Constr. /. OB=OA 1.6. Hence, OA, OB, OC are all equal, /. O is cent III. 9. And, with O as cent., rad, OB, the can be completed. Wherefore, a segment of a circle &c. Q.E.F. CoR. — If Z_DBA=l.DAB, the cent, lies in BC and the given seg* is a semicircle. If L DBA < L DAB, the cent, lies without the seg*, which is less than a semicircle. If L DBA > L DAB, the cent, lies within the seg*, which is greater than a semicircle. NOTE. The example to Prop. 1 gives a solution of this problem which is practi- cally simpler than the above, and is often substituted for it. (310) L 162 EUCLID, BOOK III. PROPOSITION XXVI. Theorem. In equal circles, equal angles stand on equal arcs, whether at the centres or circumferences. Let ABC, DEF be equal 0s, let BOG, EQF be equal /.s at the cents., and BAG, EDF at the Qces. Then shall arc BGC=arc EHP. B\ TC EX 7F Join BC, EF. Then in ^:\^ OBC, QEF, OB=QE III.Def.l. OG=QF III. Def. 1. ^BOG=Z.EQF Hyp. /. BC=EF 1.4. And, since z_BAG=Z.EDF. Hyp. /. seg* BAG is sim'^ to seg* EDF iii.Def.ll. But the St. lines BG, EF on which they stand are equal Above. /. segt BAG=seg* EDF iii. 24. And the whole OABG=whole ©DEF Hyp. /. rem» seg* BGG=remg seg* EHF. Hence, arc BGC=arc EHF. Wherefore, in equal circles &c. Q.E.D. NOTES. The following is another form of the enunciation of this proposition: — In equal circles, the arcs, on which equal angles stand, must be equal. Beginners are apt to confuse Props. 26, 27, 28, 29 with one another. In Prop. 26, prove arcs equal. Prop. 27 is converse of 26. In Prop. 28, prove arcs equal. Prop. 29 is converse of 28. The results proved in these four propositions for equal circles hold good for the same circle. It follows from this prop, that in any circle, or in equal circles, the greater angle stands upon the greater arc. PROPS. XXVI., XXVII. 163 PEOPOSITION XXVII. Theorem. In equal circles the angles which stand on equal arcs are equal, whether at the centres or circumferences. Let ABC, DEF be equal 0s, and BC, EF equal arcs. Then shall ^BOC=^EQF at the cents, and z.BAC=^EDP at the Oces. o \ \ ( / ^ BV >5 E^ For, if Z.BOO is not= Z.EQF, one must be >. If possible, suppose z.BOC> Z.EQF. AtOin OB make 2LB0G=Z.EQF 1.23. Then, if Z.BOG=Z.EQF, arc BG=arc EF III. 26. But arc BC=arc EF Hyp. /. arc BG=arc BC. i.e. the part=the whole, which is absurd. Hence, L BOC cannot be unequal to L EQF, i.e. z.BOC=^EQF. /. also z.BAC=Z-EDF in. 20. Wherefore, in equal circles &c. Q.E.D. EXERCISES. 1. Prove Prop. 26 by the method of superposition. 2. Prove Prop. 27 by the method of superposition. 3. AB, AC, AD are chords of a circle; if the angle BAC is the supplement of the angle BAD, then the sum of the arcs on which they stand is equal to the whole circumference. 164 EUCLID, BOOK III. PKOPOSITION XXVIII. Theorem. 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 ©s, and BC, EF equal chds. Then shall the major arc BAC=the major arc BDF, and the minor arc BC=the minor arc EP. Find O and Q the cents, of the 0s III. l. Join OB, OC, QE, QF. Then, in ^^s OBC, QEF, COB=QE in.Def.i. - V jOC=QF III.Def. 1. tBC=EF Hyp. /. ^BOC=z.EQF.... 1.8. Hence, arc BC=arc EF ill. 26. But, whole Oce ABC=whole Qce DEF Hyp. /. rems arc BAC=reme arc EDF. Wherefore, in equal circles &c. Q.E.D. NOTE. The enunciation may be stated thus : — In any circle, or in tico equal circles, the arcs cut off hi/ equal chords must be respectively equal to one another: the major to the major, and the minor to the minor. EXERCISES. 1. Prove Prop. 28 by superposition. 2. APC, BPD are equal chords of a circle ABCD ; prove that the arc AB is equal to the arc CD. 3. The arc AB is equal to the arc CD in the circle ABCD ; prove that the chord AD is parallel to BC. PROPS. XXVIII., XXIX. 165 PKOPOSITION XXIX. Theorem. In equal circles, equal arcs are subtended hy equal straight lines. Let ABC, DEF be equal 0s, and BC, EF equal arcs, and let BC, EF be joined. Then shall chd. BC=chd. EF. Find Oand Q the cents, of the 0s iii. i. Join OB, OC, QE, QF. Then, since arc BC=arc EF Hyp. /. Z.BOC=Z.EQF 111.27. Hence, in .^s OBC, QEF, OB = QE III. Def. 1. OC=QF III. Def. 1. Z_BOC=Z.EQF Above. /. BC=EF 1.4. Wherefore, in equal circles &c. Q.E.D. NOTE. The enunciation might be stated thus: — In any circle, or in equal circles, the chords, which cut off equal arcs, must he equal. EXERCISES. 1. Prove this prop, by superposition. 2. The straight lines which join the extremities of two parallel chords of a circle are equal to one another. 3. In the circle ACBD, the arc ACB is equal to the arc CBD ; prove that the triangle ACB is equal to the triangle CBD. 166 EUCLID, BOOK III. PKOPOSITION XXX. Problem. To bisect a given arc of a circle. Let BAG be the given arc. ^ It is req^ to bisect it. Join BC. Bisect BC at D L 10. From D draw DA at rt. Z_ s B D C to BC, and meeting the Qce at A... I. ii. Then shall arc BAG be bisected at pt. A. Join AB, AC. Then in .^s ABD, ACD, BD=CD Constr. AD is com. ^rt. Z.ADB=rt. ^ADC. /. chd. AB=chd. AC 1.4. Now AD, if prod*^ is a diam iii. i. Cor. Hence AB and AC are both minor arcs. .'. arc AB=arc AC III. 28. Wherefore, the given arc &c. Q.E.F. PEOPOSITION XXXI. Theorem. In a circle, the angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. Let ABCD be a 0, BC a diam. and AC a chd. Then shall (i) L in semicircle BAG be a rt. L; (ii) Z- in > segt aBG be < a rt. L ; (iii) z. in < segt ADG be > a rt. z.. PROPS. XXX., XXXI. 167 Find O the cent I. lo. Join BA, AD, DC, AO. Then (i) since OA=OB Rad. /.Z.OAB=Z-OBA 1.6. And, since OA=OC Rad. /. z.OAC=^OCA 1.5. /. L s OAB, OAC= L s OBA, OCA. i.e. L BAC= L s OBA, OCA. Butz.sBAC,OBA,OCA=twort.^s 1.32. /. Z.BAC is art. Z.. Again, (ii) Z-sBAC, ABCart. Z.. Wherefore, the angle in a semicircle &c. Q.E.D. Cor. — If one L of a ..^ be equal to the other two, it is a rt. z. . EXERCISES. 1. If, in a circle, one angle is the complement of another, the sum of the arcs on which they stand is half the circumference. 2. Prove that, if any other point E be taken in the arc BAC (prop. 30), the area of the triangle EEC is less than the area of the triangle ABC. 3. Prove, by a reductio ad absurdum, that if a semicircle be described on the hypotenuse of a right-angled triangle as diameter, it will pass through the right angle. 4. What is the locus of the vertices of all right-angled triangles described on the same hypotenuse? 5. ABC is a triangle; D is the middle point of BC; L, M, N are the feet of the perpendiculars from A, B, C on the sides; prove that (i) the circle, centre D and radius DB, will pass through M and N; (ii) the triangle LMN has its angles each double of the complements of those of ABC respectively. 6. The greatest rectangle that can be inscribed in a circle is a square. 7. Draw a straight line at right angles to another straight line, from its extremity, without producing the given line. 166 EUCLID, BOOK III, B D PEOPOSITION XXX. Problem. To bisect a given arc of a circle. Let BAG be the given arc. ^ It is req^ to bisect it. Join BC. Bisect BC at D L 10, From D draw DA at rt. L s to BC, and meeting the Qce at A... I. ii. Then shall arc BAG be bisected at pt. A. Join AB, AC. Then in .^s ABD, ACD, BD=CD Constr. AD is com. ^rt. Z.ADB=rt. ^ADC. .-. chd. AB=chd. AC 1.4. Now AD, if prod"^ is a diam III. i. Cor. Hence AB and AC are both minor arcs. /. arc AB=arc AC 1X1.28. Wherefore, the given arc &c. Q.E.F. PEOPOSITION XXXI. Theorem. In a circle, the angle iti a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. Let ABCD be a 0, BC a diam. and AC a chd. Then shall (i) L in semicircle BAG be a rt. z. ; (ii) Z- in > segt ABC be < a rt. L ; (iii) z. in < segt ADG be > a rt. z.. PROPS. XXX., XXXI. 167 FindO the cent I. lo. Join BA, AD, DC, AO. Then (i) since OA=OB Rad. .•.Z.OAB=Z.OBA 1.5. And, since OA=OC Rad. .-. z.OAC=^OCA 1.5. /. L s OAB, OAC= L s OB A, OCA. i.e. ^BAC=^s OBA, OCA. Butz_sBAC,OBA,OCA=twort.^s 1.32. /. Z.BAC is art. Z.. Again, (ii) z. s BAC, ABC < two rt. /.s 1. 17. But I,BAC is a rt. L Above. .*. z. ABC < art. L. Also, (iii) since ABCD is a quad^ inscribed in the O, .*. Ls ABC, ADC=two rt. Z-s iii. 22. But Z. ABC < art. L Above. /. Z_ADC>art. ^. Wherefore, the angle in a semicircle &c. Q.E.D. CoR. — If one L of a .-::\ be equal to the other two, it is a rt. z. . EXERCISES. 1. If, in a circle, one angle is the complement of another, the sum of the arcs on which they stand is half the circumference. 2. Prove that, if any other point E be taken in the arc BAG (prop. 30), the area of the triangle EBC is less than the area of the triangle ABC. 3. Prove, by a reductio ad absurdum, that if a semicircle be described on the hypotenuse of a right-angled triangle as diameter, it will pass through the right angle. 4. What is the locus of the vertices of all right-angled triangles described on the same hypotenuse? 5. ABC is a triangle; D is the middle point of BC; L, M, N are the feet of the perpendiculars from A, B, C on the sides; prove that (i) the circle, centre D and radius DB, will pass through M and N; (ii) the triangle LMN has its angles each double of the complements of those of ABC respectively. 6. The greatest rectangle that can be inscribed in a circle is a square. 7. Draw a straight line at right angles to another straight line, from its extremity, without producing the given line. 170 EUCLID, BOOK III. NOTES. The following construction for Prop. 33 is practically simpler than that in the text: — Let AB be the given line, C the given Z.. At A make Z.BAD=Z.C 1.23. Draw AE at rt. Z-S to AD 1. 11. From B draw BE at rt. L s to AB, meeting AE at E 1. 11. On AE as diam. desc. a 0. Then it can be proved, as in Ex. 3 on III. 31, that B is on the Qce, and the rest of the proof follows as before. EXAMPLE. To construct a triangle, having given the base, the vertical angle, and the sum of the sides. Let AB be the given base, C the given vertical L , D the given sum of sides. Bisect lG 1. 9. On AB desc. a seg* containing an Z.=half Z_C III. With cent. A and rad.=D, desc. an arc cutting the seg* at E. Join AE, EB. At B make Z.EBF= Z_ AEB, and let BF meet AE at F. Then FAB shall be the .^A reqC From P draw PL x to BC and PN ± to AB. Join LN, and prod. LN to meet AC prod^ at M. Then PM shall be ± to AC. Since PNB, PLB are rt. Z_s, .*. a O can be desc^ round PBLN Ex. IV. /. Z_PBN=^PLN m.2i. i.e. 2LPBA=^PLM. But z_PBA=z.PCM in.2i. .'. Z.PLM=Z.PCM, /, the pts. P, L, C, M lie on a Ex. iv. And, since PLC is a rt. Z. , .*. PMC is also a rt. Z_ ill. 22. Hence L, M, N, the feet of the J. s, lie in a st. line. Q.E.D. The above theorem is sometimes referred to as Simson^s Theorem, and the line LNM as Simson's line. 180 EUCLID, BOOK III. VII. To describe a circle which shall touch a given circle, and a given straight line at a given point. Let P be the given pt., PQ the given st. line, and ABC the given 0, cent. 0. C /^ R\ ^^^^^^^T^ A ^^ \ / ^.. \ ,' Draw OQ i. to PQ, and let OQ prod^ meet the Qce in C. Join CP cutting the Qce in B. Join OB, and from P draw PD at rt. Z_s to PQ and meeting OB prod'' inD. Then D shall be the cent, of the reqd- For, since the /. s at P, Q are rt. L s, /. PDislltoQC 1.28. /. ADPB=alt. ^OCB 1.29. = ^OBC 1.6. = Z.DBP 1.16. .-. DP=DB 1.6. Hence, the desc'i with cent. D and rad. DP will pass through B. Also, since PQ is at rt. Z. s to rad. DP. Constr .*, PQ touches this 111. 16, Cor. And, since OD joining the cents, passes through B Constr. ,*. 0ABC touches this III. 12. Q.E.F. MISCELLANEOUS EXAMPLES. 181 VIII. To draw common tanyents to two given circles. Let O be the cent, of the >, Q of the < 0. Cask 1. For the transverse pair. With cent. 0, and rad.=sum of radii of 0s O and Q, desc. a 0. From Q draw QA a tany. to this 0. Join OA, cutting the Oce of the given in B. From Q draw QC at rt. Z_s to QA, and join BO. BC shall be a com. tang-. For, since QA is a tang, to 0AA', .*, QAO is a rfc. L in.ie.Cor. But AQC is also a rt. L Constr. /. AB is II to QC I 28 Also AB=QC Constr. ,*, QA is=and II to BC 1.3.3. i.e. QCBA is a r~7 , .'. Z.QCB=^QAB L34. =a rt. L Above. and Z_OBC=^QAB L29. =art. Z_, ,*, BC is a tang, to both 0s in. le.Cor. Case 2. For the direct pair. p With cent. O, and rsid.=diff. of radii of 0s and Q, desc. a 0, and com- plete the construction and proof as before. 182 EUCLID, BOOK III. IX. To describe a circle which shall pass through two given points and touch a given circle. Let P, Q be the given pts., ABO the given 0. Through P, Q describe any cutting ABC in B, C. Join PQ, BC and let them be prod*! to meet at D. From D draw DA a tang, to 0ABC il. ir. Desc. a to pass through P, Q, A ill. i, Ex. PQA shall be the reqd 0. For, since DPQ, DBG cut 0QPBC, /. rect. DP, DQ=rect. DB, DC ill. 36, Cor. And, since DBC cuts, and DA touches 0ABG, ,*. rect. DB, DC=sq. on DA III.36. /. rect. DP, DQ=sq. on DA. /. DA touches QPA at pt. A iii.ar. Hence, since 0s QPA, ABC have a com. tang, at A, ,*. the 0s touch at A. Q.E.F. MISCELLANEOUS EXAMPLES. 183 X. To find the locus of a point from which the tangents to two circles are equal. P Case 1. When the ©s intersect. Let ABC, DEC be the ©s. Take any pt. P in their com. chd. BC prodd, and draw PA, PD tangs, to the ©s. Then, in ©ABC, rect. PB, PC=sq. on PA...in. 36. And, in ©DBC, rect. PB, PC=sq. on PD .III. 36. sq. on PA=:sq. on PD. .•. PA=PD. Hence the prod^ com. chd. is the req^ locus. Case 2. When the ©s do not intersect. Find O, Q the cents m.i. Join OQ. Divide OQ at N so that the diff. of the sqs. on ON and QN may=diff. of sqs. on the radii. (See Ex. VII. p. 124, and Ex. I. p. 176.) DrawNPatrt. Z-stoOQ i.ii- PN shall he the req^ locus. Take any pt. P in PN, Draw PA, PD tangs, to the ©s llLir. Join OP, OA, QP, QD. Then, PA2=OP2-OA2 L 47. =PN2 + ON2-OA2 Lit And PD2=QP2 - QD2 i.4r. =PN2 + QN2-QD2 L47. But ON2-QN2=OA2-QD2 Constr .'. 0N2 - OA2=QN2 - QD2. /. PN2 + 0N2 - OA2=PN2 + QN2 - QD^, i.e. PA2=PD2 Above. /. PA=PD. Hence the st. line PN is the req a diam"". It is req^ to place in QABC a chd.=D. Draw any diam. AB iii. i. Then, if AB=D, what was req*^ is done. If not, from AB cut off AE=D...i. 3. With cent. A, rad. AE, desc. a EOF cutting ABC at C. Join AC. Then shall AC=D. For, since A is cent, of 0ECF, /. AC=AE Rad. = D Constr. Wherefore, in the given circle &c. > Q.E.F, NOTES. Book IV. consists entirely of problems connected with the circle. Props. 1 to 5 deal with triangles; props. 6 to 9 with squares; and props. 11 to 16 with the inscription and circumscription of regular polygons in and about a circle, and vice versd. The polygons dealt with by Euclid are the pen- tagon, hexagon, quindecagon. The octagon, nonagon, decagon, dodecagon, &c., may also be dealt with by Euclid's methods; but it is not possible, with ruler and compasses alone, to inscribe or describe a regular seven-sided, or eleven-sided, polygon in or about a circle. EXERCISES. 1. Construct a right-angled triangle having given the hypotenuse and one side. 2. About a given straight line describe the smallest possible circle. 3. Construct a right-angled triangle whose hypotenuse shall bear to one of its sides the ratio of 3 to 1. 4. The radius of a circle is 2 inches; how many lines, each 2 inches long, can be placed in succession in the circle ? 5. Through a given point within a circle draw a chord of given length. PROPS. I., II. 195 ■ PROPOSITION II. Problem. In a given circle to insci'ibe a triangle equiangular to a given triangle. Let ABC be the given 0, and DEF the given ^^. It is req^ to inscribe in QABC a .^\ equiang^ to DEF. Draw GAH to touch the at any pt. A Iii. 17. At pt. A in AG make Z.GAB=z.DFE I. 23. At pt. A in HA make Z_HAC=Z-DEr I. 23. Join BC. Then shall ABC be the ^^ reqd. For, since GAH is a tang., and AB a chd. from its pt. of contact A Constr. .*. Z.GAB = z_ ACB in the alt. seg<^ iii. 32. But ^GAB=Z.DFE Conatr. .•.Z.ACB=z.DFE. Similarly, it may be shown that ^ABC = Z.DEF. Hence, rera^ z.BAC=rem8 Z.EDF i. 32. Wherefore, in the given circle &c. Q.E.F. EXEECISES. 1. In a given circle inscribe (i) an equilateral triangle; (ii) a right-angled isosceles triangle; (iii) an isosceles triangle which shall have each of the angles at the base equal to one -sixth of the vertical angle. 2. An equilateral triangle is inscribed in a circle whose radius is 3 inches; find the lenirth of its sides. 196 EUCLID, BOOK IV. PEOPOSITION HI. Problem. About a given circle to describe a triangle equiangular to a given triangle. Let ABC be the given 0, and DEF the given ^^.. It is req^ to describe about QABC a .^^ equiang"" to DEF. K 1^ B M Produce EF both ways to G, H. Find O the cent, of 0ABC III. i. Draw any rad. OB. At pt. O in OB make L B() A = L DEG, and on the other side of OB make Z- BOG = Z_DFH I. 23. At pts. A, B, C draw tangs, to the ill. 17. (These tangs, will, if prod*^, meet one another; for, if the tangs, at A and B do not meet they are 1| , and then OA, OB at rt. Z_s to them are || I. 29 (Ex. 5). which is impossible, since they cut at O.) Let K, L, M be the pts. at which the tangs, meet. Then shall KLM be the .^ reqd. Join OL. Then, in ^\ OAL, Z.S ALO, AOL, OAL=two rt. z_s l. 32. But OAL is a rt. L Constr. .*. Z_s ALO, AOL=a rt. L. Similarly, ^s BLO, BOL=a rt. L. Hence, z.s ALB, AOB=two rt. Z.s Ax. 2. But Z_s DEF, DEG=two rt. ^s 1. 13. /. Z.S ALB, AOB=^s DEF, DEG. But ^AOB=Z.DEG Constr. /. remsf z_ALB=reme Z.DEF. Similarly, it may be shown that z_BMC= z_DFE. /. the third Z_ AKC=the third Z.EDF ...I. 32. Wherefore, about the given circle &c. Q.E.F. PROP. III. 197 NOTES. The proof that the tangents will, if produced, meet is not given by Euclid. The following is another construction for the problem: — Let ABC be the 0, DEF the .:^. Draw any tang. GBH to the Ill, ir. Atany pt. Gin BG make Z.BGK=Z.DEF 1.23. At any pt. H in BH make Z_BHL=Z.DFE L 23. From O, the cent., draw ±s OK, OL to GK, HL 1. 12. and let these ±s, prod^ if necessary, cut the Qce in M, N. Through M, N draw tangs, to the and let these tangs, cut one another at P and the tang, at B in Q, R. Then PQR shall be the reqd. ^^. EXERCISES. 1. Prove the construction given above. 2. Prove the following construction for Prop. 3: — In the circle inscribe a triangle equiangular to the given triangle (Prop. 2), and from the centre draw radii perpendicular to the sides of the inscribed triangle; the tangents at the ends of these radii will, by their intersections, form the required triangle. 3. About a given circle circumscribe (i) an equilateral triangle; (ii) a right-angled isosceles triangle. 4. If an equilateral triangle be circumscribed about a circle, the radius of which is 8 inches, find the length of its sides. 5. About a circle describe a trapezium equiangular to a given trapezium. 6. An equilateral triangle whose side is a circumscribes a circle; find the length of the radiiis. 810) N2 198 EUCLID, BOOK IV. PEOPOSITION IV. Problem. To inscribe a circle in a given triangle. Let ABC be the given ^^. It is req'^ to inscribe a © in it. A Bisect the ^s ABC, ACB by BO, CO meeting at O i. 9. From O draw OD, OE, OF ±s to BC, AC, AB 1. 12. Then, in .^s OBD, OBF, L OBD=Z.OBF Constr. rt. L ODB=rt. Z.OFB, OB is com. /. OD=OF L 26 (ii). Similarly, it may be proved that OD=OE. Hence OD, OE, OF are all equal. .', the desc*^ with cent. O and rad. OD will pass through E and F; let this be desc<^. Then, since the z. s at D, E, F are rt. L s, .*. the touches BC, AC, AB in. lecor. Q.E.F. EXAMPLES. I. To describe a circle which shall touch one side of a triangle and the other tvx> sides produced. Let the sides BA, BC of ^^ ABC be prod a semicircle, and is < a rt. L ill. 3i. When the cent, falls on a side of the -=::::\, the L opp. to this side is in a semicircle, and is a rt. L III. 31. When the cent, falls without the .^r^, the L opp. to the side beyond which the cent, lies, is in a seg*^ < a semi- circle, and is > a rt. L III. 81. PROP. V. 201 And, conversely, if the ^^^ is acute-angled, the cent, of the falls within it; if the ^^ is right-angled, the cent, falls on the hypot.; if the ^^ is obtuse-angled, the cent, falls beyond the side which subtends the obtuse angle. NOTE. Props. 4 and 5, and Example I. on Prop, 4, are very important. For particular cases of these props, see pages 86, 89, 94 {Ex. 71), 135 [Note and Ex. 1). EXAMPLE. If is the centre of the circle circumscribing the triangle ABC, and if E, F are the feet of the perpendiculars from B, C on the opposite sides, then OA cuts EF at right angles. Let OA and EF cut at K. A Then shall AKE he art. L. For, since OA=OC Rad. .-. ^OAE=Z.OCA 1.5. X /\^S^E /.Z-AOC, and twice Z.OAE =two rt. Z.S 1.32. But Z.AOC at cent, is double of Z.ABC at Qce, .*. twice Z_s ABC, OAE=two rt. Z.s. /. Z_s ABC, OAE=:a rt. L. But, since each of the L s BFC, BEC is a rt. Z. , .*. a ©'will go round B, F, E, C Ex. IV. page 177. /. i_AEF=Z_FBC Ex. I. page 159. Hence Z.s AEF, OAE=a rt. L. .*. theremg Z.AKE of .-^ AKE is art. L 1.32. Q.E.D. EXERCISES. 1. If the centres of the inscribed and circumscribed circles of a triangle coincide, the triangle is equilateral. 2. The diameter of the circle circumscribing an equilateral triangle is double of the diameter of the inscribed circle. 3. If, in the figure of IV. 5, OF be drawn perpendicular to BC, then BC is bisected at F. 4. If, in the figure of IV. 5, OD be produced to meet the circumference in G, then the angle GCB is half of the angle ACB. 5. If the perpendiculars from the angles A, B, C of a triangle ABC to the opposite sides be produced to meet the circumference of the circum- scribed circle in D, E, F respectively, then the arcs EF, FD, DE are bisected at A, B, C respectively. 6. In the figure of the example above, prove that, if D be the foot of the perpendicular from A on BC, then ED is at right angles to OC. 202 EUCLID, BOOK IV. J^EOPOSITION VI. Problem. To inscribe a square in a given circle. Let ABCD be the given 0. It is req^ to inscribe a sq. in it. A C Find O the cent in. \. Draw the diams. AOC, BOD at rt. Z.s to each other l ii. Join AB, BC, CD, DA. Then ABCD shaU be the sq. reqd. For, in .^s AOB, COB, ( AO=CO Rad. •,• -| OB is com. irt. L AOB=rt. L COB Constr. /. AB=CB L4. Similarly, BC=CD, CD=DA. Hence, fig. ABCD is equilat. And, since AC is a diam. /. Z. ABC in a semicircle is a rt, L in. 3i. Hence, fig. ABCD is a sq i. 46 Cor. Wherefore, has been inscribed &c. Q.E.F. EXERCISES. 1. The area of the inscribed square is double of the square on the radius of the circle. 2. If any points E, F, G, H be taken in the arcs AB, BC, CD, DA respec- tively, the sum of the angles AEB, BFC, CGD, DHA is six right angles. 3. If the radius of the circle be 5 inches, find the length of the side of the inscribed square. 4. In a given circle inscribe a regular octagon. 5. If a regular octagon be inscribed in a circle of 3 inches radius, find the length of one of its sides. PROPS. VI., VII. 203 PEOPOSITION VII. Problem. To describe a square about a given circle. Let ABCD be the given 0. It is req^ to desc. a sq. about it. ffi H Find O the cent III. i. Draw the diams. AOC, BOD at rt. Z-S to each other I. ii. At A, B, C, D draw tangs, to the O, cutting at E, F, G, H III. ir. Then EFG-H shaU be the reqd sq. For, since EF and HG are tangs. .*. Z.S at B and D are rt. Z.s III. 18. But Z. s at O are also rt. Z. s Constr. Hence, EF, AC, HG are || I. 29. Similarly, EH, BD, FG are || . .', all the quad^^ in the fig. are /~7s. Hence, EH=FG, and EF=HG i. 34. Also, EH=BD I. 34. =AC, =EF 1.34. /. the fig. EFGH is equilat. And, since EBOA is a n, :. z.AEB=Z-AOB 1.34. =a rt. L Constr. /. the fig. EFGH is rectangular l. 46 Cor. Hence EFGH is a sq. Wherefore, about the given circle &c. Q.E.F. EXEKCISES. 1. Prove that the tangents at A and D must meet each other. 2. The square described about, is double that inscribed in, a circle. 3. About a given circle describe a rhombus two of the sides of which shall include a given angle. 4. If, in the figure of this proposition, HF be joined, prove that HF passes throucjh O. 204 EUCLID, BOOK IV, PKOPOSITION VIII. Problem. To inscribe a circle in a given square. Let ABCD be the given sq. It is req^ to inscribe a i?i it. A E D H O B G C Bisect AD, AB at E, F i. lo. Through E, F draw EG, FH || to AB, AD and cutting at 0...I. 3i. Then all the quad^^ in the fig. are Os. Hence AE=FO, and ED=OH I. 34. But AE=ED Constr. /. FO=OH. Similarly, EO=OG. And, since AD=AB Hyp. and E, F are their mid. pts Constr. /. AE=AF. Hence, OE, OF, OG, OH are all equal. With cent. O, rad. OE, desc. a 0; this will pass through the pts. F, G, H. And, since EG is || to AB, /. z.DEO=z.DAB 1.29. ^a rt. L Constr. Similarly, the Z. s at F, G, H are rt. L s. /. the EFGH touches the sides of the sq. "Wherefore, in the given sq. &c. Q.E.E EXERCISES. 1. Inscribe a circle in a given rhombus. 2. Prove that, if in the figure above, AO, DO be joined, ADD is a right angle. 3. Can a circle be inscribed in a rhomboid? PROPS. VIII., IX. 205 PROPOSITION IX. Problem. To describe a circle about a given square. Let ABCD be the given sq. It is req^ to desc. a aboiit it. Join AC, BD, cutting at O. Then, in .=^s ABC, ADC, AB=AD Hyp. AC is com. BC=DC Hyp. .-. 2.BAC=Z.DAC 1.8. =half ^BAD, =half a rt. L Hyp. Similarly, L ABD is half a rt. L. /. ^BAO=Z.ABO. /. AO=BO I. 6. Similarly, BO=CO, CO=DO, DO=AO. Hence, AO, BO, CO, DO are all equal. With cent. O, rad. OA, desc. a O; this will pass through the pts. B, C, D, and circumscribe the sq. Wherefore, about the given sq. &c. Q.E.F. EXEKCISES. 1. Describe a circle about a given rectangle. 2. Describe, when possible, a circle about a given quadrilateral. When is this impossible? 3. If the side of a square be 3 inches long, find the length of the radius of the circumscribed circle. 206 EUCLID, BOOK IV. PEOPOSITION X. Problem. To describe an isosceles triangle having each of the angles at the base double of the third angle. B^^ "D Take any st. line AB. Divide AB at C so that rect. AB, BC=sq. on AC li. ii. With cent. A and rad. AB desc. a ©BDE. In 0BDE place chd. BD=AC iv. i. Join AD. Then ABD shall be the .-^ reqd. Join DC. About .^ACD desc. a ©ACD iv. 5. Then, since rect. AB, BC=sq. on AC Constr. =sq. on BD Constr. /, BD touches the ©ACD in. 37. And, since DC is drawn from the pt. of contact D, .*. L BDC= L DAC in alt. segt iii. 32. A.dd L CDA to each, /. whole z.BDA=z.s DAC, CDA. But ^DCB=z_sDAC, CDA 1.32. .•.Z.BDA=aDCB. But, since AD=AB Rad. .•.Z.BDA=^ABD 1.5. Hence L DCB= L ABD. /. CD=BD L6. ^ AC Constr. .•.Z-CDA=z.DAC 1.5. Hence Z-DCB is double of Z.DAC. But Z-BDAand Z. ABD each = Z-DCB Above. /. L BDA and L ABD are each double of L DAC. Wherefore, ha^ been described &c. Q.E.F. N.B. — This prop, is a very important one. PROP. X. 207 EXAMPLES. I. Find the number of degrees in each of the angles of the triangle in IV. 10. Since the three I.s of ^:lABB=two rt. Z.s 1.32. and each of the Z.s at B and D is double of I. A, .*. Z. A is one-fifth of two rt. L s, or L A=| rt. Z. = I X 90°= 36°. Also, Z-B and LD each=| rt. L^IT. Q.E.D. II. Describe an isosceles triangle having each of the angles at the base equal to three-quarters of the vertical angle. [Since the three Z. s of a -"^^ = 2 rt. ^ s, the req'i -ii^will be such that its vertj L with twice | of its vert' Z_ = 2 rt. /. s, i.e. ^ ofitsverti Z- = 2rt. Z.s. .', its vert' Z- = | of a rt. L, which is the magnitude of one of the base 2_ s in IV. 10. From the above analysis (as it is called) of the problem we arrive at the following construction:—] Make an isos. ^^ABC having each of the Z-S at B and C double of the L at A IV. lO. From BA cut off BD=BC 1.3. A Join DC. Then BOD shall be the reqd ^^. For, since BC=BD Constr. ^^ .\ Z_BCD = Z.BDC 1.5. But Z.DBC=4rt. Z. = -|of 2 rt.Z.S Ex. I. above. /.Z-S BCD, BDC together = | of 2 rt. Z. s . . . I. 32. /. L BCD=| of a rt. Z. = I of Z. DBC. B C Q.E.D. EXERCISES. 1. Construct angles of (i) 18°, (ii) 54°, (iii) 126°, (iv) 12°. 2. Divide a right angle into (i) five, (ii) fifteen, equal parts. 3. Describe an isosceles triangle having each of the angles at the base equal to one-eighth of the vertical angle. 4. In the figure of IV. 10, (i) If DE be joined DE=BD. (ii) What portion of the circumference of the smaller circle does the larger circle intercept? (iii) If O be centre of circle ACD and the diameter COF be drawn, prove that the triangle CEF has one of its angles four times the size of another, (iv) If AE meets BD produced in F, prove that FAB is another isosceles triangle of the same kind, (v) Arc BD is one-tenth of the whole circumference. Xvi) If AE meets BD produced at F, then CDFE is a parallelogram. "jiOa EUCLID, BOOK IV. PEOPOSITION XI. Problem. To inscribe an equilateral arid equiangular pentagon in a given circle. Let ABODE be the given 0. * It is req^ to insc. a reg. ^pentagon in it. A Make an isos. ^^FGH having z. s at G, H each double of L F...IV. lo. In ABODE insc. a .^AOD equiang"^ to .^FGH iv. 2. having /_ at A= z. at F. Bisect Z.S AOD, ADO by CE, BD meeting the Occ at B, E....I. 9. Join AB, BO, DE, EA. Then ABODE shall be the pentagon reqd For, since Z.s AOD, ADO are each double of Z.OAD Constr. and are bisected by OE, BD Constr /. the Z_s ADB, BDO, OAD, DOE, EOA are all equal. /. the arcs AB, BO, OD, DE, EA are all equal iii. 26. /. thechds.AB, BO, OD, DE, EA are all equal m. 29. /. pentagon ABODE is equilat^ Again, arc AB=arc DE Above. Add arc BOD to each, /. whole arc ABOD=whole arc BODE. .*. L AED which stands on arc ABOD= L BAE on arc BCDE...III. 27. Similarly, each of the Z.s ABO, BOD, ODE= Z.BAE. /. pentagon ABODE is equiang^. Hence, ABODE is a reg. pentagon. Wherefore, an equilateral &c. Q.E.F. EXERCISES. 1. Find, in degrees, the magnitude of the angle subtended at the centre by a side of the regular pentagon. 2. If all the diagonals of the pentagon be drawn, prove that the diagonals form, by their intersections, another regular pentagon. PROPS. XL, Xll. 209 PKOPOSTTION XII. Problem. To describe an equilateral and equiangular pentagon about a given circle. Let ABODE be the given 0. L It is req^ to desc. a reg. pentago7i .^>-^-^-<:^ about it. A^^Z _^ In ABODE inscribe a reg, \ \ xv- pentagon ABODE IV. \\. ^^ ' ' ^ ^ ^*^ At the pts. A, B, 0, D, E draw tangs, to the cutting at t\- -*P) r,G,H,K,L m. 17. Then FGHKL shall be the pentagon reqd. q — ^ ^ — II Join BD. Then, since OB=CD Constr. .•.aOBD=z.ODB 1.5. And, since GH is a tang., and OD a chd. from the pt. of contact 0, /.Z.HOD=Z.OBD in alt. seg* m. 32. Also, since HK is a tang., and DO a chd. from the pt. of contact D, .•.Z-ODH=Z.OBD in alt. seg* in. 32. .\Z.HOD=^HDO. /. HO=HD 1.6. Similarly, GO=GB. Again, since GH is a tang, and OB a chd. from the pt. of contact 0, /. L GOB= L ODB in alt. seg* III. 32. = Z_OBD Above. = Z.HOD Above. Hence, in ^^^ BGC, DHO, AGB0=Z.HD0 Above. Z.GOB=A.HOD Above. OB =0D Constr. .•.CG=CH ) ^^^ and L at G= /_ at H ) Similarly it may be shown that any other side FG is bisected at the point of contact B, and that Z. at F= Z_ at G, and so on. Hence pentagon FGHKL is equiang^ Also, since GO=GB Above. /. GH=GF Ax. 6. Similarly it may be shown that GF=FL, and so on. Hence pentagon FGHKL is equilat^ /. FGHKL is a reg. pentagon. Wherefore, an equilateral &c. Q.E.F. 210 EUCLID, BOOK IV. PEOPOSITION XIII. Problem. To inscribe a circle in a give7i equilateral and equiangular pen- tagon. Let ABCDE be the given reg. pentagon. It is req^ to insc. a O in it. A C N D Bisect LB BCD, CDE by CO, DO meeting at O i. 9. Join OB. From O draw ±s OM, ON to BC, CD 1. 12. Then, in ^^s OBC, ODC, BC=CD Hyp. OC is com. LBCO^LBCO.. Constr. .•.^CBO=z.CDO 1.4. . =half ^CDE Constr. =half ^CBA Hyp. Similarly, if OA, OE be drawn, it may be shown that they bisect the z_ s at A and E. Again, in .^rXs OCM, OCN, ^OCM=Z.OCN Constr. ^OMC=^ONC Ax. 11. OC is com. .-. OM=ON I. 26. Similarly, if ± s be drawn from O to B A, AE, ED they may be shown to be each equal to OM. With cent. O and rad. OM desc. a ; then since this passes through the feet of the other J_ s, ,'. it touches the sides of the pentagon in. I6. i.e. a is inscribed in the pentagon .....iv. def. 5. Wherefore, a circle &c. Q.E.F. PROPS. XIII., XIV. 211 PROPOSITION XIV. Problem. To describe a circle about a given equilateral and equiangular pentagon. Let ABODE be the given reg. pentagon. It is req^ to desc. a about it. A c^^:r ^^-^D Bisect Z_s BCD, ODE by CO, DO meeting at O i. 9. Join OB. Then, in -^s OBC, ODC, BC =CD Hyp. OC is com. Z_BCO=Z.DCO Constr. .•.Z.CBO=Z-CDO 1.4. =half Z.CDE Constr. =half Z.CBA Hyp. Similarly, if OA, OE be drawn, it may be shown that they bisect the z_ s at A and E. And, since Z-CBO=half one of the Z_s of the pentagon... Above, and Z-BCO=half one of the Z_s of the pentagon... Constr. .•.Z.CBO=^BCO Ax.r. /. OB=OC 1.6. Similarly it may be shown that OA, OE, OD are each=OC. With cent. O, and rad. OC desc. a ; then this passes through B, A, E, D, and /. circumscribes the pentagon iv. def. ( Wherefore, a circle &c. Q.E.F. EXERCISES. 1. Prove that a circle can be described about any regular polygon. 2. Prove that a circle can be inscribed in any regular polygon. 3. If any regular polygon be inscribed in a circle, the tangents drawn to the circle at its points of contact form a regular polygon. 212 EUCLID, BOOK IV. PKOPOSITION XV. Problem. To inscribe an equilateral and equiangular hexagon in a given circle. Let ABC be the given 0. It is req'^ to insc. a reg. hexagon in it. Find O the cent ill. i. Draw the diam. AOC. With cent. C, rad. CO desc. 0BOD cutting 0ABC at B, D Draw the diams. BOF, DOE. Join AE, EB, BC, CD, DF, FA. Then ABBGDF shall be the hexagon req*^. For, since ^^^ DOC, BOC are equilat Constr. /. each of the /.s DOC, BOC=one-third of two rt.z_s i. 32. /. whole Z.DOB= two-thirds of two rt. z_s. /. rem« Z.DOF= one-third of two rt. Z.s I. 13. Hence, Z.S FOA, AOE, EOB each = one-third of two rt.Z.s 1. 15. i.e. the six z_s at O are all equal. /, the six arcs on which they stand are all equal III. 26. /. the six chds. AE, EB, BC, CD, DF, FA are all equal... iii. 29. i.e. the hexagon is equilat. Again, since the sum of the four arcs AE, EB, BC, CD =the sum of the four arcs EB, BC, CD, DF Ax. 2. .•.Z_DFA=^FAE III. 27. Similarly, z.FAE= Z.EBC, and so on. i.e. the hexagon is equiang. Wherefore, AEBCDF is a reg. hexagon. Q.E.F. Cor. — The side of a reg. hexagon insc*^. in a is equal to the rad. EXERCISES. 1. Opposite sides of a regular hexagon are parallel. 2. Inscribe a circle in a given regular hexagon. 3. If the radius of the circle be 3 inches, find the length of the chord AB. 4. If the radius of the circle be a, find the length of the perpendicular from the centre on a side of the hexagon. 5. If the radius be a feet, find the area of the hexagon. 6. The difference between the areas of the inscribed hexagon and dodecagon in a circle of radius 2 inches, is 6 (2 - V3 ) square inches. PROPS. XV., XVI. 213 PEOPOSITION XVI. Problem. To inscribe an equilateral and equiangular quindecagon in a given circle. Let ABC be the given 0. It is req^ to inscribe a reg. quindecagon in it. Inscribe an equilat. .^rAABC in the TV. 2. and a reg. pentagon ADEFG iv. ii. Then, of fifteen equal parts into which the whole Oce is to be divided, the arc AB contains five, and the arc AD contains three: .*, the rem^ arc DB contains two. Bisect the arc DB at H III. 30. /, arcs DH, HB are each one-fifteenth of the whole Oce. Join DH, HB, and in the place, in succession, chds. equal to either of these lines. These chds. will form a quindecagon, which may be proved to be regular by the method used in the preceding props. Wherefore, has been inscribed &c. Q. E.F. NOTE. A regular hexagon, or quindecagon, may be described about a circle by drawing tangents to the circle through the angular points of the inscribed figure, as was shown in the ease of the pentagon. And, by the same method as was used for the pentagon, a circle may be in- scribed in, or circumscribed about, a regular hexagon, or quindecagon. Also, by bisecting the arcs which are cut off by the sides of any one of the figures dealt with in Book IV., a regular figure of twice that number of sides may be inscribed in the circle. Hence, by the methods used in Book IV., regular polygons of 3, 6, 12, 24 &c.; 4, 8, 16, 32 &c..: 5, 10, 20, 40 &c.; 15, 30, 60, 120 &c., sides may be inscribed in, or de- scribed about, a circle. 214 EUCLID, BOOK IV. MISCELLANEOUS EXEECISES. 1. Find the magnitude of the angle subtended at the centre by the part of any tangent intercepted by the square circumscribing the circle. 2. ABC is a triangle ; if any tangent to that part of the circumference of the inscribed circle which is convex to the point A, meet AB, AC in D, E, then the difference of the perimeters of the triangles ABC, ADE is twice BC. 3. Any equilateral figure inscribed in a circle is also equiangular. 4. The area of a regular octagon inscribed in a circle is equal to that of the rectangle contained by the sides of the inscribed and circum- scribed squares. 5. Inscribe a circle in a given quadrant. 6. Find the locus of the centres of the circles inscribed in all right- angled triangles standing on the same hypotenuse as base. 7. Circles are described, each touching one side of the triangle and the other two sides produced. Show that the straight line joining any two of the centres of these circles passes through an angular point of the triangle, and that a circle can be described passing through these two centres and the other two angular points of the triangle. •8. The perpendicular from A, upon BC, meets the circumference of the circumscribed circle in G. If P be the point in which the perpen- diculars from the angles upon the opposite sides intersect, prove that PG is bisected by the side BC of the triangle ABC. 9. If the circle inscribed in a triangle ABC touch the sides AB, AC in the points D, E, and a straight line be drawn from A to the centre of the circle, meeting the circumference in G, show that the i^oint G is the centre of the circle inscribed in the triangle ADE. 10. If it be possible to describe a triangle with its angular points on tho outer of two concentric circles and its sides tangents to the inner, the radius of one circle must be double that of the other, and the triangle so described must be equilateral. 11. P is a point on the circumference of the circle circumscribing a given triangle ABC. The sides of a triangle DEF are parallel to the straight lines PA, PB, PC. Prove that the triangle DEF is equi- angular to the triangle ABC. 12. If with one of the angular points of a regular pentagon as centre and one of its diagonals as radius a circle be described ; a side of the pen- tagon will be equal to a side of the regular decagon inscribed in tho circle. MISCELLANEOUS EXERCISES. 215 13. If I, be the centres of the inscribed and circumscribed circles of the triangle ABC, and if AI be produced to meet the circumscribed circle in D, prove that OD bisects BO. 14. If ABC be a triangle, show that the circle through B, C, and the centre of the escribed circle touching BC, passes through the centre of the inscribed circle. 15. A, B, C, D, E, F are successive angular points of a regular decagon inscribed in a circle of which O is the centre. OC cuts AD in G. Prove that AE bisects OG- at right angles. 16. If the line bisecting the angle A of a triangle ABC meet the lines bisecting internally and externally the angle C in E and F, and the circle described about the triangle ABC in 0, then EO = FO. 17. Describe an isosceles triangle having each of the angles at the base one-third of the vertical angle. 18. If one side of a regular pentagon be produced, trisect the external angle. 19. If ABODE be an equilateral and equiangular pentagon inscribed in a circle, and if P be the middle point of the arc AB, prove that AP together with the radius of the circle is equal to PC. 20. Show that the circles, each of which touches two sides of an equilateral and equiangular pentagon inscribed in a circle at the extremities of a third, meet in a point. 21. In a given circle inscribe a triangle of given area having its vertex at a fixed point in the circumference and its vertical angle equal to a given rectilineal angle. 22. If DA be one side of a regular hexagon inscribed in a circle, AB a tangent equal in length to AD and making an obtuse angle with it, C the centre of the circle, and if BD meet the circle in E and BC meet the nearer part of the circumference in F, prove that AE and EF are equal to sides of regular polygons in the circle of twelve and twenty-four sides respectively. 23. Two equilateral triangles are described about the same circle; show that the intersections will form a hexagon, equilateral but not gener- ally equiangular. 24. If a regular pentagon, hexagon, and decagon are inscribed in the same circle, the square of a side of the pentagon is equal to the square of a side of the hexagon together with the square of a side of the decagon. 25. Triangles are constructed on the same base, with equal vertical angles ; prove that the locus of the centres of the escribed circles, each of which touches one of the sides externally and the other side and base produced, is an arc of a circle, the centre of which is on the circum- ference of the circle circumscribing the triangles. 216 EUCLID, BOOK IV. 26. Describe a circle touching the side BC of the triangle ABC and the other two sides produced, and prove that the distance between the points of contact of the side BC with the inscribed circle, and the latter circle, is equal to the difference between the sides AB and AC. 27. Having given an angular point of a triangle, the circumscribed circle, and the centre of the inscribed circle, construct the triangle. 28. A line drawn parallel to the base BC of a triangle ABC meets the other sides in D and E respectively. Show that the circles circum- scribing the triangles ABC, ADE have a common tangent. 29. If two equiangular triangles be circumscribed about the same circle, a circle will pass through any two corresponding angular points and the intersections of the lines containing them. 30. ABCDE is a regular pentagon. F is the middle point of the side CD Show that the pentagon is equal in area to a rectangle, one of whose sides is AF, and the other the excess of AC over CF. 31. Having given the length, but not the position, of one side of a triangle the centres of the inscribed circle and of the circle which touches the given side and the other two sides produced, and the position of the vertex opposite to the given side, construct the triangle. 32. If, in a triangle ABC, straight lines from B and C, perpendicular to the opposite sides, meet in L, and B', C be the centres of the circlea described round the triangles CLA, ALB, then B'C will be equal and parallel to BC. 33. If O be the centre of the circle inscribed in the triangle ABC and AO, BO be produced to meet the opposite sides in E, F ; prove that, if a circle can be described round the quadrilateral CEOF, then tho angle C must be equal to one-third of two right angles. 34. A circle B passes through the centre of another circle A; a triangle is described round A having two of its angular points on Bj prove that the third angular point is on the line of centres. 35. The perpendiculars from the centres of the escribed circles of a triangle upon the corresponding sides meet at a point when produced. 36. If the inscribed circle touch the sides of the triangle in D, E, F, and the diameter which passes through A meet FD in M and DE in N ; show that CM, BN, EF are parallel. 37. A triangle is inscribed in a given circle so as to have its centre of per- pendiculars at a given point; prove that the middle points of its sides lie on a fixed circle. 38. Given the vertical angle, and the radii of the inscribed and circum- scribed circles, construct the triangle. ^ BOOK V, DEFINITIONS. 1. A less magnitude is said to be a part of a greater 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 certain number of times exactly. 3. Ratio is the mutual relation of two magnitudes of the same kind to one another in respect of quantity. NOTES. Book V. treats of Proportion. The first four Books deal with the absolute equality or inequality of Geometrical magnitudes; in Book VI., however, relative greatness is con- sidered; this necessitates a definition of Proportion holding good for all Geometrical magnitudes, whether commensurable or incommensurable, and either the proof, or assumption, of the principles of the theory. Only those parts of Book V. which supply this necessary introduction to Book VI. are given here, and then the enunciations only of the propositions. The proofs are excluded because Book V. seldom forms part of a course of elementary Geometry, and the theory of Proportion is now generally studied in Algebra. But it should be remembered that Algebraical proofs must not be regarded as complete when applied to Geometrical magnitudes, since they assume the commensurability of those magnitudes. In def. 1, Euclid uses the word part in the restricted sense aliquot part, or sub-multiple; i.e. a part which is contained an exact number of times in the whole. In def. 3, the word quantity must be understood to mean the number of times the one contains the other; this is sometimes expressed by the word quantuplicity. (310) O 218 EUCLID, BOOK V. 5. Equal ratios. The first of four magnitudes is said to have the same ratio to the second that the third has to the fourth, when, any equimultiples whatever of i\iQ first and third being taken, and any equimultiples whatever of the second and fourth, the multiple of the drst is greater than, equal to, or less than that of the secondj according as the multiple of the third is greater than, equal to, or less than that of the fourth. And the four magnitudes are then called proportionals. [For example: if A, B, C, D be four magnitudes, wA, mC equimultiples of A, C, 71 B, nJ) equimultiples of B, D, and if, whatever integral values be given to m, n, wC is always >rtD, when m A>'mB, mC is always =nD, when mA=yiB, mC is always < wD, when mA< nB, then the ratio of A to B is equal to the ratio of C to D, and A, B, C, D are proportionals. This is expressed by saying "A is to B as C is to D." 1 10. When three magnitudes are proportionals the first is said to have to the third the duplicate ratio of that which it has to the second. And the second is said to be a mean proportional be- tween the first and third. 11. When four magnitudes are proportionals the first is said to have to the fourth the triplicate ratio of that which it has to the second; and so on. Compound Ratio. When there are any number of magnitudes of the same kind, the first is said to have to the last the ratio which is com- pounded of the ratios of the first to the second, the second to the third, and so on. [ For example : if A, B, C, D are four magnitudes, the ratio A to D is compounded of the ratios of A to B, B to C, and C to D. ] 12. In proportionals, the antecedent terms are said to be homo- logous to one another; as also the consequents to one another. DEFINITIONS. 219 ADDITIONAL DEFINITIONS. Equimultiples of magnitudes contain them the same number of times. When the two terms of a ratio are equal it is called a ratio of equality. When the first term of the ratio is greater than the second it is called a ratio of greater inequality. When the first term of the ratio is less than the second it is called a ratio of less inequality. The first term of a ratio is called the antecedent, and the second term the consequent. If A, B, C, D are proportionals, A and D, the first and last terms, are called the extremes; and B and C, the second and third terms, are called the means. Three magnitudes are said to be in continued proportion when the first is to the second as the second is to the third. The ratio of B to A is called the reciprocal of the ratio of A to B. NOTES. The definition of proportion is of the greatest importance, and should be carefully studied by the student before he proceeds further. The statement of the proportion A is to B as C is to D is often expressed thus : — A : B : : C : D, or thus, A : B = C : D. A The Algebraic method of expressing a ratio, - , being a very convenient one, will also be found in the Examples, where it should be regarded as a symbol for the words the ratio of A to B, and not as implying the operation of division; it should not be used for book-ivork {see Preface). In the "book- work" of this edition none of these forms have been used, as the words themselves are so short that they scarcely seem to need abbreviating. Euclid's definitions of duplicate and triplicate ratio can be easily shown to agree with those of Algebra ; for instance, if a, b, c are in continued pro- portion, a:b=h: c, or J = ^ /. f . ^=^ . ^, or 4= -» which shows be h b b c ¥ c that Euclid's definition of duplicate ratio is in agreement with the Algebraic one that ^„- is the duplicate ratio of r« 6^ 6 220 EUCLID, BOOK V. Theorems of Book V referred to in Book VI. PROPOSITION B. If four magnitudes are proportionals, they are proportionals when taken inversely. [i.e., If A : B : : C :D; Then B : A : : D : C. ] Invertendo. PROPOSITION D. If the first term of a proportion he a multiple, or a part, of the second, the third is the same multiple, or part, of the fourth. [i.e., If A :B : :C :D, and if A=mB; then C=wD. ] PROPOSITION VII. Equal magnitudes have the same ratio to the same magnitude; and conversely. [ i.e., If A=B ; then A : C : : B : C, and C : A : : C : B. ] PROPOSITION IX. Magnitudes which have the same ratio to the same magnitude are equal to one another; and conversely. [ i.e., If A : C : : B : C, or, if C : A : : C : B; then A=B. ] PROPOSITION XL Ratios that are equal to the same ratio are equal to one another. [ i.e., If A : B : : C : D, and if E : F : : C : D ; then A : B : : E : F. ] PROPOSITION XII. If any number of magnitudes he proportionals, as one of the ante- cedents is to its consequent, so is the sum of all the a7itecedents to the su7n of all the consequents. [ i.e., If A : B : : C : D, and C : D : : E : F ; then A : B : : A + C + E : B + D + F.] PROPOSITION XIV. If the first teryn of a proportion is greater than the third, the second is greater than the fourth; if equal, equal; and if less, less. [ i.e., If A : B : : C : D, andif A>C; then B>D; if A=C; then B=D; if Aro\}OYiioji when the second is a mean proportional between !the first and third. 1 NOTES. No use is made by EucHd of def. 2 as it stands, but the sida, of such figures are said to be reciprocally proportional. In this book, references to the postulates, axioms, and propositions of Book I. will not be inserted in the constructions. PROP. I. 223 PKOPOSITION I. Theorem. Triangles and parallelograms of the same altitude are to one another as their bases. Let ABC, ACD be two .^s having the same alt. Then shall ^:::^ABC be to ^^ACT> as BC to CD. I I I F ^ B C D G H K Produce BD both ways to F, K. From BF cut off any no. of parts BE, EF each=BC. From DK cut off any other no. of parts DG, GH, HK each = CD. Join AE, AF, AG, AH, AK. Then, since FE, EB, BC are all equal, .'. .^sAFE, AEB, ABC are all equal I. 38, note. /. ^^AFC is the same mult, of ..^ABC that FC is of BC. Similarly, it may be shown that .^ ACK is the same mult, of .^ ACD that CK is of CD. Hence, ^^AFC and FC are equimults. of ..^rrXABC and BC, the first and third, and ^i::\ACK and CK are equimults. of ^^ACD and CD, the second and fourth, of the four mags. .^^ABC, .^AACD, base BC, base CD; also, if FC=CK, .:^AFC=.^ACK 1.38. if FOCK, ^AFO^ACK, ) ^ ,„ ^ if FCP^ ^ 'c ir^ -^ Join BE, CD. Then, since DE is || to BC Hyp. /.^aBDE=.^CDE 1.37. /..-rxBDE is to .^ADE as ^^CDE is to ^^ADE...v. 7. But .^BDE is to .^AADE as BD is to DA vi. 1. And^^CDE isto .^ADE as CE is to EA vi. 1. BD is to DA as CE is to EA v. 11. Part II.— Let BD be to DA as CE to EA; Then shall DB be 1| to BO. Join BE, CD. Then, ^r^BDE is to -^^ADE as BD is to DA vi. 1. And .^CDE is to ^rriADE as CE is -to EA vi. 1. But BD is to DA as CE is to EA Hyp. /. .^S,BDE is to .-rr^ADE as ^^CDE is to ^^ADE...v. 11. .•..^BDE=.=^CDE V. 9. and they stand upon the same base DE, .-.DE is II to BC I. 39. Wherefore, if a straight line &c. Q.E.D. PROP. II. 225 NOTE. Since BD is to DA as CE is to EA. ,*, sum of BD, DA is to DA as sum of CE, EA is to EA Compo. i.e., AB is to AD as AC is to AE. A form of the result which is often useful. EXAMPLE. If, in the figure of VI. 2y BE and CD cut at 0, then AO, or AO pro- duced, will bisect BC. Let AO meet BC at F. Then, .^DBC=^^EBC I. 37. Take away the com. ^-^OBC, /. .^^ODB=.-:AOEC. ^ODB_DB ^j^ BA EC ~CA ^riOEC But .^OBA' .VI. 2. ^:::\OCA .^ODB .VI. •Above. .^OCA /. .-::\OBA=.^OCA v. 9. and they stand on the same base OA, ,*, their alts, are equal... Ex. 4, p. 67. i.e., ±BM=J.CN. Hence in .^s BFM, CFN, |'^BFM=Z.CrN, V J Z.BMF=Z.CNF, [ BM=CN. ,*, BF=CF 1.26. Q.E.D. EXERCISES. 1. If two straight lines are cut by three parallels, they are cut proportion- ally. Prove this and the converse. 2. If a quadrilateral have two of its sides parallel, the line joining the middle points of these sides will pass through the point of intersection of its diagonals. 3. If O is a fixed point, and the ratio OP : OQ is constant, then if the locus of P is a straight line, so also is that of Q. 4. If from any point in the diagonal AC of a quadrilateral ABCD, OE, OF be drawn parallel to AB, AD and meeting BC, CD in E, F ; then EF shall be parallel to BD. 5. From a given point P in the side AB of a triangle ABC, draw a straight line to AC produced which shall be bisected by BC. 226 EUCLID, BOOK VI. PEOPOSITION III. Theorem. If the vertical angle of a triangle he bisected hy a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another; and, conversely, if the segments of the base have the same ratio which the other sides of the triayigle have to one another, the straight line drawn from the vertex to the point of section shall bisect the vertical angle. Part I.— Let Z.BAC of .^^kBC be bisected by AD meeting BC at D. Then shall BD be to DC as BA to AC. B D C Through C draw CE 1| to AD, meeting BA prod^ at E. Then since AD, CE are || , /. z.BAD=z_AEC i.29,ii and Z-DAC=z.ACE i.29,i. But z.BAD=z.DAC Hyp. /. ^AEC = Z-ACE. .-. AE=AC 1.6. ButBD is to DC as BA is to AE vi. 2. /. BD is to DC as BA is to AC v. 7. Part II.— Let BD be to DC as BA to AC; Then shall AD bisect Z.BAC. Through C draw CE H to AD, meeting BA prod Hyp. /. ^AEC = ^ACE. /. AE=AC 1.6. ButBD is to DC as BA is to AE vi. 2. .'. BD is to DC as BA is to AC v. 7. Part II.— Let BD be to DC as BA to AC; Then shall AD bisect lCA¥. Through C draw CE |1 to AD, meeting AB at E. Then, since BD is to DC as BA is to AC Hyp. andBD is to DC as BA is to AE vi. 2. /. BA is to AC as BA is to AE v. 11. /. AC =AE V.9. /.^ACE = z.AEC.. 1.5. But Z-ACE = z.CAD and Z-AEC=Z.FAD /. iLCAD=^FAD. i.e., AD bisects L CAF. Wherefore, if the exterior «Stc. Q.E.D. 228 EUCLID, BOOK VI. NOTES. Prop. A is not Euclid's ; it was added by Simson. Props. III. and A might be included in one enunciation, thus : — If the interior and exterior angles at the vertex of any triangle he bisected by straight lines which meet the base and the base produced, they divide the base, internally and externally, into segments which have the same ratio as the sides of the triangle; and, conversely. EXAMPLES. I. If the vertical angle of a triangle be bisected, both internally and ex- ternally, the bisectors divide the base harmonically. Let the bisectors meet the base in D, E. BD_BA DC~AO BE "^EC . BD_DC •* BE—EC _ BC— BD — BE— BC* /, BD, BC, BE are in harmonic proportion def. p. 222. Q.E.D. Then .VI. 3. .VT. A. 11. Construct a triangle, having given the base, the vertical angle and the ratio of the sides. Let AB be the given base, C the given vert^ Z., ^ the given ratio of the sides. Bisect Z.C. From A draw a st. line making any L with AB, and from it cut off AF=M, FG=N. Join GB. Draw FD || to GB and meeting AB at D. On AD desc. a seg* of a containing an L=h C HI- 33. On DB desc. a seg* of a containing an L =i C, meeting the other seg* at E. JoinAE.EB. Then AEB shall be the ^^ reqd For, since Z.AED=i C=Z.DEB Constr. /.whole ^AEB=C, N Also, AE is to EB as AD is to DB vi.3. as AF is to FG vi. 2. rtS M is to N ... Constr. Q.E.F. PROPS. III., A. 229 III. Find the locus of a point which moves so that the ratio of its distances froin two fixed points is constant. Let A, B be the fixed pts. and P the moving pt. Case i. If the const, ratio is one of equality, i.e., if, PA=PB, the locus is evi- dently a St. line ± to AB through its mid. pt. M . Case ii. If the const, ratio — is not one of equality ; Divide AB at C in the given ratio ^."^^ (see preceding Ex.). ^ ^ ^ ^^^j. Join PC. A C B D Draw PD at rt. Z. s to PC meeting AB prod^ at D. Then, since AC is to CB as M is to N Constr. as AP is to PB Hyp. /.PC bisects Z.APB vi. 3. and .*,PD at rt. Z. s to PC bisects the ext^ L BPE Ex. 6, p. 36. /.AD is to DB as M is to N vi. A. .*. C and D are both fixed pts. for all positions of P. Hence the locus of P is that of the vertex of & rt. L^ ^:^CPD whose hypot. CD is fixed, and this is a on CD as diam Ex. 4, p. 167. EXERCISES. 1. ABC is a triangle with the base BC bisected at D. DE, DF bisect the angles ADC, ADB, meeting AC, AB at E, F. Prove that EF is parallel to BC. 2. Trisect a line by the help of vi. 3. 3. The sides of a triangle are 3, 4, and 5 inches ; find the lengths of the segments into which the bisector of the opposite angle divides the longest side. 4. A, B, C are points in a line and D is a point at which AB, BC subtend equal angles ; show that the locus of D is a circle. 5. The Z-C of a ..^ABC is bisected by CF meeting AB at F. Z_B is bisected by BG meeting CF at G. Prove that AF : FG : : AC : CG. 6. Two circles touch internally at O. A straight line touches the inner circle at C and meets the outer circle at A, B ; and OA, OB meet the inner circle at P, Q ; prove that OP : OQ=AC : CB. 7. Interpret the result when the two sides in vi. A. are equal. 8. The straight lines bisecting one interior and two exterior angles of a triangle are concurrent. 9. If, in the figure of Example I., O is the middle point of BC, then OB is a mean proportional between OD and OE. 230 EUCLID, BOOK VI. PEOPOSITION IV. Theorem. The sides about the equal angles of triangles which are equiangular to one another are proportionals ; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or the consequents of the ratios. Let ^^s ABC, DCE have aABC=z_DCE, Z.ACB=2.DEC, and z.BAC=Z_CDE: B C Then shall BA be to BC as CD to OE, BO be to AC as CB to DE, and BA be to AC as CD to DB. Let the ^^s be so placed that their bases BC, CE lie in one con- tinuous st. line with their vertices A, D on the same side of it. Then, since z.ACB=z.DEC Hyp. and z_s ABC, ACB p Apply .:^ ABC to DEF, with pt. A on D and AB along DE, E F Then, since L A= L D, AC must lie along DF. Let G, H be the positions of the pts. B, C. Join GH. Then, since L G= L B= L E, /. GHis II toEF 1.28. /. DG is to EG as DH is to HF vi.2. .*.DG is to DE as DH is to DF Compo. or DG is to DH as DE is to DF Altem. i.e. AB is to AC as DE is to DF. Similarly, by applying the .-^s with pt. B on E, or pt. C on F, it may be shown that the sides about these L s are : : is. ,*, the ^::ls are similar VI. def. i. Q.E.D. EXEECISES. 1. Show that the line joining the middle points of the sides of a triangle is half the base. 2. The shadow of a stick 3 feet long is 5 feet when the shadow of a tower is 40 yards ; find the height of the tower. 3. In the figure of VI. 2, F is the middle point of BC, show that AF bisects DE, 4. AB, CD are two parallel straight lines ; E is the middle point of CD ; AC and BE meet at F, and AE and BD at G : show that FG is parallel to AB. 5. ABC is a triangle with angle ACB double of ABC ; the bisector of angle ACB meets AB at D ; prove that AC is a mean proportional between AD and AB. 6. The diagonals of a trapezium, two of whose sides are parallel and one of these double the other, cut one another at a point of trisection. 7. In the figure of I. 43, show that GE, FH will meet, if produced, on CA produced. 232 EUCLID, BOOK VI. PEOPOSITION Y. Theorem. If the sides of two triangles about each of their angles, he propor- tionals, the triangles shall he equiangular to one another, and shall have those angles equal which are opposite to the homologous sides. Let the .^^ ABC, DEF have their sides : :i«, namely, AB to BC as DE to EF, BC to AC as EF to DF, andAB to AC as DE to DF; Then shall -^ ABC be equiangr to ^^DBP. D At pt. E in EF, make ^FEG= ^ ABC. At pt. F in EF, make L EFG= L ACB. /. rems ^EGF=remg ^BAC 1.32. i.e. .^^ABC is equiang'^ to .^GEF. /. AB isto BC as GE is to EF 1.4. ButAB isto BC as DE isto EF Hyp. /. GE isto EF as DE isto EF v. ii. /. GE=DE V.9. Similarly it may be shown that GF=DF. Hence in .^^ GEF, DEF, ( GE=DE, •.• j GF=DF, V EF is com. /. ^EGF=^EDF 1.8. Similarly ^GEF=Z-DEF, and ^GFE=z.DFE. i.e. .^GEF is equiang^ to ^^ADEF, But .=::^ABC is equiang'"tO ^r^GEF Above. /. .^ABC is equiang'- to ^:XDEF. Wherefore, if the sides &c. Q.E.D. NOTE. Tt follows from def. 1, that the ^^s are similar. This proposition is the converse of Prop. 4. PROPS, v., VI. 233 PROPOSITION VI. Theorem. 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 equiangular to one another, and shall have those angles equal which are opposite to the homologous sides. Let .^s ABC, DEF have L ABC= L DEF, and let AB be to BC as DE to EF; Then shall ^^ABC be equiangr to ^:^DEP. D Ef ^F B C G At pt. E in EF make ^FEG= A ABC, or DEF. At pt. F in EF make L EFG= L ACB. .'. reniK ^ EGF=remK z. BAC 1.32. i.e. .i^ABC is equiang'" to ^^G¥jF. /. AB is to BC as GE is to EF vi. 4. ButAB is to BC as DE is to EF Hyp. .-. GE is to EF as DE is to EF v. 11. .-. GE=DE V. 9. Hence in ^^s GEF, DEF, GE = DE EF is com. L GEF= L DEF. /. .::^GEF=.-:^DEF in all respects i. 4. But .-:::AABC is equiang^ to .i::AGEF Aboye. .*. .^ABC is eqiiiang'- to .^^DEF. Wherefore, if two triangles &c. Q.E.D. NOTE. It follows from prop. 4, since the ^^\^ are equiangr to one another, that their sides about their equal angles are proportionals, and consequently, from def. 1, that the triangles are similar. Props, 5 and 6 can, like prop. 4, be easily proved by superposition. (310) P 234 EITCLID, BOOK VI. PEOPOSITION VII. Theorem. If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles 'proportionals; then, if each of the remaining angles he either less, or not less, than a right angle, or if one of them be a right angle, the triangles shall be equiangular to 07ie another, and shall have those angles equal about which the sides are propoi'tionals. Let ^r^s ABC, DEF have L ABC= L DEF, and let BA be to AC as ED to DF, also, (i) let Z.s at C and F be either both acute or both obtuse; or, (ii) let L at C be a rt. z. ; Then shall -^ ABC be equiangr to ^^ DEF. (i) For, if Z.BAC be not= Z.EDF, one must be the >. If possible, suppose L BAC > L EDF. At pt. D in ED make L^J)G= L BAC. and let DG meet EF prod^ at G. Then, since z.ABC=z_DEG Hyp. and z.BAC=Z.EDG Constr. .*. rems z.ACB=remg Z.DGE i. 32. i.e. ^^ ABC is equiang"" to ^^ DEG. /. BA is to AC as ED is to DG vi. 4. ButBA is to AC as ED is to DF Hyp. /.ED is to DG as ED is to DF v. 11. .•.DG=DF V.9. .-. ^DGF=^DFG 1.5. But ^DGF=Z.ACB Above. /. ^DFG=z_ACB. But i_sDFE, DFG=twort. ^s 1. 13. .*. 2_s DFE, ACB=two rt. Z.s, which is impossible, since they are, by hyp., either both <, or both >, a rt. z.. PROP. VII. 235 Hence, L BAG is not unequal to L EDF, i.e. L BAC= L EDF. But ^ABC=z.DEF Hyp. .*. remg z.ACB=rems ^DFE i. 32. i.e. ^^ ABC is equiang"^ to .^ DEF. (ii) Again, if the L at C be a rt. Z_ , then it may be shown, as in the preceding case, that Z.S ACB, DFE together=two rt. Z.s. and .*., in this case, Z_DFE must also be a rt. Z., and, consequently, ^:^ ABC be equiaug^ to ^^^ DEF. Wherefore, if two triangles &c. Q.E.D. NOTES. From this proposition it is manifest that — If two triangles have one angle of the one equal to one angle of the other, and the sides, taken in order, about one other angle in each proportionals; then the third angles are either equal, or supplementary; and that, in the first case, the triangles are similar. The enunciations of VI. 5 and VI. 7 are faulty: — In VI. 5 it should be stated that the sides must be taken in order about the angles; and, in VI. 7, that, of the sides which are proportionals, those which subtend the equal angles must be homologous. EXERCISES. 1. A point D is taken within a triangle ABC; on EC, without the triangle, is constructed a triangle BEC similar to BDA; prove that the triangle DBE is similar to ABC. 2. The straight lines which join corresponding angular points of two similar triangles whose homologous sides are parallel will, if produced, meet in one point. 3. Distinguish between "an equiangular triangle" and "a triangle equi- angular to another." 4. ABCD is a parallelogram; P,Q are points in a straight line parallel to AB; PA and QB m^t at R, and PD, QC at S. Show that PvS is parallel to AD. 6. In the sides AB, AC of a triangle ABC two points D, E are taken such that BD=CE; DE, BC are produced to meet at F, show that AB : AC : : EF : DF. 6. ABC is a common tangent to two circles whose centres are O, Q. ABC meets OQ at A, and through A a secant is drawn cutting the circles; prove that the radii to corresponding points of section are parallel. 236 EUCLID, BOOK VI. PROPOSITION VIII. Theorem. In a right-angled triangle, if a perpendicular be 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 rt.-angled .^^ having the rt. L at A, and let AD be drawn ± to BC; Then shall ^^s DBA, DAG be simr to ^^KQG and to each other. A D C For, in ^\^ DBA, ABC, since rt. Z.BDA=rt. Z.BAC, and Z. at B is com. : ,\ rem?? Z.DAB=rems Z.ACB 1.32. i.e. .^riADBA is equiang^ to ^:::AABC, .*. these .^rAs are sim'^ VI. 4. In like manner it may be shown that .i^s. DAC, ABC are equiang'^ to one another, and /, sim"^. Hence, since ^^^ DBA, DAC are each equiaug'^ to .^ABC, .*. ..-::ADBA is equiang"^ to ..=::::\DAC. .*. these -=^s are sim"^ vi. 4. Wherefore, in a right-angled triangle &c. Q.E.D. CoR. — From this prop, it is manifest (i) that the J. from the rt. L to the hypot. is a mean : : ^ between the seg*^ of the base. For, since ^::\^ DBA, DAC aresim"'", /. BD is to DA as DA is to DC (ii) that each side is a mean : : ^ between the base and the seg* of the base adj. to that side. For, since ^^^ DBA, ABC are sim'^, .*. BD is to BA as BA is to BC; and, since ,-^A s DAC, ABC are sim'", - /.DC is to CA as CA is to CB. def. 1. PROPS. VIII., IX. 237 PROPOSITION IX. Problem. From a given straight line to cut off any part required. Let AB be the given st. line: It is req*^ to cut off from it a certain aliquot part. ■t n From A draw a st. line AC making any L with AB. In AC take any pt. D. From DC cut off, in succession, parts each = AD until the whole AG contains AD as many times as AB contains the req^ part. Join GB. Through D draw DH || to GB, meeting AB at H. Then shall AH be the part reqd. For, since DH and GB are || , /.AD is to DG as AH is to HB vi. 2. /.AD is to AG as AH is to AB Compo. But AG is a mult, of AD Constr. /, AB is the same mult, of AH V. D. i.e. AH is the req'^ part of AB. Wherefore, f7^07n a given &c. Q.E.F. EXERCISES. 1. The sides of a right-angled triangle are 3 and 4 inches ; find (1) the length of the perpendicular to the hypotenuse, * (ii) the length of the segments of the hypotenuse. 2. Prove that the three sides and the perpendicular are four proportionals. 3. Prove that the segments of the base are to one another in the duplicate ratio of the sides, assuming the theorem on p. 249. 4. The radius of a circle is a mean proportional between the segments into which the part of any tangent intercepted by two other parallel tan- gents, is divided at its point of contact. 5. Trisect a given straight line. Indicate three other constructions by which this problem may be solved. 6. Given a line -| of an inch long, draw a line 1 inch long. 7. Cut off from a triangle one-seventh of its area. 238 EUCLID, BOOK VI. PEOPOSITION X. Problem. To divide a straight line similarly to a given divided straight line; that is, into parts which shall have the same ratios to one another that the parts of the given divided line have. _ Let AB, AC be the given st. lines, AC being divided at D, E ; It is req^ to divide AB similarly to AC. Let AB, AC be so placed as D>^1^L to contain any L . Join BC. Through D, E, draw DF, ^ jr q g EG II to CB meeting AB in F, G. Then AB shall be divided similarly to AC at pts. P, G. Through D draw DHK |i to AB. Then DG and HB are Os. .-. DH=FG, and HK=GB 1.34. But, since EH, CK are || , Constr. .-. DH is to HK as DE is to EC vi. 2. ^■.e. FG is to GB as DE is to EC Above. Also, since DF, EG are || Constr. .*. AF is to FG as AD is to DE vi. 2. Hence, AF is to GB as AD is to EC... Ex. seq. Wherefore, the given straight line &c. Q.E.F. NOTE. The following is an important special case of the problem. To divide a given straight line into ttvo parts which shall be in a given ratio. Let AB be the given st. line, and P to Q the given ratio. From A draw any st. line AC. From AC cut off AD=P. Case i : Internally; From DC cut off DE=Q. Join EB. Through D draw DF || to EB. Case ii : Externally; From DA cut off DE=Q. Join EB. Through D draw DF || to EB, meeting AB prod^ at F. Then, in both cases, since EB, DF are || , A .*. AF is to FB as AD is to as P is to PROPS. X., XI., XII. PEOPOSITION XI. Problem. To find a third proportiorial to two given straight lines. Let A, B, be the two given st. lines. It is req'^ to find a third : :^ to A and B. O- B- C H JC E Draw two st. lines CD, CE containing any L . From CD cut off CF=A, FG=B. From CE cut off CH=B. Join FH. Through G draw GK || to FH, meeting CE at K. Then shall HK be the reqd third : :i. For, since FH, GK are || Constr. /.CF is to FG as CH is to HK vi. 2. i.e., A is to B as B is to HK Constr. Wherefore, has been found &c. Q.E.F. PEOPOSITION XII. Problem. To find a fourth proportional to three given straight lines. Let A, B, L be the three given st lines ; It is req^ to find a fourth : :^ to A, B, L. B L C H K" E Draw two st. lines CD, CE containing any L . From CD ciit off CF=A, FG=B. From CE cut off CH=L. Join FH. Through G draw GK |I to FH, meeting CE at K. Then shall HK be the reqd fourth : :i. For, since FH, GK are || Constr. /. CF is to FG as CH is to HK VL 2. i.e., A is to B as L is to HK Constr. Wherefore, has been found &c. Q.E.F. 240 EUCLID, BOOK VI. PEOPOSITION XIII. Problem. To find a mean proportional between two given straight lines. Let AB, EC be the two given st. lines ; It is required to find a mean : :' between them. Place AB, BC in a st. line. On AC desc. a semicircle ADC. From B draw BD at rt. Z_ s to AC. Then BD shall be the reqd mean \'J. Join AD, DC. Then Z. ADC, in a semicircle, is a rt. L IIL 3i. And, since DB is the ± from the rt. L to the base, /. DB is a mean : :^ between AB, BC, the seg*^ of the base vi. 8, cor. Wherefore, has been found &c. Q.E.F. EXERCISES. 1. The perpendicular let fall from any point in the circumference of a circle on any diameter is a mean proportional between the perpendiculars let fall from the point on the tangents at the extremities of that diameter. 2. ABC is a triangle. At A a straight line AD is drawn, making the angle CAD equal to CBA, and at C a straight line CD is drawn making the angle ACD equal to BAC. Show that AD is a fourth proportional to AB, BC, CA. 3. GAB is a triangle. Any straight line through O cuts AB at G, and a parallel to OB, drawn through A, at X, and a parallel to OA, drawn through B, at Y. Show that GO is a mean proportional between GX and GY. 4. If two circles touch externally and also touch a straight line, the part of the line intercepted between the points of contact is a mean propor- tional between the diameters. 5. A, B, C are three points in a straight line: find a point P in the line such that PB may be a mean proportional between PA and PC. PROPS. XIII., XIV. 241 PROPOSITION XIV. Theorem. Equal 'parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional ; and, conversely, parallelograms which have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to 07ie another. Part I.— Let AC, OF be=Os, having ^BC.D=z.ECG; Then shall BC be to CG as BO to CD. Place the Os with their sides BC, CG in a st. line, the /Z7s standing on opp. sides of it. Then EC, CD are also in a st. line i. u. Complete ODC D E F Then, since OAC=OCF Hyp. /. OAC is to ODO as OCF is to [JJ)Q v. r. But O AC is to or>G as BC is to CG vi. i. And OCF is to ODG as EC is to CD vi. i. /, BC is to CG as EC is to CD v. ii. Part IL— Let z.BCD=l.ECG, and let BC be to CG as EC to CD; Then shall OAC=OCF. Place the Os with their sides BC, CG in a st. line, and the Os on opp. sides of this line. Then EC, CD are also in a st. line 1. 14. Complete ODG. Then BC is to CG as EC is to CD .Hyp. But BC is to CG as OAC is to ODG vi. i. and EC is to CD as OCF is to ODG v. i. /.OAC is to ODG as OCF is to ODG v. ii. .-. OAC=OCF V. 9. Wherefore, equal parallelograms &c. Q.E.D. 242 EUCLID, BOOK VI. PEOPOSITION XV. Theorem. Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally pro- portional; and, conversely, triangles which have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another. Part I.— Let ABC, CDE be=.=^s, having z_BCA=z.DCE; Then shall BC be to CB as DC to CA. Place the ..irris with their sides BC, CE in a St. line, and their vertices A, D, on opp. sides of this line. Then DC, CA are also a St. line 1. 14. Join AE. Then, since ^^ABC=^:2^CDE Hyp. .-. ^^ABC is to .:::XACE as .^CDE is to .-^ACE ...v. r. But ^^ ABC is to ^^ACE as BC is to CE vi. i. And .^ CDE is to .^ACE as DC is to CA vi. i. /.BC is to CE as DC is to CA v.ii. Part II.— Let .^s ABC, CDE have L BCA= L DCE, and let BC be to CE as DC to CA; Then shall .-:^ABC=^^CDB. Place the -^^s with their sides BC, CE in a st. line, and their vertices A, D on opp. sides of this line. Then DC, CA are also in a st. line i. u. Join AE. ThenBC is to CE as DC is to CA Hyp. But BC is to CE as ^^ABC is to ^^ACE vi. i. and DC is to CA as ^^CDE is to .^ACE vi. i. /. ^^ABC is to ^:AACE as .^CDE is to .^ACE v. ii. /. .^ABC=.^CDE V. 9. Wherefore, equal triangles (&c. Q.E.D. PROP. XV. 243 NOTE. Prop. 14 might have been combined with prop. 15, and the truth of the prop, for parallelograms deduced from that for triangles, as in prop. 1. Both props, have a second converse not dealt with by Euclid ; the follow- ing is a proof of this converse of prop. 16. JEqual triangles ivhich have the sides about a pair of angles reciprocally proportional, have that pair of angles either equals or supplementary. Let ABC, DOE be=.^s ^ having BC to CE as DC to CA. ^ u^ p Then shall (i) either Z.ACB=/-DCB, or ^ (ii) La ACE, DCB=two rt. Ls. Place the ^:::As with BC, CE in a st. line j) and their vertices on opp. sides of this line, A Then (i.) if DC, CA are in a st. line ^,^<-^ \r g Z.ACB=Z.DCE 1.15. B 7t But (ii.) if not, produce AC to F, make Cr=CD, and join EF, DF. ^f F Then, since BC is to CE as DC is to CA Hyp. /.BO is to CE as OF is to CA Above. AlsoZ.rCE=^ACB 1.15. : , ^^^GK=.^^ABG VI. 15. But ^^DCE=.-:::\ABC Hyp. /. .^FCE=^^DCE. /. DFis II to BE 1.39. /. Z.ACB=Z.CFD 1.29. = ^CDF 1.6. /.Z-s ACB, DCE=^CDF, DOE, =two rt. L s I. 29. Q.E.D. EXEECISES. 1. In the figure of VI. 15, prove that the triangles BCD, ACE are similar. 2. If, in the figure of VI. 15, BA, DE are produced and meet at F, then OF will when produced bisect BD. 3. If from the ends A, B, of the hypotenuse of a right-angled triangle, per- pendiculars AE, BD be drawn meeting BC, AC produced at E, D, then triangle ACB is equal to triangle ECD. 4. Construct an isosceles triangle, the area and vertical angle being given. 5. P is any point on the side AC of the triangle ABC ; CQ is parallel to BP and meets AB produced at Q ; AN, AM are mean proportionals between AB, AQ and AC, AP. Prove that the triangles ANM, ABC are equal. 244 EUCLID, BOOK VI. PEOPOSITION XVI. Theorem. If four straight lines he proportionals, the rectangle contained by the extremes is equal to the rectangle contained hy the means; and, conversely, if the rectangle contained hy the extremes he equal to the rectangle contained hy the means, the four straight lines are propor- tionals. Part I.— Let the four st. lines AB, CD, EF, GH be : \^\ so that AB is to CD as EF to GH; Then shall rect. AB. GH=rect. CD, EP. L H A B C Draw AK, CL at rt. ^ s to AB, CD. Cut off AK=GH, and CL=EF. Complete the rectangles KB, LD. Then, since AB is to CD as EF is to GH Hyp. .'. AB is to CD as CL is to AK v. 7. i.e. the sides about the = Z. s at A and C of the Os KB, LD are reciprocally : :^, .-. O KB=OLD VI. 14. i.e. rect. AB, AK=rect. CD, CL ii. def. i. .', rect. AB, GH=rect. CD, EF Above. Part II.— Let rect. AB, GH=rect. CD, EF; Then shall AB be to CD as EP to GH. Draw AK, CL at rt. z_s to AB, CD. Cut off AK=GH, and CL=EF. Complete the rectangles KB, LD. Then, since rect. AB, GH=rect. CD, EF Hyp. /. rect. AB, AK=rect. CD, CL Above. i.e. EJ KB=ZI7 LD. .*. the sides about the = z. s at A, C are reciprocally : : ^, or, AB is to CD as CL is to AK vi. u. ^.e. AB is to CD as EF is to GH v. 7. Wherefore, if four straight lines Q.E.D. PROPS. XVI., XVII. 245 PEOPOSITION XVII. Theorem. If three straight lines he proportionals, the rectangle contained hy the extremes is equal to the square on the mean; and, conversely/, if the rectangle contained hy the extremes he equal to the square on the mean, the three straight lines are proportioivxls. Part I.— Let the three st. lines AB, CD, EF be : '}^, so that AB is to CD as CD to EF; Then shall rect. AB, BF=sq. on CD. C D G H Draw GH=CD. Then, since AB is to CD as CD is to EF Hyp. /. AB is to CD as GH is to EF v. 7. /. rect. AB, EF=rect. CD, GH vi. i6. =rect. CD, CD Above. =sq. on CD. Part II.— Let rect. AB, EF=sq. on CD, Then shall AB be to CD as CD to EF. Draw GH=CD. Then, since rect. AB, EF=sq. on CD Hyp. =rect. CD, GH Above. /. AB is to CD as GH is to EF vi. le. orAB is to CD as CD is to EF v. 7. Wherefore, if three straight lines &c. Q.E.D. NOTE. VI, 17 is merely one special case of VI. 16, that in which the two means are equal. EXERCISES. 1. If the perpendicular from the right angle in a right-angled triangle divide the hypotenuse so that one of its segments is equal to one of the sides of the triangle, the hypotenuse is divided in medial section. 2. ABCD is a parallelogram ; from B a straight line is drawn cutting the diagonal AC at F, the side DC at G, and the side AD produced at E; show that the rectangle EF, FG is equal to the square on BF. 246 EUCLID, BOOK VI. PKOPOSITION XVIII. Problem. On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure. Let AB be the given st. line and CDEFG the given fig. It is req^ to desc. on AB a fig. simT and similarly situated to CDEFG. K H^ \ / q ^ C D A ^B Join FD, GD. At pt. A in AB make Z-BAH= Z.DCG. At pt. B in AB make A. ABH= ^CDG. /. rems z.AHB=rems Z-CGD 1.32. and .i^ HAB is equiang'" to ^^ GCD. At pt. H in HB make L BHK= L DGF. At pt. B in HB make L HBK= L GDF. .'. rem"? z.HKB=rem& Z.GFD l 32. and ^^ KHB is equiang^ to ^^ FGD. At pt. K in KB make L BKL= L DFE. At pt. B in KB make iLKBL=z.FDE. .*. rems z.KLB=reme /.FED l 32. and .^^ LKB is equiang'^ to ..^ EFB. Let all these Z. s be drawn towards the same parts as the corresponding z. s in the given fig. Then shaU fig. ABLKH be simr and similarly situated to CDEFG. For, since Z_AHB = z.CGD ) and Z.BHK=z.DGF 1 '^^^*'^- /. whole z.AHK= whole Z_CGF. PROP. XVIII. 247 Similarly, L HKL= L GFE, andz.LBA=^EDC. Hence, fig. ABLKH is equiang^ to CDEFG. Again, since each pair of ^^::\^ in the fig. are respectively equiang"^ to one another, .-. AB is to AH as CD is to CG vi. 4. i.e. the sides about theZ.s at A, C are : :K Also AH is to BH as CG is to DG vi. 4. and BH is to KH as DG is to FG vi. 4. /.AH is to KH as CG is to FG Ex. *q. i.e. the sides about the z.s at H, G are : :K Similarly the sides about the Z.s at K, F, and L, E, are : :^^ Lastly, since LB is to KB as ED is to FD vi. 4. and KB is to HB as FD is to GD vi. 4. /.LB is to HB as ED is to GD Ex. seq. ButHB is to AB as GD is to CD vi. 4. .'.LB is to AB as ED is to CD Ex. »q. i.e. the sides about the Z.s at B, D, are : :^^ .-. fig. ABLKH is sim"^ to CDEFG. Wherefore, on the given straight line &c. Q.E.F. NOTE. The problem has been solved for the case of a five -sided figure; this includes the case of a triangle or quadrilateral, and might, obviously, be extended for a figure of six, or any number of sides. EXERCISES. 1. If HL, GE be joined, prove that the triangles KHL, FGE are similar. 2. On a given straight line AB, describe a quadrilateral similar but oppositely situated to the given quadrilateral CDEF. 3. In the figure of VI. 18, AB, CD, being parallel, prove that HG, KF, LC, if joined and produced, meet in one point. 4. If the middle points of adjacent sides of the figures ABLKH, CDEFG be joined, the figures so formed are similar. 248 EUCLID, BOOK VI. PROPOSITION XIX. Theorem. Similar triangles are to one another in the duplicate ratio of their homologou& sides. Let ABC, DEF be sim'^ .-:As, having Z. B= z. E and sides BC, EF homologous : Then shall .=^ABC be to ^^rriDBP in the dup. ratio of BC to BF. ^^ ^ iC Y.'^ *F EVom BC cut off BG a third : :i to BC and EF vi. ii. Join AG. Then, since ..^^KBC, DEF are sim'" Hyp. /.AB is to BC as DE is to EF vi. def. i. or, AB is to DE as BC is to EF Aitem. But BC is to EF as EF is to BG Constr. /. AB is to DE as EF is to BG v. ii. i.e., the sides about the = Z_ s at B, E, of .i::::\sABG, DEF, are reciprocally : •.\ /. ^^ABG=^::^DEF VI. 15. Again, since BC is to EF as EF is to BG Constr. /. BC is to BG in the dup. ratio of BC to EF v. def. lo. But .-:x ABC is to .^ABG as BC is to BG vi. i. /. ^r^ABC is to ^^ABG in the dup. ratio of BC to EF v. ii. /. .^ABC is to .-^DEF in the dup. ratio of BC to EF v. 7. Wherefore, similar triangles &c. Q.E.D. CoR. From this it is manifest that, if three st. lines be : :^^, as the first is to the third, so is any ^^ desc*^ on the first to a sim'', and similarly desc*^ ..^^ on the second. PROP. XIX. 249 NOTE. Prop. 19 is of great importance. The following theorem will be required for some of the exercises. If two ratios are equal, their duplicates are also equal. Let A, B, C, D be : :^% such that A is to B as C is to D ; Then shall the dup, ratio of A to B=dup. ratio of C to D. To A, B, take a third : A M, and to C, D, a third : -J N vi. u. Then, A is to B as B is to M. Constr. and C is to D as D is to N Constr. But A is to B as C is to D Hyp. /. B isto M as D is to N V. ii. Hence, A isto M as C isto N Ex seq. But, since A, B, M, are in cont<* prop^, ,\ A is to M in dup. ratio of A to B V. def. lo. and, since C, D, N, are in cont^ prop*^, .*. C is to N in dup. ratio of C to D V. def. lO. ,', the dup. ratio of A to B=the dup. ratio of C to D. Q.E.D. EXERCISES. 1. State and prove the converse of the theorem given above. 2. Prove that similar triangles are to one another in the duplicate ratio of (i) their altitudes; (ii) their corresponding medians; (iii) the radii of their inscribed circles; (iv) the radii of their circumscribed circles. 3. ABC is a triangle with angle A greater than B : if D be taken in BC such that angle CAD=B, then CD : CB in the duplicate ratio of AD to AB. 4. Any two equilateral triangles, described on the sides of an acute-angled triangle, are together greater than the third. 5. ABC is a given triangle: construct a similar triangle of double the area. 6. ABC is a triangle, AB is produced to E, AD is a line meeting BC in D, BF is parallel to ED and meets AD in F; determine a triangle similar to ABC and equal to AEF. 7. ABC is a triangle inscribed in a circle, AD, AE, are drawn parallel to the tangents at B, C, to meet the base in D, E. Prove that BD is to CE in the duplicate ratio of AB to AC. ( 310 ) Q 250 EUCLID, BOOK VI. PEOPOSITION XX. Theorem. Similar polygons may he divided into the same number of similar triangles, having the same ratio to one another that the polygons have; and the polygons are to one another in the duplicate ratio of their homologous sides. Let ABODE, FGHKL, be sim>- polygons, with the side AB homologous to FG. Join AC, AD, FH, FK. Then shall (i) the polyg-ons be divided into the same number of simr -^r^s ; (ii) each pair of .-r^s have the same ratio to one another that the polygons have; (iii) polygon ABODE be to polygon FGHKL in the dup. ratio of AB to FG. A ^i^ = ^^ jj. .^ (1) For, since polygon ABCDE is sim'^ to polygon FGHKL... Hyp. /.^ABC=^FGH, ) ^,^^,^ and AB is to BC as FG is to GH ) :, .^r^ABC is sim^ to .-::\FGH Ti. 6. /. z.BCA=^GHF vi.def.i. But z.BCD=^GHK Hyp. .*. rem« 2LACD=remg Z.FHK. And, since .^^AABC is sim^ to .-rrxFGH Above. /.AC is to BC as FH is to GH vi. def. i. But BC is to CD as GH is to HK Hyp. /.AC is to CD as FH is to HK Ex^q. i.e. sides about= L s ACD, FHK, of -^s ACD, FHK are : :«, /. .:ilACD is sim-^ to .-^i FHK vi.6. Similarly, it may be shown that .r^ADE is sim'- to .^r^FKL. (ii) Again, since ^^ABC is sim^ to ^AFGH Above. .'. ^^ABC is to .^r^FGH in the dup. ratio of AC to FH...V1. 19. PROP. XX. 251 And, since ..^^ACD is sim"" to -^^FHK Above. .*. ^^ACD is to .^FHK in the dup. ratio of AC to FH...VI. 19. /. .-^ABC is to .^FGH as .-rr^ACD is to .-riFHK v. ii. Similarly, it may be shown that .^ACD is to ^:::?iFHK as .^^ADE is to .^FKL. Hence, -^^ABC is to ..^FGH as sum of ..^^ris ABC, ACD, ADE is to sum of .^s FGH, FHK, FKL V. 12. le., .^ABC is to .^FGH as fig. ABCDE is to fig. FGHKL. (iii) Also, since ^^ABC is sim'^ to ^^FGH Above. /. ^^ABC is to ^^FGH in the dup. ratio of AB to FG...VI. 19. Hence, polygon ABCDE is to polygon FGHKL in the dup. ratio of AB to FG...V. 11. Wherefore, similar polygons &c. Q.E.D, CoR. To AB, FG let a third : -.^ MN be taken. Then AB is to MN in the dup. ratio of AB to FG V. def. 10. But, any polygon on AB is to the sim'^ and similarly desc*^ fig. on FG in the dup. ratio of A B to FG Above. .*. AB is to MN as fig. on AB is to fig. on FG v. 11. i.e., If three st. lines he : ;% as the first is to the third, so is the rect^ fig. desC^ on the first to a sim/" and similarly desc^ fig. on the second. NOTES. By the help of this proposition it can be shown that (1) Similar triangles are to one another as the squares on their corre- sponding sides. A -T) For, let ABC, DEF be simr ..^s and on BC, EF letsqs. BG, EH be descd. Then, ^^ABC is to .^DEF in the dup. ratio of BC to EF VI. 19. Also, since the sqs. are sim^ figs,, .*. sq. BG is to sq. EH in the dup. ratio of BC to EF VI. 20. /. .^XABC is to ^r^DEF as sq. on BC is to sq. on EF V. 11. In the same way it may be shown that (ii) Similar polygons are to one another in the ratio of the squares on their corresponding sides. Also, since ..^ABO is to .^DEF in the dup. ratio of BC to EF, and ^:^ABC is to.=^DEF as sq. on BC is to sq. on EF, it follows that (iii) The duplicate ratio of BC to EF is the same as the ratio of the squares on BC and EF. (See note on page 219.) D 252 EUCLID, BOOK VI. PROPOSITION XXL Theorem. Rectilineal figures which are similar to the same rectilineal figure, are also similar to each other. Let each of the rect^ figs. A and B be sim^ to the rect^ fig. C. Then shall A be simr to B. /h For, since A is sim^ to C Hyp. /, A is equiang. to C, and the sides about their= L s are : : ^^ VI. def. i. And, since B is sim'^ to C Hyp. .*. B is equiang. to C, and the sides about their= L s are : : ^^ vi. def. i. Hence, A is equiang. to B, and the sides about their= Z_s are : :^^ v. ii. /. A is sim^" to B VI. def. i. Wherefore, rectilineal figures &c. Q.E.D. PROPOSITION XXII. Theorem. If four straight lines be proportionals, the similar rectilineal figures similarly described on them shall also be proportionals; and, con- versely, if the similar rectilineal figures similarly described on four straight lines be proportionals, those straight lines shall be propor- tionals. Part I.— Let AB, CD, EF, GH be four st. lines, such that AB is to CD as EF is to GH, and on AB, CD, let the sim'' rect^ figs. KAB, LCD, be similarly desc*^, and on EF, GH, let the sim^ rect^ figs. MF, NH, be similarly desc^. Then shall fig. KAB be to fig. LCD as fig. MP to fig. NH. To AB and CD find a third : :i X, and to EF and GH find a third : :^ Y ....vi. n. PROPS. XXI. XXII. 253 B C^ ^D M. n R Y F G H Q . Coustr. Then, since AB is to CD as CD is to X and EF is to GH as GH is to Y Also, AB is to CD as EF is to GH Hyp. .-.CD is to X as GH is to Y v. ii. Hence, AB is to X as EF is to Y Exteq. But AB is to X as fig. KAB is to fig. LCD ) ^^ ^o cor and EF is to Y as fig. MF is to fig. NH ) /. fig. KAB is to fig. LCD as fig. MF is to fig. NH....V. ii. Part II.— Let fig. KAB be to fig. LCD as fig. MF is to fig. NH. Then shall AB be to CD as EF to GH. To AB, CD, EF find a fourth : :^ PQ vi. 12. On PQ desc. a fig. RQ sini'^ and similarly situated to MF, or NH vi. is. Then, since AB is to CD as EF is to PQ Constr. .-. fig. KAB is to fig. LCD as fig. MF is to fig. RQ...Part 1. But fig. KAB is to fig. LCD as fig. MF is to fig. NH..Hyp. /. fig. RQ = fig. NH V. 9. and RQ, NH are sim'^ figs, similarly situated, .-. PQ = GH.* But AB is to CD as EF is to PQ Constr. .-. AB is to CD as EF is to GH v. r. Wherefore, if four straight lines &c. Q.E.D, NOTE. *In this prop. Euclid assumes that — Equal, similar and similarly described rectilineal figures have their homo- logous sides equal. This is easily proved; for, by prop. 20, the figures are to one another in the dup. ratio of their homologous sides; that is, in the ratio of the squares on their homologous sides VI. 20, note, and, if the squares on them are equal, the sides are themselves equal. 254 EUCLID, BOOK VI. PEOPOSITION XXIII. Theorem. Parallelograms which are equiangular to one another have to one another the ratio tohich is compounded of the ratios of their sides. Let OAC be equiang. to OCF, having ^BCD=z.ECG. Then shall OAC be to OOP in the ratio com- pounded of the ratios of their sides. A D / B n — /G E" T Let the Os be so placed that BC, CG are in a st. line and the Os on opp. sides of this line. .*. DC, CE are in a st. line 1. 14. Complete ODC. Take any st. line K. To BC, CG, and K find a fourth : :i L VI. 12. To DC, CE, and L find a fourth : :^ M VI. 12. Then BC is to CG as K is to L and DC is to CE as L is to Mi But K is to M in the ratio compounded of the ratios of K to L, and L to M v. def. 11. .'. K is to M in the ratio compounded of the ratios of BC to CG, and DC to CE. But OAC is to ODG as BC is to CG vi. 1. as K is to L Constr. and ODG is to OCF as DC is to CE vi. 1. as L is to M Constr. .*. OAC is to OCF as K is to M Exjeq. .*. OAC is to OCF in the ratio compounded of the ratios of BC to CG, and DC to CE ...Above. 'Vlh&YQioYe, parallelograms kc. Q.E.D. PROPS. XXIII., XXIV. 255 PEOPOSITION XXIV. Theorem. Parallelograms about the diameter of any parallelogram are similar to the whole parallelogram and to one anoth&r. Let ABCD be a O^ AC a diam., and EG, HK, Os about the diam. Then shall Os EG, HK, be simr to OABCD, and to one another. B H ^ For, since DC is 1| to GF Hyp. /. aADC=Z.AGF 1.29. and, since BC is || to EF Hyp. /. z.ABC=aAEF I. 29. also, /.BCD, and Z.EFG, are each =/. BAD 1.34. /. ^BCD=z.EFG. Hence OABCD is equiang. to OAEFG. Again, in .^s ABC, AEF, since Z-ABC=Z.AEF Above. and L B AC is com. .'. -.^ABC is equiang. to ^rr^AEF 1.32. .'. AB is to BC as AE is to EF vi. 4. But BC=AD, and EF=AG 1.34. Hence, AB is to AD as AE is to AG v. 7. Also DC=AB, and GF=AE 1.34. Hence, DC is to BC as GF is to EF ^ ^ ^ and DC is to AD as GF is to AG f i.e. sides about = Z-S of Os ABCD, AEFG are : :^ /. OABCD is sim'^ to OAEFG vi. def. In the same way it may be shown that OABCD is sim' to OFHCK. /. OAEFG is sim'' to OFHCK vi. 21. Wherefore, parallelograms &c. Q.E.D. 256 EUCLID, BOOK VI. PEOPOSITION XXV. Problem. To describe a rectilineal figure which shall be similar to one, and equal to another given rectilineal figure. Let ABCD be one, and E the other, given rect^ fig. It is req'^ to desc. a rect'' fig. sim^ to ABCD and = E. A "^ - b4 ^c ^K E L' ^M F' G ^H To BC apply a OBFGC=fig. ABCD. To GC apply a OCGHK=fig. E, having z.GCK= ^FBC. Then BC and CK are in one st. line ) and FG and GH are in one st. line ) Between BC and CK find a mean : -.^ LM vi. 13. On LM desc. a rect^ fig. NM sim^ and similarly situated to fig. ABCD VI. 18. Then NM shall be the fig. reqd. For, since BC is to LM as LM is to CK Constr. /. B C is to CK as fig. AC is to fig. NM VI. 20, cor. ButBC is to CK as OBG is to £JCH VL 1. .-.fig. AC is to fig. NM as OBG is to OCH v. 11. But fig. AC =OBG Constr. /. fig. NM=OCH V.9. =fig. E Constr. And fig. NM is also sim^ to fig. AC Constr. Wherefore, has been described &c. - Q.E.F, NOTE. This prop., inserted as it is between VI. 24 and its converse VI. 26, appears somewhat out of place; it might well have followed prop. 26. It may be thus enunciated: — To make a rectilineal figure having the size of one, and the shape of another, given rectilineal figure. PROPS. XXV., XXVI. 257 PEOPOSITION XXVI. Theorem. If two similar parallelograms have a common angle, and he simi- larly situated, they are about the same diameter. Let the Os ABCD, AEFG be sim'^ and similarly situated, and have the com. L BAD. Then shall Os ABCD, ABPG, be about the same diam. B* *C Join AC. Then, if the diam. AC of OABCD do not pass through F, let it, if possible, cut FG, or FG prod arc EM, ^BOH> ^EQM,) if arc BHarc EM, sect. OBH > sect. QEM, if arc BH were inserted by Simson. EXERCISES. 1. State and prove the converse of Prop. B. 2. Prove that, if the exterior angle at A be bisected by AD meeting the base produced at D, then rect. AB, AC = rect. BD, DC - AD^. 3. Prove that in equal circles equal sectors stand on equal arcs. 4. Prove, what is assumed in Prop. 33, that (i) in equal circles the greater angle stands upon the greater arc. (ii) in equal circles the greater sector stands upon the greater arc. PROPS. B, C. PEOPOSITION C. Theorem. If from the vertical angle of a triangle a straight line he 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 described about the triangle. Let ABC be a .^^, with AD ± to BC. Then shall rect. BA, AC = rect. contd by AD and the dianar of the desc^ about the ^^. E Desc. a about -,^ABC iv. 5. Draw diam^ AE. Join EC. Then, Z.ECA in a semicircle is a rt. Z. III. 31. And, since rt. z.BDA=rt. z_ECA, and L ABD= z_ AEC in same seg* ill. 21. /. ^i^AABD is equiang^ to ^::AAEC I. 32. /. BA is to AD as EA is to AC vi. 4. /. rect. BA, AC=rect. EA, AD vi. 16. "Wherefore, if from the vertical &c. Q.E.D. EXERCISES. 1. State and prove the converse of prop. C. 2. If AB and AC are equal, what form does the proposition assume? 3. Construct a triangle having given the base, the radius of the circum- scribed circle, and the rectangle contained by the sides. 264 EUCLID, BOOK VI. PROPOSITION D. Theorem.- The rectangle contained hy the diagonals of a quadrilateral figure inscribed in a circle is equal to both the rectangles contained bj/ its opposite sides. Let ABCD be a quad^ insc^ in a 0, and let AC, BD be its diags. Then shall rect. BD, OA=sum of rects. AB, CD and BO, DA. At A in BA make L BAE= l DAC, and let AE meet BD at E. Then, since ^BAE=2_DAC Constr. Add L EAC to each. /. ^BAC=^EAD. But z. BC A= L EDA in same seg* Hence, .^^ABC is equiang"^ to ^^AW) .. /. BC is to CA as ED is to DA., /. rect. BC, DA=rect. ED, CA... Again, since Z.BAE=z.DAC and Z.ABE=z.ACD in same ,'. .i::^ABE is equiang'^ to ^:::\ACD ..... . /. AB is to BE as AC is to CD.., /. rect. AB, CD=rect. BE, CA.... But rect. BC, DA=rect. ED, CA .. /. sum of rects. AB, CD and BC, DA =sum of rects. BE, CA and ED, CA. =rect. BD, CA ii. i. Wherefore, the rectangle contained &c. Q.E.D. seg ..III. 21. ..I. 32. ..VI. 4. ..VI. 16. ..Constr. ..III. 21. ..T. 32. ..VI. 4. ..VI. 16. . .Above. PROP. D. 255 NOTES. The important proposition D is sometimes referred to as "Ptolemy's Theorem," as it occurs in his work on Astronomy. (Ptolemy lived at Alexandria about a.d. 70.) It is the special case of the following theorem — The rectangle contained by the diagonals of any quadrilateral is less than the sum of the rectangles contained by its opposite sides, unless the quadri- lateral can be inscribed in a circle. Let ABCD be a quad^ which cannot be insc^ in a 0, i.e. having Z_ABD not=Z_ACD. At B in AB make Z.ABE=Z.ACD. At A in AB make L BAE= L CAD. Join ED. Then, since ^;:::XABE is equiang'' to ^^^ACD...Constr. /. AB is to BE as AC is to CD vi. 4. /. rect. AB, CD=rect. BE, AC vi. 16. Again, since Z.BAE=Z.CAD Constr. Add L EAC to each, /. Z.BAC=Z.EAD. B" Also, since .i^KS>¥i is equiang^ to ^^\ ACD Above. .*. BA is to AE as CA is to AD vi. 4. or, BA is to CA as AE is to AD Altern. Hence .=^ABC is equiang^ to .-riXAED VI. 6 .*. BC is to CA as ED is to DA VI. 4. /. rect. BC, DA=rect. ED, AC VI. 16. Hence, sum of rects. AB, CD and BC, DA =sum of rects. BE, AC, and ED, AC =rect. contd by AC and sum of BE, ED >rect. cont either r^ or r^. "With cent. O2 and rad.=r2 - fi desc. a 0. With cent. O3, and rad.=r3-ri desc. a 0. Desc. a to pass through pt. Oi and touch these two 0s at K, L . Find O its cent. Join OOi cutting the Oce in M. With cent. O rad. OM desc. a ; this will be the reqd. NOTE. Examples XI., XII., XIII., XIV. belong to a class of problems known as The Tangencies; i.e., to describe a circle which shall, in the case of a point, pass through, and, in the case of a line or circle, touch any given three of the following nine — three points, three lines, three circles. These can be arranged in ten different sets; of which four are the examples mentioned above, two others are Exs. 4, 5 below, and for the remaining four see page 198, Prop. 4; page 200, Prop. 5; page 182, Ex. IX.; page 188, Ex. 42. EXERCISES. 1. Draw the three other figures which fulfil the conditions of the problem (i) in Ex. XII. ; (ii) in Ex. XIII. 2. Prove Ex. XIV. 3. In the figure of Ex. XIV. the circle has been described to touch all the given circles externally; how many possible solutions are there if the circle touch any one or more of the given circles internally ? 4. Describe a circle which shall touch two given circles and a given straight line. 6. Describe a circle to touch two given straight lines and a given circle. 6. The Tangency solved in Ex. XII. may be indicated shortly thus : — "point, line, circle;" indicate the remaining nine in a similar way. 276 EUCLID, BOOK VI. Def. — If opposite pairs of sides of a quadrilateral be produced to meet, and their points of intersection be joined, the line which joins these points is called the third diagonal of the quadrilateral ; and the figure thus formed is called a complete quadrilateral. XV. The middle points of the three diagonals of a complete quadrilateral lie in one straight line. Let ABCD be a quad^ with BA, CD, prod^ to meet at E, and AD, BC, prod'^ to meet at F, and let the three diags. AC, BD, EF be bisected at 0, Q, R. Then shall OQR be a St. line. .E Complete EJb EBGD, AHCE, and let DG, AH meet BC at K, L. Then ^ = ^^ = IK ^^^ CL DA KB . FC _ CL^ _ CH ^ •* FK "" KB ~" KG ■ i.e., the sides about the = Z-S at C, K, of .^r^s FCH, FKG are : :\ :, ^^FGH is equiang. to ^^^'KGr VI. 6. Hence, the side GF of ^r^FKG must pass through H, i.e., GHF is a st. line. But, since the diags. of a O bisect each other Ex. r, p. 67. .*. the diag. EG of O EBGD is bisected at O, and diag. EH of O EAHC is bisected at Q. /. OQ is II to GH Ex. p. 69. Similarly, QE, is 1| to HF. But GHF is a st. line Above. ,', OQR is also a st. line. Q.E.D EXERCISES. 1. The circles described on the three diagonals of a complete quadri- lateral have the same radical axis. 2. The four circles round the four triangles formed by four straight lines have a common point. The method of proof used in the above prop, is due to the Rev. C. Taylor, D.D., and is inserted here with his permission. MISCELLANEOUS EXAMPLES. 277 XVI. If he the centre, and R the radius, of the circumscribed circle of a triangle A RC; Q the centre, and r the radius of the inscribed circle, then 0Q'^=Ii!^-2]ir. D Join AQ, and prod. AQ to meet the circumscribed at D. Join DO, and prod. DO to meet the same at E. Join BD, BE, BQ, OQ, and prod. OQ to meet Oce at F. G. Draw QN ± to AB. Then Z_ BED = /_ NAQ in same seg* III. 21. andrt. Z_ANQ=rt. Z.EBD in semicircle EBD in. 31. Hence .^^^As EBD, ANQ are simr. .*. ED is to DB as AQ is to QN VI. 4. /. ED.QN = DB.AQ VI. le. i.e., 2R.r= DB . AQ. But ^DQB=^sQBA + BAQ L 32. = 4 Z.S A + B. And L DBQ = L s DBG + CBQ. = Z.sDAC + CBQ m. 2L = 1 Z_s A + B. /. DB = Q.D L6. Hence, 2 Rr = AQ . QD. = GQ.QF m.36. = (Oa-OQ).(OF + OQ). = (R-OQ).(E + OQ). = R2_OQ2 n.5,Cor. *.e., OQ2=R^-2Rr. q j, j^ EXERCISE. Prove that if S be the centre and r^ the radius of the escribed circle to the side BC, then OS2 = R2 + 2Rra. The above theorem is known as Euler's Theorem. 278 EUCLID, BOOK VI. MISCELLANEOUS EXERCISES. 1. Triangles on equal bases are to one another as their altitudes. 2. If ACB, BCD be equal angles and BD be perpendicular to BC and BA to AC, prove that the triangle DBC is to the triangle ABC as DC is to CA. 3. D is a point in the side AC of a triangle ABC, E a point in AB. If BD, CE divide each other into parts in the ratio 4 : 1, then D, E divide CA, BA in the ratio 3:1. 4. Through a given point O two straight lines are drawn meeting two fixed lines which intersect in A; one of the lines BC is bisected in O, and the other DE makes equal angles with the fixed lines: prove that AB + AC=AD + AE. 5. Enunciate the propositions which prove that in the case of triangles the conditions of similarity are not independent. 6. The side BC of a triangle ABC is produced to D, so that the triangles ABD, ACD are similar. Prove that AD touches the circle described about the triangle ABC. 7. If A' B' C are respectively the feet of the perpendiculars from A, B, C on the sides of the triangle ABC, show that the triangle AB' C is similar to ABC. 8. If D be the middle point of the side BC of the triangle ABC and if any straight line be drawn through C, meeting AD in E and AB in F, then the ratio of AE to ED will be double of the ratio of AF to FB. 9. If the vertical angle of a triangle be bisected, and if two lines be drawn through the vertex equally inclined to the bisector, one of which meets the base and the other the circumscribing circle of the triangle, the rectangle contained by them is constant. 10. ABCD is a parallelogram, and APQ is drawn cutting BC and DC produced in P and Q. If the angle ABP' be made equal to the angle ADQ', BP'=BP, and DQ'=DQ, show that the angles PBP', QDQ', P'AQ' will be equal, and that the ratio of AP' to AQ' will be equal to the ratio of AP to AQ. 11. ABC is a triangle; the angle A is bisected by AD meeting BC at D ; in AC a point E is taken such that AE is a third proportional to AB, AD : prove that angle CDE is half of angle BAC. 12. The angle B of a triangle ABC is bisected. Lines AD, CE are drawn from A, C at right angles to the bisector. Prove that rect. AD, BE =rect. BD, CE. 13. Draw a straight line such that the perpendiculars let fall from any point in it on two given straight lines may be in a given ratio. MISCELLANEOUS EXERCISES. 279 14. In a triangle ABC, BC is bisected in D, AD is bisected in E, and BE is produced to meet AC in F : show that AF=i AC, and EF=| BF. 15. From two fixed points A, B perpendiculars AC, BD are drawn on a fixed line CD, Find, when possible, a point P in CD such that the rectangle CP, PD may be equal to the rectangle AC, BD. 16. ABC is a triangle having the angle ACB double of the angle ABC ; the bisector of the angle ACB meets AB in D: prove that the square on AC is equal to the rectangle AD, AB. 17. ACB is a diameter of a circle, CD a radius perpendicular to it. The chord AD is bisected in E ; BE meets CD in F and the circle in G. Show that three times the rectangle contained by BF, EG- is equal to the square on the radius. 18. A tangent to a circle at the point A, intersects two parallel tangents in B, C, the points of contact of which with the circle are D, E respectively: show that if BE, CD intersect in F, AF is parallel to the tangents BD, CE. 19. Find a point in the side of a triangle from which two lines, drawn one to the opposite angle, and the other parallel to the base, shall cut off, towards the vertex and towards the base, equal triangles. 20. In the triangle ABC, AC=2 BC, CD, CE bisect angle C, and the exterior angle formed by producing AC: prove that the triangles CBD, ACD, ABC, CDE have their areas as 1 : 2 : 3 : 4. 21. If a point O be taken within a parallelogram ABCD such that the angles OBA, ODA are equal, prove that the angles OAD, OCD are equal. 22. The opposite sides BA, CD of a quadrilateral ABCD, which can be inscribed in a circle, meet, when produced, at E ; F is the point of intersection of the diagonals, and EF meets AD at G : prove that the rectangle EA, AB is to the rectangle ED, DC as AGT is to G-D. 23. From the angular points of a parallelogram ABCD perpendiculars are drawn on the diagonals meeting them in E, F, G, and H respectively; prove that EFGH is a parallelogram similar to ABCD. 24. In a given square inscribe a square of which the area shall be equal to three-fourths of that of the given square. 25. Given the base and the ratio of the sides of a triangle, find the locus of the vertex. 26. From a given point without a circle draw a straight line cutting the circle, and dividing the diameter perpendicular to it in a given ratio. 27. AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced, if necessary; if from any point C of AB a perpendicular be drawn to AB, meeting AP and BP in points D and E respectively, and the circumference of the circle in a point F, show that CD is a third proportional to CE and CF, 280 EUCLID, BOOK VI. 28. In a given circle place a straight line parallel to a given straight line and having a given ratio to it; the ratio being not greater than that of the diameter to the given line. 29. ABC being a given triangle, show how to construct a similar triangle which has double the area of ABC. 30. EA, EA' are diameters of two circles touching each other externally at E ; a chord AB of the former circle when produced touches the latter at C, while a chord A'B' of the latter touches the former at C: prove that the rectangle contained by AB, A' B' is four times as great as that contained by BC, B'C. 31. Three lines AA', BB', CC drawn from the angles of a triangle to meet the opposite sides in A', B', C meet in a point P such that the ratios AP : PA', BP : PB', and CP : PC are all equal: find the position of P. 32. Draw through a given point a straight line, so that the part of it inter- cepted between a given straight line and a given circle may be divided at the given point in a given ratio. Between what limits must the ratio lie in order that a solution may be possible? 33. AB is a fixed chord of a circle; PQ is any chord bisected by AB : prove that the locus of the point of intersection of the tangents at P and Q is a circle. 34. Two circles touch one another at C. Any double chord PCQ is drawn through C. If P' be the other extremity of the diameter through P, show that P' Q passes through a fixed point. 35. A, B, C, D are pts. in a st. line; AC is the harmonic mean between AB, AD. Prove that if AC is bisected in O, OC is the geometric mean between OB, OD. 36. Construct an equilateral triangle equal in area to a given triangle. 37. From a given point outside two given circles, which do not meet, to draw a straight line such that the portions of it intercepted by each circle shall be respectively proportional to their radii. 38. The diagonals AC, BD of a quadrilateral figure inscribed in a circle cut at E : show that rect. AB, BC : rect. AD, DC : : BE : ED. 39. A and B are fixed points on the circumference of a circle, and C is the middle point of the arc AB; show that, if D be any other point on the circumference, then DA + DB is to DC in a constant ratio. 40. The straight lines EAB, EDC and FDA, FCB form four triangles in one plane ; O is the common point of intersection of the circles circumscribing these triangles : prove that the rectangle contained by OA and OC is equal to the rectangle contained by OE and OF. BOOK XL DEFINITIONS. 1. A solid is tliat which has length, breadth, and thickness. 2. That which bounds a solid is a sur- face. 3. A straight line is perpendicular to a plane whe« it is perpen- dicular to every straight line which can be drawn in that plane to meet it. 4. A plane is perpendicular to an- other plane when straight lines, drawn in the one plane perpen- dicular to the common section of the two planes, are also perpendicular to the other plane. Xl NOTES. The first six Books of Euclid treat of Plane Geometry; in other words, all the lines composing the figures, the properties of which are to be con- sidered, are supposed to lie in one plane; consequently, the figures can be easily represented on paper. But Book XI. treats of Solid Geometry, or Geometry in Space, and, the lines of the figures lying in different planes, these can only be represented on paper by perspective sketches. This con- stitutes the chief difficulty of the beginner in Solid Geometry, (especially if he has had no preliminary training in Drawing), as equal lines or angles are no longer necessarily represented by equal lengths or angles on the paper, as was the case in Plane Geometry. To represent a figure in Solid Geometry with the same degreee of correctness as those of Plane Geometry, it would be necessary to construct models with sheets of card -board for the different planes and wires for the lines. In the diagrams which follow, the lines used to indicate the position of the different planes are varied, a darker line being used to denote the edge of any plane, or figiare, nearest to the observer. Def . 1 amounts to the statement that space has three dimensions. (310) S 282 EUCLID, BOOK XI. 6. The inclination of a straight line to a plane is the acute angle between that line and the line joining the point at which it meets the plane to the point at which the per- pendicular, drawn from any point in it to the plane, meets the plane. 6. The inclination of a plane to a plane is the acute angle between two straight lines drawn at right angles to the common section of the planes, from any point in it, one in one plane and one in the other. 7. Two planes are said to have the same inclination to one another which two other planes have, when their angles of inclination are equal. 8. Parallel planes are such as do not meet when produced. ADDITIONAL DEFINITIONS. The line in which two planes cut one another is called their common section. The point at which a perpendicular, drawn to a plane, meets the plane, is called it% foot. The projection of a point on a plane is the foot of the perpendicular drawn from the point to the plane. The projection of a line on a plane is the locus of the feet of the perpendiculars drawn from all points in the line to the plane. Hence, the inclination of a line to a plane may be defined as the angle between the line and its ^ ^ projection on the plane. The inclination of a plane to a plane is called a dihedral angle. A line drawn at right angles to a plane is said to be normal to the plane. r DEFINITIONS. 9. A solid angle is contained by three, or more, plane angles, which have a common vertex (V), but are not in the same plane. 10. Equal and similar solid figures are such as are contained by similar planes equal in number and magnitude. 11. Similar solid figures are such as have all their solid angles equal, and are contained by the same number of similar planes. 12. A pyramid is a solid figure contained by planes, of which all but one (the base) pass through the same point (the ve7'tex). 13. A prism, is a solid figure contained by plane figures, two of which (the bases) are equal, similar, and simi- larly situated ; and the others, con- sequently, are parallelograms. ADDITIONAL DEFINITIONS. A trihedral angle is a solid angle contained by three plane angles. A polyhedral angle is a solid angle contained by more than three plane angles. The plane surfaces of a solid figure are called its /aces; the straight lines in which these surfaces intersect, its edges; the solid angles its corners. A pyramid on a triangular base is called a tetrahedron. A prism, the sides of which are rectangles, is called a right prism. The altitude of a pyramid is the perpendicular drawn from its vertex to its base ; and the altitude of a prism is the perpendicular distance between its bases. NOTE. The solid figures dealt with by Euclid are all convex figures; i.e. they have no re-entrant angles (see p. 60, note). Without this restriction, Def. 10 would not hold good. 284 EUCLID, BOOK XI. 14. A sphere is a solid figure described by the revolution of a semi- circle about its diameter, which remains fixed. 15. The axis of a sphere is the fixed straight line about which the semi-circle revolves. 16. The ceyitre of a sphere is that of the semi-circle. 17. A diameter of a sphere is a straight line drawn through the centre, terminated both ways by the surface of the sphere. 18. 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. The cone is called right-angled, obtuse-angled, or acute- angled, according as the triangle has its fixed side equal to, less than, or greater than the other of the two sides which contain the right angle. 19. The axis of a cone is the fixed side about which the triangle revolves. 20. The base of a cone is the circle described by that side, (of the two containing the right angle), which revolves. ADDITIONAL DEFINITIONS. A radius of a sphere is a straight line drawn from the centre to any point in the surface of the sphere. The solid figure of Def. 18 is often called a right cone; and a portion of it, ADE, cut oiF by a plane not parallel to the base, an oblique cone. The point A is called the vertex. The portion FBCG included between two parallel planes is called a The altitude of a cone is the perpendicular distance of the vertex from the plane of the base. The hypotenuse AB of the right-angled triangle is called a gener- ating line of the cone, or its slant side. DEFINITIONS. 285 DEFINITIONS. 21. A cylinder is a solid figure described by the revolution of a rectangle about one of its sides, which remains fixed. 22. The axis of a cylinder is the fixed side about which the rectangle revolves. 23. The bases of a cylinder are the circles de- scribed by those opposite sides of the rec- tangle which revolve. 24. Similar cones, and cylinders, are those which have their axes, and the diameters of their bases, proportionals. 25. A cube is a solid figure, contained by six equal squares. 26. A regular tetrahedron is a solid figure contained by four equal, equilateral tri- angles. on. A regular octahedron is a solid figure contained by eight equal, equilateral tri- angles. 28. A regular dodecahedron is a solid figure contained by tivelve equal, regular pentagons. 29. A regular icosahedron is a solid figure contained by twenty equal, equilateral tri- angles. 286 EUCLID, BOOK XI. ADDITIONAL DEFINITIONS. A parallelepiped is a solid figure contained by six quadrilaterals of which each opposite pair are parallel. A rectangular parallelepiped has all its faces rectangles. » I ^>. 1" yc, A line, AC, joining opposite corners a parallelepiped is called a dia- gonal. LtS / 7 -->.^ / of jj----""p A polyhedron is any solid figure bounded by planes. A regular polyhedron has all its faces equal and similar regular polygons. Hexahedron, octahedron, &c., are names for polyphedra of 6, 8, &c., faces. The solid figure of Def. 21 is often called a right cylinder; and a portion of it, ABCD, cut ofi* by parallel planes, not parallel to its bases, is called an oblique cylinder. The altitude of a cylinder is the perpendicular distance between its bases. The side of the rectangle parallel to the fixed side is called a generating line of the cy- linder. A straight line is said to be parallel to a plane when they do not meet if produced. The angle between two lines which do not meet is the angle between two other lines parallel to them respectively, which meet at any point. The volume of a solid figure is the space enclosed by its surfaces. DEFINITIONS. 237 NOTES. The regular tetrahedron, cube, regular octahedron, regular dodecahedron, and regular icosahedron are called the five regular solids. It is shown here- after (see page 316) that there can be but five. The sphere, cone, and cylinder are called solids of revolution. The parallelepiped is a prism whose bases are parallelograms, and the cube a parallelepiped whose faces are squares. A plane can be supposed to revolve round the straight line joining any two points in it while this line remains unmoved, and can thus occupy any position in space — hence we see that an infinite number of planes can be drawn through two fixed points, but only one plane through three points which are not all in a straight line. EXERCISES. 1. How many faces, edges, corners, has (i) a tetrahedron, (ii) a cube, (iii) a hexahedron, (iv) an octahedron, (v) a dodecahedron, (vi) an icosahedron? 2. How many diagonals can be drawn in (i) a tetrahedron; (ii) a parallelepiped? 3. What is the least number of faces that (i) a pyramid; (ii) a prism, can have? 4. Find the length of a generating line of a cone whose altitude is 5 inches, and the diameter of its base 2 feet. 5. A line, one extremity of which is fixed, moves freely in space; what is the locus of the other extremity? 6. A line 3 inches long is inclined to a plane at an angle of (i) 30°, (ii) 45°, (iii) 60°; find its projection on the plane. 7. Find the height of a frustum of a right cone, the radii of the ends of which are 11 and 17 inches, and the slant side of which is 10 inches. 8. The height of a frustum of a right cone is 8 inches, and the radii of its ends are 9 and 12 inches; find the height of the cone. 9. How many plane angles form each solid angle of an icosahedron? 10. Find the height of a pyramid on a square base, whose sides are equi- lateral triangles the sides of which are each 3 inches long. 288 EUCLID, BOOK XI. PEOPOSITION I. Theorem. One part of a straight line cannot he in a plane and another part without it. If it be possible, let the part AB, of the st. line ABC, lie in the plane EF, and the part BC without it. ;H Then, since the st. line AB is in the plane EF, it can be prod*^ in that plane.* Prod. AB to any pt. D in the plane EF. Let a plane GH, which passes through ABD, be turned about ABD until it passes through the pt. C. Then, since B and C are pts. in this plane, /. the st. line BC lies in it, I. def. 7. At B in this plane draw any st. line BK. Then, since ABC is a st. line, Hyp. .*. Z.S ABK, KBC=two rt. /.s, Lis. And, since ABD is a st. line Constr. /. Z.S ABK, KBD=twort. z.s 1.13. /. L s ABK, KBC = L s ABK, KBD. Take away the com. L ABK. /. rems L KBC=rem& Z.KBD. or, the part=the whole, which is absurd. Wherefore, one part of a straight line «&c. Q.E.D. NOTE. The proof of this proposition is made to depend upon that of the follow- ing : Two straight lines cannot have a common part, or segment, a proof of which was inserted by Simson as a corollary to Prop. 11 of Book I., but of which he made no use before Prop. 1 of Book XI. It hardly seems to need formal proof, and any necessity might be removed by some such extension of Axiom 10 as follows: — "Two straight lines cannot enclose a space; and if two st. lines meet in more than one point, they must coincide throughout their length; also, if produced, they will continue to be coincident." * This follows from Post. 2 and the definition of a plane, and is tacitly assumed all through the first Six Books. PROPS. I., II., III. 289 PEOPOSITION II. Theorem. Two straight lines which cut one another are in one plane; and three straight lines which meet one another are in one plane. Let the st. lines AB, CD cut at E, and in AB, CD let any pts. B, C be joined. Then shall (i) AB, CD be in one plane; (ii) AB, BC, CD be in one plane. Let any plane through AB be turned about AB until it passes through C. Then (i), since the pts. C and E are in this plane, ^.^ ^^^ ^^^^^ ^^ j.^^ ^^j^ .^ -^^ -^ ^^ ^ i.e., AB and CD are in one plane. Again, (ii), since the pts. B and C are in this plane, /. St. line BC is in it I.,def. 7. i.e., AB, BC, CD are in one plane. Wherefore, two straight lines &c. Q.E.D. PEOPOSITION III. Theorem. If two planes cut one another their common section is a straight line. Let the planes AB, CD cut one another in the line EF. Then shall EF be a st. line. For, if not, if possible, draw the st. line EGF in plane AB, and the st. line EHF in plane CD. Then the st. lines EGF, EHF, meeting at E and F, enclose a space, which is im- possible. Hence EF cannot be other than a st. line. Wherefore, if two playies cut &c. Q.E.D. EXERCISES. Show that but one plane can be made to pass through (i) two straight lines which cut; (ii) the sides of a triangle; (iii) three points not collincar [i.e., not in the same straight line); (iv) two parallel straight lines. 290 EUCLID, BOOK XI. PEOPOSITION IV. Theorem. If a straight line he 'perpendicvlar to each of two straight lines at their point of intersection, it is perpendicular to the plane which passes through them, that is, in which they are. Let the st. line PN be ± to AB and CD at their pt. of intersec- tion N. Then shall PN be j. to plane BP in which AB, CD lie. P E^ s: Cut off NA, NB, NC, ND all equal to one another. Join AC, BD. Through N draw any st. line GNH, cutting AC, BD, at G, H. Take any pt. P in PN. Join PA, PB, PC, PD, PG, PH. Then in ^\^ ACN, BDN, AN=BN Constr. 'CN=DN" Constr. ^ANC=z.BND 1.15. /. AC=BD, and 2LCAN=aDBN 1.4. Hence, in .^s AGN, BHN, /LGAN=Z.HBN Above. IGNA=^HNB 1.15. AN=BN Constr. /. AG=BH, and GN=HN I. 26. Again, in .^s PAN, PBN, AN=BN Constr. PN is com. rt. Z-PNA=rt.^PNB Hyp. /. PA=PB 1.4. Similarly, PC=PD. PROP. IV. 291 Hence, in -^s PAC, PBD, AC=BD ^ PA=PB I Above. PC=PD ) .-. z.PAC=Z.PBD 1.8. And, in .:^s PAG, PBH, PA=PB ^ AG=BH I Above. z.pag=^pbhJ /. PG=PH 1.4. Now in ^s PGN, PHN, GN=HN Above. PN is com. PG=PH Abovt,. .-. z.PNG=^PNH 1.8. /. PN is ± to GH I. def.io. Similarly it may be shown that PN is ± to every st. line through N in the plane EF. /. PN is A. to the plane EF xi.def.3. Wherefore, if a straight line &c. Q.E.D. NOTE. The above is Euclid's proof of this important proposition; a shorter proof will be found on page 328. EXERCISES. 1. If a line be drawn through the centre of a circle perpendicular to the plane of the circle, any point in this line is equidistant from all points in the circumference. 2. At a given point in a given plane but one straight line can be drawn at right angles to the plane. 3. The edges of a rectangular parallelepiped are 3, 4, and 12 inches long respectively; find the length of a diagonal. 4. Prove XI. 4 from the following construction: — Let PN be perpendicular to AN, EN; in any line NC in the plane of AN, BN take a point C, and through C draw ACB such that AB is bisected at C; join PA, PB, PC. (Apply the theorem given on page 120.) 292 EUCLID, BOOK XI. PEOPOSITION V. Theorem. If three straight lines meet at one 'point, and a straight line stand at right angles to each of them at that point, the three straight lines shall be in one and the same plane. Let PN be at rt. Z.s to AN, BN, and CN, at their pt. of inter- section N. Then shall AN, BN, and CN be in the same plane. For if not, if possible, suppose CN to lie without the plane EF in which AB and CD are. Let a plane through PN be turned about PN until it passes through the pt. C. This plane will intersect the plane EF in a st. line ND XI. 3. And, since PN is ± to AN and BN, /, PN is ± to the st. line DN in the same plane xi. 4. i.e. L PND is a rt. L. But Z.PNC is a rt. L Hyp. .•.Z.PND=^PNC, or, the whole = its part, which is absurd. Hence, CN cannot lie without the plane EF. i.e. AN, BN, CN are in the same plane. Wherefore, if three straight lines &c. Q.E.D. PEOPOSITION VI. Theorem. If two straight lines be at right lines to the same plane, they shall be parallel to one another. Let AB and CD be each ± to the plane EF. Then shall AB be || to CD. Let AB, CD meet the plane EF at the pts. B, D. Join BD. In plane EF draw DG at rt. As to DB. CutoffDG=AB. Join BG, AG, AD. PROPS, v., VI. Then, since AB is ± to plane EF Hyp. /. each of the Ls ABD, ABG is a rt. L XI. def. 3. Similarly, each of the L s CDB, CDG is a rt. Z- . Hence, in .^s ABD, GDB, AB=GD Constr. BD is com. rt. ^ABD=rt. ^GDB. /. AD=BG 1.4. And, in .^a ABG, GDA, AB=GD Constr. BG=DA Above. ^AG is com. /. z.ABG=Z.GDA 1.8. But z. ABG is a rt. L Above. /. L GDA is a rt. L. Hence GD is at rt. z. s to each of the st. lines BD, AD, CD, at their pt. of iiitersection D. /, BD, AD, CD are in the same plane XI. 5. But AB lies in the same plane with BD and AD xi. 2. ,*, AB and CD are in the same plane. Also, Z.S ABD, CDB are two rt. /.s Hyp. /. ABis II to CD 1.28. Wherefore, if two straight lines &c. Q.E.D. NOTE. It ia important to notice that AB and CD must be shown to lie in the same plane before I. 28 can be applied to prove them parallel. 294 EUCLID, BOOK XI. PEOPOSITION VII. Theorem. If two straight lines be parallel, the straight line drawn from any point in one to any point in the other, is in the same plane with the parallels. Let AB be || to CD, and let E, F be any pts. in them. Then shall st. line EF lie in the plane of AB, CD. Join EF. A E B Then, since AB is || to CD Hyp. .*. AB and CD are in the same plane. . .1. def. 35. But, E and F are pts. in this plane, C F D /. the St. line EF lies wholly in this plane I. def. 7. Wherefore, if two straight lines &c. Q.E.D. Z PEOPOSITION VIII. Theorem. If two straight lines he parallel, and one of them be at right angles to a plane, the other shall also be at right angles to the same plane. Let AB be || to CD, and let AB bexto the plane EF. Then shall CD be also a. to the plane EF. Ai^ C Let B, D, be the pts. in which AB, CD, meet plane EF. Join BD. In plane EF, draw DG at rt. z_ s to BD. Cut off DG=AB. Join BG, AG, AD. Then, since AB is ± to plane EF Hyp. /. each of the L^ ABD, ABG is a rt. L XI. def. 3. And, since AB is || to CD Hyp. /. z_s ABD, CDB=twort. ^s 1.29. But ^ ABD is a rt. L Abore. /. L CDB is also a rt. L. Again, in ^^s ABD, GDB, ( AB=GD Constr. •/ -[ BD is com. irt. Z-ABD=rt. z_GDB, .-. AD=GB 1.4. PROPS. VII., VIII. 295 And, in .^s ABG, GDA, AB=GD Constr. GB=AD Above. AG is com. /. ^ABG=Z.GDA 1.8. But A ABG is a rt. L Above. .*, L GDA is also a rt. L. Hence, since each of the L s GDB, GDA is a rt. z. , /. GD is ± to the plane in which BD, AD lie xi. 4. But, BD, AD are both in the plane of the || s XI. 7. /. Z.CDG is art. L XL def. 1. But Z.CDB is art. L Above. i.e. CD is ± to both BD and GD, /. CD is ± to the plane EF, in which they are.... XL 4. Wherefore, if two straight lines &c. Q.E.D. EXAMPLE. The locus of points equidistant from two fixed points is the plane which bisects at right angles the line joining the two points. Let A, B be the fixed pts., C the mid. pt. of AB, DE the plane bisecting AB at rt. L s. Take any pt. P in plane DE. Join PA, PB, PC. Then, in ^:^s PAC, PBC, AC=BC Hyp. PC is com. jt. L PCA=rt. L PCB XL def. 3 /. PA=PB L4. i.e. any pt. in plane DE is equidist. from A and B. ,*, plane DE is the locus of such pts. Q.E.D. EXERCISES. 1. Find the locus of straight lines which are at right angles to a fixed straight line at a fixed point. 2. The perpendicular is the shortest line that can be drawn to a plane from a given external point; and of any others, those which make equal angles with the perpendicular are equal; and conversely. 296 EUCLID, BOOK XI. PEOPOSITION IX. Theorem. 2^wo straight lines, which are each of them parallel to a third straight line not in the same plane with them, are parallel to one another. Let AB and CD be each II to EF. Then shall AB be || to CD. In EF take any pt. G. In the plane containing AB and EF, draw GH at rt. z.s to EF, meeting AB at H. In the plane containing CD and EF, draw GK at rt. z.s to EF, meeting CD at K. Then, since EF is at rt. /.s to GH and GK Constr. /. EF is J. to the plane HKG xi. 4. ButABis II toEF Hyp. /. AB is also ± to the plane HKG XI. 8. Similarly, CD is j. to the plane HKG. i.e., AB and CD are both J. to the plane HKG, /. AB is II to CD XL 6. Wherefore, two straight lines &c. Q.E.D. PEOPOSITION X. Theorem. If two straight lines meeting one another he parallel to two others that meet hut are not in the same plane with the first two, the first two and the other two shall contain equal angles. Let A B, AC be || to DE, DF respectively. Then shall l BAC= l EDP. Cut off AB, AC, DE, DF all equal. Join AD, BE, CF, BC, EF. Then, since AB is = and || to DE Constr. /.BE is = and || to AD.... I. Similarly, CF is = and || to AD. Hence BE is = and || to CF.. .*. BC = EF Hence, in ^^s ABC, DEF, ( AB=DE ) V < AC=DF ] \ BC =EF Above. /. ^BAC=^EDF ....1.8. Wherefore, if two straight lines &c. Q.E.D. .Constr. PROPS. IX.. X., XI. 297 PROPOSITION XI. Problem. To draw a straight line perpendicular to a given plane from a given point without it. Let EF be the given plane, and P the given external pt. It is req^ to draw from P a st. line ± to the plane EF. \V -.F Draw any st. line AB in the plane EF. From P draw PM ± to AB 1. 12. Then if PM is ± to, plane EF, what was reqK For, if not, the st. lines EF, GH in the same plane will meet if prod'i either towards E, G, or F, H I. del 35. Let them, if possible, meet when prod'^ towards F, H, at the pt. K. Then, since EF, a part of st. line EFK, is in plane AB Hyp. /. K is in plane AB xi. i. Similarly it may be shown that K is in plane CD. .*. the planes AB and CD, if prod'^, will meet, which is impossible, for they are || Hyp. Hence, EF, GH cannot meet, when prod^ towards F, H. Similarly it may be shown that they cannot meet, when prod^ towards E, G. • EFislItoGH I.def.35. Wherefore, if two parallel planes &c. Q.E.D. NOTE. The following are other forms of the enunciation of Proposition XV. : — If one pair of intersecting straight lines be parallel to another pair in a different plane, the plane of the first pair shall be parallel to the plane of the second pair; or, If two angles, in different planes, have their arms respectively parallel, the planes are also parallel. 302 EUCLID, BOOK XI. PKOPOSITION XVII. Theorem. If two straight lines he cut hy parallel planes, they shall he cut in the same ratio. Let the st. lines AB, CD be cut by the || planes GH, KL, MN, at the pts. A, E, B and C, F, D. Then shaU AB be to BB as CP to PD. B _33 ^l^ Join AC, AD, BD. Let AD cut the plane KL at O. Join EO, OF. Then, since the plane ABD cuts the || planes KL, MN, /. the com. sections EO, BD are || XL 16. And, since the plane DAC cuts the || planes GH, KL, /. the com. sections OF, AC are || XL 16. Hence, since EO is 1| to BD, a side of ^iABD, /. AE isto EB as AO is to OD VL2. And, since OF is || to AC, a side of .^DAC, /.AG isto OD as CF isto FD VL2. Hence AE isto EB as CF isto FD v.n. Wherefore, if two straight lines &c. Q.E.D. NOTE. A more perfect form of the enunciation is the following: — If two straight lines be cut by three parallel planes they shall be cut proportionally, and those segments shall be homologous which are intercepted between the same planes. PROPS. XVII., XVIII. 303 PKOPOSITION XVIII. Theorem. If a straight line he at right angles to a plane, every 'plane which passes through it shall be at right angles to that plane. Let PN be ± to the plane AB. Then shall every plane through PN be ± to plane AB. Let CD be a plane passing through PN and having a com. section ED with the plane AB. Take any pt. F in ED, and draw EG, in plane CD, at rt. z_stoED 111. Then, since PN is ± to plane AB Hyp. /. Z. PNF is a rt. Z. xi. def. 3. But Z.GFN is a rt. L Constr. ,\ Ls PNF, GFN=two rt. Ls. /. GFislltoPN 1.28. But PN is i. to plane AB Hyp. /. GF is also ± to plane AB xi. 8. Hence, since any st. line GF, in plane CD, drawn at rt. L s to their com. section ED, is J_ to plane AB, /, plane CD is J. to plane AB xi. def. 4. Wherefore, if a straight line &c. Q.E.D. EXERCISES. 1. If one plane is perpendicular to a second, the second is perpendicular to the first. 2. Prove the converse of Prop. XVIII., — that if one plane be perpendicular to another, the perpendicular to the second, drawn from any point in their common section, lies in the first plane. 304 EUCLID, BOOK XI. PKOPOSITION XIX. Theorem. If two planes which cut one another be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane. Let the planes AB and BD be each ± to plane EF. Then shall their com. section BC be j. to plane EF. - E- Let AC, CD be the com. sections of planes AB, BD with plane EF. Then, if BC be not ± to plane EF, if possible, draw, in plane AB, from pt. C, the st. line CG at rt. Z.s to AC; and, in the plane BD, from pt. C, the st. line CH at rt. Z.s to CD I. ii. Then, since plane AB is ± to plane EF Hyp. and CG is drawn in plane AB ± to their com. section Constr. /. CG is J. to plane EF XI. def. 4. Similarly, CH is ± to plane EF. i.e., from the same pt, C two straight lines have been drawn, each A. to plane EF, which is impossible XI. 13. Hence BC cannot be otherwise than ± to plane EF. Whereiore, if two planes &LC. Q.E.D. PROPOSITION XX. Theorem. If a solid angle be contained by three plane angles, any tiDO of them are together greater than the third. Let the solid z. at O be cont*^ by the three plane L s AOB, BOC, CO A. Then shall any two of these z_s be together > the third. PROPS. XIX., XX. 305 Case 1. If z.s AOB, BOC, CO A are all equal, it is evident that any two are > the third. Case 2. But, if z. s AOB, BOC, COA are not all equal, let L AOB be that which is not < either of the others. ^ 3 At pt. O, in the plane AOB, make z.AOD=z_COA 1.23. Cut off OD=OC. Through D draw st. line ADB, in the plane AOB, meeting OA, OB at A, B. Join CA CB. Then, in .-rXs AOC, AOD, ( OC=OD Constr. */ I OA is com. iz.COA=Z.DOA Constr. .-. AC=AD 1.4. But AC, CB>AB 1.20. /. CB>DB Ax.3. Hence, in ^^% BOC, BOD, j'OC=OD Coustr. •/ -| OB is com. iCB>DB Above. /. z.BOC>^BOD 1.25. But^COA=Z.AOD Constr. /. /.s BOC, COA> ^s BOD, AOD. i.e. L9. BOC, COA > z. AOB. Also, since L AOB is not < either of the others, .*. L AOB, together with either of the others, > the third. Wherefore, if a solid angle &c. Q.E.D. 306 EUCLID, BOOK XI. PKOPOSITION XXI. Theorem. The plane angles, which contain a solid angle, are together less than four right angles. Let the solid angle at O be cont<^ by the plane Z-s AOB, BOC, COD, DOE, EOA. Then shall these z_s be together < four rt. lb. Let a plane FG make the com. sections AB, BC, CD, DE, EA, with the planes in which the L s are. Take any pt. Q within the polygon ABODE. Join QA, QB, QC, QD, QE, dividing the polygon into as many ^^^s as it has sides, that is, as there are plane L s forming the solid L at O. Then, since the solid z. at B is cont* by three plane L s, /. Z-sOBA, OBC >z_ABC xi. 20. i.e. L s OBA, OBC > Z_ s QBA, QBC, and so on, for the other corners C, D &c. Hence, all the base Ls oi all the ^iirAs with vertex O > all the base Z.s of all the ^^s with vertex Q. But all the Z_s of all the ^^s with vertex O = all the Z.S of all the ^^s with vertex Q i. 32. .*, all the vertical Z-S of all the ..^^s with vertex O < all the vert^ Z. s of all the ..^^s with vertex Q. But the Z-S at Q=four rt. z.s I. 15 cor. /, the z_ s at O < four rt. L s. Wherefore, the plane angles &c. Q.E.D. NOTE. This proposition was only proved for a trihedral angle by Euclid. Simson added to Euclid's proof a second case dealing with a polyhedral angle. As, however, the above proof is quite independent of the number of angles which form the solid angle, the first case has not been inserted in the text PROP. XXI. 307 NOTES. The following was Euclid's proof ; Let the solid angle at be cont^ by the three plane Z.s AOB, BOC, COA. Then shall these three Ls be together < four rt. L s. Take any pts. A, B, C in OA, OB, OC. n Join AB, BC, CA. Then since solid Z. A is cont** by three plane Ls, .*. Ls BAO, OAC >Z.BAC XL 20. Similarly Ls ABO, OBC > Z. ABC, and Ls BCO, OCA > Z.BCA. ♦ /. the six LS BAO, OAC, ABO, OBC, BCO, OCA> Z-s BAC, ABC, BCA. But LS BAC, ABC, BCA=two rt. Ls I. 32. /. the six Ls BAO, OAC, ABO, OBC, BCO, OCA >two rt. Z.s. But the three Z-S of each of the ^i^s OAB, OBC, OCA=two rt.Z.s...I. 32. .'. the nine Ls BAO, ABO, AOB, OBC, BCO, BOC, OCA, OAC, COA =six rt. L s. But, of these, the six BAO, OAC, ABO, OBC, BCO, OCA >two rt. Ls Above. /. the reme three Ls AOB, BOC, COA < four rt. Ls. It should be noticed that Prop. XXI. only holds good for a convex solid angle, i.e. the polygon ABODE in the figure must have no re-entrant angle. [It is not customary to read the remaining nineteen propositions of the Eleventh Book, or those of Book XII. These contain some very important results concerning the volumes of solids, but the proofs are often long and difficult, and the modern methods by which these results are obtained are beyond the scope of the present work.] EXERCISES. 1. If in the fig. of Prop. XX., OQ be drawn within the solid angle at O and not in the same plane with any two of the lines OA, OB, OC, then (i) the sum of the Lb AOQ, BOQ, COQ > half the sum of the Z-8 AOB, BOC, COA. (ii) the sum of the Ls AOQ, BOQ, COQ < sum of the Ls AOB, BOC, COA. 2. If, in the figure of Prop. XXI., OQ be joined, prove that the sum of the angles AOQ, BOQ &c. is greater than half the sum of the angles AOB, BOO &c. 308 lUCLID, BOOK XI. MISCELLANEOUS EXAMPLES. I, To draiv a straight line perpendicular to each of two given straight lines which are not in the same plane. Let AB and CD be the two given st. lines. In CD take any pt. E. Through E draw EF || to AB. Let GH be the plane containing CD and EF. Take any pt. K in AB. From K draw KN ± to plane GH XI. ii. Through N draw NQ || to EF, meeting CD at Q I. 3i. Then, since QN is || to EF, and EF is |1 to AB Constr. /. QN is II to AB XI. 9. and, AB, QN, KN are in the same plane xi. r. Through Q draw, in this plane, QP || to KN, meeting AB at P I. 3i. Then shall PQ be ± to both AB and CD. For, since KN is ± to plane GH, and PQ is || to KN Constr. /. PQ is ± to plane GH XI. 8. /, PQ is ± to CD in that plane XL def. i. Also, since PQNK is a £7 Constr. /. ^KPQ=Z.KNQ 1.34. But Z.KNQ is art. L XL def. i. /. Z_KPQ is also a rt. L i.e. PQ is ± to AB. Q.E.F. II. Find a point in a given straight line such that the sum of its distances from two given points, not in the plane of the line, may he a minimum. Let P, Q be the pts., AB the line, yiQ DrawPM, QN ±stoAB XL n. In plane PMN draw NR at rt. Z_ s to AB, and = QN. Join PR cutting AB at O, and join OQ. Then PO + OQ shall be a min. (The proof is similar that of Ex. III. page 82.) MISCELLANEOUS EXAMPLES. 309 III. The faces of a parallelepiped are parallelograms^ and opposite faces are equal. Let ABCD, EFGH be opp. faces of a || p^^. E ihi \ \ . __ - - F B^^i;:: \ / ^ g Then, (i) since plane AH is || to BG .*. Def. p. 286. and plane AC meets them, ,', the com. sections AD, BC are || XL 16. Similarly, AB, DC are || . Hence, ABCD is a O Def. p. 14. In the same way it may be shown that the other faces are /~7s. Again, (ii) Join AC, EGr. Then, since ABFE is a O Above. /. AB=EF L34. Similarly, DC=HG. Also, since AD is || to EH, and DC is || to HG- Above. /. Z_ADC=^EHG XL 10. Hence, in .^^a ADC, EHG, AD=EH j DC=HG [ Above. Z_ADC=Z.EHG) /. .-::iADC=.^EHG l 4. Hence, ZI7ABCD=OErGH L 34. In the same way it may shown that the other pairs of opp. faces are equal. Q.E.D. EXERCISES. 1. In the figure of Ex. 1 prove that — (i) PQ is the shortest distance between the lines; (ii) no other line but PQ can be perpendicular to both lines . 2. In the figure of Ex. Ill,, prove that — (i) the sum of the squares on the diagonals AG, EC is double the sum of the squares on AC, CG; (ii) if BD, AC, cut at Q, and EG, EH at R, then QR is equal and parallel to CG. 3. Points are taken one on each of two adjacent walls of a room; find the shortest line that can be drawn on the walls between the points. 310 EUCLID, BOOK XI. IV. The diagonals of a parallelepiped meet at a point, and bisect one another. Let ACGE be a || ped. Join AC, EG. Then, since AE is=and || to DH ) and DH is=and II to CG ( ,'. AE is=and |I to CG Ax. landXl. 9. /.ACGE is a O 1.33. Hence its diags. AG, CE bisect one another at O ...Ex. r, p. hi. Similarly, DCFE is a CJ, and its diags. DF, CE bisect one another. i.e. the diag. DF also passes through O. Similarly, ABGH is a 0> and its diags. AG, BH bisect one another at O. i.e. the four diags. of the || P^d AG, BH, CE, DF, are concurrent and bisect each other. Q.E.D. V. Straight lines joining the middle points of opposite edges of a parallele- piped are bisected by the point of intersection of its diagonals. For if K be mid. pt. of AE, in fig. of Ex. IV., and KO be joined and prodd it will meet CG, which lies in the same plane AG, at some pt. L. Then, in .^^s AKO, GLO, AO=OG Ex. IV. Z_KAO=Z.LGO Ex. IV. and I. 29. Z.AOK=Z-GOL 1.15. .•.AK=LG, andKO=OL 1.26. But AK=half AE =half CG Ex. IV. /. LG=half CG. i.e. KOL joins the mid. pts. of opp. sides, and is bisected at O. Similarly for any other pair of sides. Q.E.D. MISCELLANEOUS EXAMPLES. 311 VI. TTie section of a parallelepided hy any plane which cuts two pairs of opposite edges, but does not cut the remaining pair, is a parallelogram. Let PQRS be such a section of the |1 ped ACGE. A E Then since plane PR cuts the || planes AC, EG-, ,*, the com. sections PQ and SR are || .... Similarly, PS is || to QR. /. fig. PQRS is a O. .XI. 16. Q.E.D. VII. The square on a diagonal of a rectangular parallelepiped is equal to the sum of the squares on three adjacent edges. Let ACGE be a reef || ped Join BG, AG. Then, since each of the L s ABC, ABE is a rt. L Def. p. 286. .*, AB is ± to plane BG XL 4. /. Z.ABG is art. Z_. H D B _^(. Hence, AG2=AB2 + BG2 1. 47. =AB2 + BC2 + CG« L47. =AB2 + BC2 + Br2 L34. Q.E.D. EXERCISES. 1. The diagonals of a rectangular parallelepiped are equal to one another. 2. Cut a cube by a plane so that the section may be (i) an equilateral triangle; (ii) a trapezium; (iii) a regular hexagon. 3. The section of a parallelepiped by a plane parallel to an edge is a paral- lelogram. 4. The sum of the squares on the four diagonals of any parallelepiped is equal to four times the sum of the squares on three adjacent edges. 312 EUCLID, BOOK XI. VIII. The straight lines joining the vertices of a tetrahedron to the points of intersection of the medians of the opposite faces meet at a point, and divide each other in the ratio of d to 1. Let VABC be any tef^ . Bisect BC at D. Join AD, VD. From DA cut off DE=4 of AD. From DV cut off Dr=^ of VD. Join VE, AF, EF vi. 9. Then, since VE, AF lie in one plane VAD XI.2. ,*. these lines cut at some pt. G. AE 2 VF And, since ;=— = - =:i:r?: Constr. ED 1 FD /. FE is II to VA vi.2(ii.) TT VGr AGr _^, „ /,» "^"^^'GE = GF • ^""' But Z!^=X!^ (bysimf As VAG.GFE) ^ hVI. 4 . (by simr A s VAD, FED) * ...Constr. GE EF _VD ~FD' _3 I'" i.e. VE, AF cut at G, and divide one another in the ratio of 3 to 1. Similarly, it may be shown that VE, BK, and VE, CL (where K, L, are the points of intersection of the medians of .=^8 VAC, VAB)^ cut at the same point G. Q.E.D. EXERCISES. In any tetrahedron the straight lines joining the middle points of oppo- site edges meet at a point. If abc be a section of the tetrahedron by a plane parallel to ABC, then VE passes through the intersection of the medians of abc. MISCELLANEOUS EXAMPLES. 313 IX. The section of a tetrahedron by a plane parallel to each of tico opposite edges is a parallelogram. Let VABC be any tet", and let DEFG be the section made by a plane DF II to the opp. edges VA and BC. Then shall DBFG be a O- For, since plane DF isjlto BC Hyp. ,*, they cannot meet if prod<^ ....Def. p. /, No St. line in plane DF can meet BC. /, EF cannot meet BC. But EF and BC are in the same plane ABC. /. EFis||toBC Ldef.35 Similarly, DG is || to BC. /. DG is II to EF XI. 9. In the same way it may be shown that DE is || to GF. /. DEFG is a O Def. p. u. Q.E.D. EXERCISES. 1. In a regular tetrahedron (i) The foot of perpendicular from the vertex on the base is the centre of the circle circumscribing the base. (ii) The square of this perpendicular is § of the square on an edge. (iii) The sum of the perpendiculars on the faces from any point in the base is equal to the altitude. (iv) The lines joining the middle points of opposite sides meet at a point, are at right angles to one another, and bisect each other. (v) Find the height when the edges are each 3 inches long, (vi) Find the height, and area of surface, when the edge is a. 2. Cut a regialar tetrahedron so that the section may be a square. 3. Find the locus of a point equidistant from the angles of an equilateral triangle. (310) U 314 EUCLID, BOOK XI. X. Sections of a pyramid made by planes parallel to the base are similar. Let VABCDE be a pyr., and abode a, section || to the base. Tlien shall abode be simr to ABODE. For since plane aci is I| to plane AD, and plane VBC meets them, .*. the com. sections be, BC are || xi. 16. Similarly, c(Z is || to CD. Hence Lbcd=LBCD XI. lo. Also,-^=l:'. ' BC VC cd ~CD* VI. 4. i.e. the Z. s at c, C are =, and the sides about these Z.s are : -J^. Similarly, for the other Z. s of the polygons. jg q /, the polygons abode, ABCDE are simi^ XI. def. l. Q.E.D XI. The areas of the sections of a pyramid, made by planes parallel to the base, are to one another in the duplicate ratio of their altitudes. Join BN, bn. Draw VN J_ to plane AD, meeting plane ad at n...Xl. ii. Then, since VN is ± to plane AD, ,*. it is J. to the || plane ad XI. 14 (conv.). And, since the plain VBN meets the || planes ad, AD, ,*, the com. sections bn, BN are || XI. 16. Hence, ^^s Ybn, VBN are sim'". And, since fig. abode is sim^ to fig. ABCDE Ex. IX. , Gs. abode bc^ Yb^ Yn^ „ „, • • fig.ABCDE =BC^=VB^=W-^^- ''' "°^^' ^'^^ ""'■ '• Q.E.D. XII. Every plane section of a sphere is a circle. Let the plane EF cut the sphere of rad. r, and whose cent, is O. Then shall their com. section BCD be a ©. From O draw ON J. to plane EF. Take any pt. C in BCD and join OC, CN. MISCELLANEOUS EXERCISES. 315 Then, since ON is jl to plane EF, .*. Z_ONC is a rt.Z. XL dof. 3. /. CN2=OC2 - 0N» L4r. =r2-0N=» i.e. any pt. C in the curve BCD is at a constant distance from the fixed pt. N. E' ,*, locus of C is a ©. Q.E.D. XIII. To find the centre of a sphere which shall pass through four given points not in the same plane. Let A, B, C, D, be the four pts. Find M, N, the cents, of the 0s circumscribing ^^s ABC, ABD...IV. 6. Draw MP at rt. Z. s to plane ABC, and NQ at rt. Z-S to plane ABD ...xi. 12. Then, since MP is a ± to plane ABC drawn from cent, of circum- scribing Constr. ,*, any point in MP is equidist. from A and B Ex. p. 29: Similarly, any pt. in NQ is equidist. from A, B. /. MP, NQ, both lie in the plane which bisects AB at rt. Z.s...Ex. p. 296. But MP, NQ, being ± to different planes, are not || . ,*, they will meet at some pt. 0. Then, since O is equidist. from A, B, C, and also from A, B, D, .*, O is equidist. from the four pts. A, B, C, D, and is the req'^ cent. Q.E.F. NOTE. If the plane EF passes through the centre of the sphere, the section is called a great circle of the sphere ; if not, a small circle. EXERCISES. 1. The common section of two spheres is a circle, and its plane is perpen- dicular to the line joining their centres, which it divides into parts the diff. of whose squares is equal to the difference of the squares of the radii. {Note. This is called the radical plane of the spheres.) 2. All tangents to a sphere from a given external point are equal. 3. The planes of small circles of equal radii are equidistant from the centre of the sphere. 4. Find the radii of the inscribed and circumscribed spheres of a regular tetrahedron whose edge is a. 316 EUCLID, BOOK XI. XIV. There can he hut five regular solids. For, since the sum of the plane L s forming any solid L must be < 360° XL 2L and an L of an equilat. ^^ is 60°, ,*, three, four, or five, but not more, equilat. ^^s can form a solid /.. Also, an Z. of a sq. is 90°, .*, three, but not more, sqs. can form a solid L • And an Z. of a reg. pentagon is 108° ,', three, but not more, reg. pentagons can form a solid /_ . Again, an /_ of a reg. hexagon is 120°, and three such L s are not < 360°, ,*. three, or more, reg. hexagons cannot form a solid L . Neither, for like reasons, can reg. heptagons, octagons, &c. Hence there can be hwt five reg. solids, namely — (i) the reg. tetrahedron, having each solid L formed by 3 equilat. ^^s; (ii) „ hexahedron (or cube), „ „ ,, 3 squares; (iii) „ octahedron, „ ,, „ 4 equilat. ^^s; (iv) „ dodecahedron, „ „ ,, 3 reg. pentagons; (v) „ icosahedron „ „ „ 5 equilat. ^^s. XV. If 'E, he the number of edges, F the numher of faces, and S the number of solid angles of any polyhedron, then S + F = E + 2. Suppose the polyhedron to be gradually built up by fitting together n polygons, taken one at a time; and let E, Ej, E3, &c., stand for the number of edges when the first, second, third, &c. polygon is added on; and so on. MISCELLANEOUS EXERCISES. 317 Then, for the first polygon, there are as many edges as corners. /. Ei = Si, or Ei=:Si + (l-l). And, when the second polygon is added, one edge, and Uvo corners will coincide with those of the first polygon, /. E2 = S2 + 1, or E2=S2 + {2-l). Also, when the third polygon is added, two edges, and three corners will coincide with those of the first and second polygons, /.Eg = 83 + 1 + 1, or E3 = S3 + (3-l). And so on. Hence, when the last but one of the polygons is added, En-i = S„-i+(^^-l). But the last polygon adds no fresh edges, nor corners, i.e. E is E„_i, S is Sn-i, and F is n. .-. E = S + (F^-1), or E + 2 = S + r. MISCELLANEOUS EXERCISES. 1. The area of the surface of a cube is double of the square on one of its diagonals. 2. Prove that if every straight line which can be drawn meeting two given straight lines meet a third given straight line, the three given lines all lie in one plane. 3. If two straight lines are parallel, their projections on any plane are also parallel. 4. From a point P, outside a given plane, two perpendiculars are drawn, one to the plane, and the other to a given straight line in the plane; prove that the line joining their feet is at right angles to the given line. 5. Two perpendiculars are let fall from any point on two given planes: show that the angle between the perpendiculai-s is equal to the angle of inclination of the planes to one another. 6. Three lines, OP, OQ, OR, intersect at right angles in the point O, and OS is drawn perpendicular to QR. Show that, if PS is joined, PS will be perpendicular to QR. 7. Find the locus of the foot of the perpendicular from a given point upon a plane which passes through a given straight line. 318 EUCLID, BOOK XI. 8. AB, CD are two straight lines, of which AB lies in a plane to which CD is perpendicular. Show that the perpendiculars drawn to AB from the different points of CD all pass through a fixed point. 9. The projections of parallel straight lines on any plane are proportional to the lines. 10. ABCD is a regular tetrahedron, and, from the vertex A, a perpendicular is drawn to the base BCD, meeting it in O : show that three times the square on AO is equal to twice the square on AB. 11. The angle which a line makes with its projection on a plane, is less than that which it makes with any other line in that plane. 12. Draw a straight line, which shall be equally inclined to three straight lines, which meet at a point, but are not in the same plane. 13. If a prism be cut by parallel planes the sections will be equal. 14. Show that the shortest distance between two opposite edges of a regular tetrahedron is equal to half the diagonal of the square described on an edge. 15. Straight lines are drawn from two points, not in that plane, to meet each other in a given plane. Find when their sum is a minimum. 16. OA, OB, OC are three straight lines, not in the same plane, planes through OB, OC perpendicular to OBC intersect in Oa, planes through OC, OA perpendicular to OCA intersect in Oh, and planes through OA, OB perpendicular to OAB intersect in Oc; prove that OA, OB, OC are perpendicular to the planes 06c, Oca, Oab respec- tively. 17. If two straight lines in one plane, be equally inclined to another plane, they will be equally inclined to the common section of the two planes. 18. If in the three edges, which meet at one angle of a cube, three points A, B, C be taken at equal distances from the angle, the area of the triangle ABC formed by joining these points with each other is -—- a^, where a is the distance OA. 19. OA and OB are two intersecting straight lines. 00 is any straight line through their point of intersection, but not in their plane, which is such that the angle COA is equal to the angle COB. Show that the projection of the straight line OC on the plane AOB bisects the angle AOB. 20. A pyramid is constructed on a square base, having all its edges equal to one another; find the inclination of two of the triangular faces to one another. MISCELLANEOUS EXERCISES. 319 21. Of all the planes which can be drawn through a given line, find that which has least inclination to a given plane. 22. On a given equilateral triangle as base, describe a regular tetrahedron. 23. If any point be taken within a cube, the square of its distance from a corner of the cube is equal to the sum of the squares of the perpen- diculars from the point on the three faces containing that corner. 24. The lines joining the middle points of adjacent sides of a quadrilateral, the sides of which are not all in the same plane, form a parallelogram. {Note. — Such a quadrilateral is called gauche.) 25. Find the locus of a point in space equidistant from three fixed points. 26. From the centre of the circle circumscribing a triangle ABC, a perpen- dicular to its plane is drawn of length equal to the side of the square inscribed in that circle ; show that the radius of the sphere which passes through A, B, C and the extremity of the perpendicular is three-fourths of the perpendicular. 27. If P be a point equidistant from the angles A, B, C of a right-angled triangle of which A is the right angle and D the middle point of BC, prove that PD is at right angles to the plane of ABC. 28. A pyramid on a square base is cut by planes parallel to the base; show that the points of intersection of the diagonals of the sections all lie in a straight line. 29. The sum of the squares on the four diagonals of an oblique parallele- piped is equal to the sum of the squares on its twelve edges. 30. Given three straight lines, draw through a given point a straight line equally inclined to the three. 31. If a regular tetrahedron be cut by a plane parallel to two edges which do not meet, the perimeter of the parallelogram in which it is cut shall be double of an edge of the tetrahedron. 32. The angles of inclination ©f two faces of a regular tetrahedron and regular octahedron are supplementary. 33. If the middle points of the edges of a regular tetrahedron be joined, the figure formed is a regular octahedron. 34. Make a trihedral angle equal to a given trihedral angle. 35. Find the locus of points in space the difference of the squares of the distances of which from two given points is constant. 320 APPENDIX. APPENDIX I. TRANSVERSALS; HARMONIC SECTION; POLE AND POLAR. Note. — In modern geometry many important results are obtained by combining Euclid's methods with those of arithmetic and algebra. Def. — A line which cuts a system of lines is called a transversal. I. If three concurrent straight lines, drawn through the angles of a triangle, meet the sides, or sides produced, the products of alternate segments of the sides, taken in order, are equal. Let ABC be a ^rA, and AOD, BOE, COF be concurrent lines meeting the sides or sides prod, AC in E, and BC produced in F; it also intersects AC, AB, CB, all produced in G, H, and K, respectively; show that AE + BK -f CG= AD + BH H- CF. Find the corresponding result, when all the vertices of an equilateral triangle are inside a circle, which intersects the sides of the triangle produced both ways. FOURTH STAGE. 1. Inscribe a regular pentagon in a given circle, and in the pentagon in- scribe a regular figure of ten sides, having its alternate sides coincident with the sides of the pentagon. 2. In the triangles ABC, DEF, the angle ABC is equal to the angle DEF and the sides about these angles are proportional: show that the angle BAC is equal to one of the two angles EDF, DFE, and give the ratio of the third sides of the two triangles. MPQ, NPR, are two intersecting circles such that the sum of the squares on their radii is equal to the square on the distance between their centres. Show how to draw a straight line MPN such that the rectangle MP, PN shall be equal to a given square. 338 APPENDIX. 3. When are quantities said to be in continued proportion? Divide a given finite straight line into two parts such that the whole line and the two parts shall be in continued proportion. The perpendicular let fall from the right angle to the hypotenuse of a right-angled triangle divides the hypotenuse in mean and extreme ratio: show that the three sides of the triangle are in continued propor- tion. 4. When is a line said to be perpendicular to a plane, and when parallel to the plane? From a given point, how many lines can be drawn perpendicular to a given plane, and how many parallel to that plane? The angle B of the triangle ABC is a right angle : P is a point, not in the plane of the triangle, and equidistant from A, B, and C : if a straight line PD bisect AC, prove that PD is perpendicular to the plane ABC. Mat, 1888. FIRST STAGE. 1. Two angles of a triangle are equal; show that the sides opposite to these angles are equal. In the equal sides AB, AC of an isosceles triangle ABC points D and E are taken, so that AD is equal to AE, and CD and BE are drawn intersecting in F ; show that the triangle BFC is isosceles. 2. Show how to divide a given angle into two equal parts. If a diagonal AC of a quadrilateral ABCD bisects the angles at A and C, show that it is at right angles to the other diagonal BD. 3. If one angle of a triangle is greater than another, show that the side "which is opposite the greater angle is longer than the side which is opposite the less angle. ABC is a triangle having an acute angle at B, which is greater than the angle at A; the side AB is produced to D, and BE is drawn to meet AC produced in E in such a way that the angle DBE is equal to the angle ABC ; show that BE is longer than BC. 4. Show that the complements of the parallelograms which are about the diagonals of any parallelogram are equal to one another. If one of the parallelograms about the diagonal of any parallelogram is equal to half one of the complements, show that the complement is equal to half the other parallelogram. 6. ABC is a given triangle and P a given point; show how to draw through P a straight line to cut AB and AC (or those sides either or both produced) in Q and R, so that AQR may be an isosceles triangle. Within what space must P be situated if Q and R are on the sides (not the sides produced) respectively? 6. In the triangle ABC the angles at A and C are each one-fourth of a right angle; CD is drawn cutting AB produced at right angles in D; show that the square on AB is double the square on BD. SECOND STAGE. 1. Show that if a straight line be divided into any two parts, the squares on the whole line and on one of the parts are equal to twice the rectangle EXAMINATION PAPERS. 339 contained by the whole line and that part together with the square on the other part. Let ABC be a triangle having a right angle at C; from D, any point in AC, draw DE at right angles to AB ; without using any property of the circle, show that the rectangle CA.AD is equal to the rectangle BA.AE. 2. AB is a chord of a circle ; C the middle point of AB ; show that the straight line drawn through C perpendicular to AB passes through the centre of the circle. Show that all circles whose centres lie on a given straight line and whose circumferences pass through a given point have a common chord of intersection. 3. Define similar segments of a circle, and show that on the same straight line and on the same side of it there cannot be two similar segments of circles not coinciding with each other. A straight line is drawn through the point of contact of two circles touching each other internally; show that the segments cut off on the same side of the line are similar. 4. Show that in equal circles the arcs which subtend equal angles, whether at the centre or circumference, are equal. ABCD is a quadrilateral inscribed in a circle whose centre is ; if the angles BAD and BOD are together equal to two right angles, show that the arc BAD is double the arc BCD. 6. Show that a tangent to a circle makes with a chord through the point of contact angles equal to those in the alternate segments of the circle. In the hypotenuse AB of a right-angled triangle ABC a point E is taken, so that AE equals AC; CD is drawn to meet AB at right angles in D; show that the line joining C and E bisects the angle BCD. 6. Given a circle and two parallel chords, show how to draw a circle to touch both chords, and the circle internally. THIRD STAGE. 1. A line of given length moves between two lines at right angles to each other, having one of its ends on each of them ; find when its centre is at a minimum distance from a given line. 2. A, B, C are three points on the circumference of a circle; D is the middle point of the arc AB, and E is the middle point of the arc BC; draw the chord CE ; join AE and CD intersecting in E; show that CE is equal toEF. HONOURS. 1. A line AB is divided in C so that CB is double AC, and circles are described on AC and CB as diameters; show how to draw through A a line such that the chords intercepted by the two circles may be equal. 2. Show how to describe the greatest equilateral triangle, each side of which passes through a vertex of a given triangle. 3. Through C, the middle point of the arc ACB of a given circle, any chord CD is drawn cutting the straight line AB in E ; find the locus of the centre of the circle passing through B, D and E. 340 APPENDIX. FOURTH STAGE. 1. Give the necessary and sufficient conditions that two triangles may be similar: also that two polygons may be similar. Show that the areas of similar triangles are as the squares on cor- responding sides. A triangle and a parallelogram have a common angle, and the area of the triangle is three times that of the parallelogram; give the rela- tion connecting those sides of the two figures which contain the com- mon angle. 2. Show how to trisect the arc of a circle whose chord is the side of a regular pentagon inscribed in the circle. If regular figures of five, six, and ten sides respectively be inscribed in the same circle, show that the square on a side of the hexagon and the square on a side of the decagon are together equal to the square on a side of the pentagon. 3. If four parallel lines are given in position, show that they cut off on any transverse line segments which are in a constant ratio. Similar and similarly situated polygons are described on AB and CD, which are given unequal parallel lines. Show that all the lines joining corresponding corners meet at a point. 4. If two lines which intersect are respectively parallel to two other in- tersecting lines, show that the plane of the former pair is parallel to the plane of the latter pair, or coincident with it. A pyramid on a four-sided base is cut by any number of parallel planes ; show that the sections are similar figures, and that the points of intersection of the diagonals of all the sections are in a straight line. LONDON UNIVERSITY MATRICULATION EXAMINATION. January, 1889. Euclid, Books I. -IV. 1. Prove that, if two triangles have the sides of the one respectively equal to the sides of the other, they are eqiial in all respects. 2. Two triangles on the same base are such that one lies wholly inside the other; prove that the inner one has the smaller perimeter. Extend this proposition to two polygons on the same base, of which the inner one has no re-entrant angle. 3. Prove that the sum of the angles of any triangle is equal to two right angles. 4. Show that the middle points of the sides of any quadrilateral are the vertices of a parallelogram. Prove that the area of this parallelogram is half the area of the quadrilateral. EXAMINATION PAPERS. 341 6. Prove by a geometrical construction that, if a straight line is divided into two segments, the square described on the whole line is equal to the squares described on the segments together with twice the rectangle con- tained by the segments. 6. Show that, if two circles touch each other, the line joining their centres passes through the point of contact. 7. A segment of a circle is described on a straight line AB, at any point P on it the tangent PT is drawn meeting AB produced in T ; prove that the angle which PT makes with AB is equal to the difference of the angles PAB and PBA. 8. Chords of a circle are drawn through a given point inside it; prove that the rectangles contained by their segments are all equal. Investigate also for what other points these rectangles have the same area as for the given point. 9. A number of triangles with equal vertical angles are inscribed in the same circle; show that their bases are all tangents to a circle. 10. Show how to inscribe a regular polygon of fifteen sides in a given circle. June, 1889. 1. In the triangles ABC, DEF, it is given that AB = DE, the angle ABC = the angle DEF, and the angle BCA = the angle EFD. Prove that the triangles are equal in all respects. 2. Prove that the complements of the parallelograms about the diagonal of a parallelogram are equal. 3. Show how to divide a given straight line into five equal parts. 4. In an obtuse -angled triangle, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle exceeds the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by the side on which, when produced, the perpendiciilar falls, and the straight line inter- cepted without the triangle, between the perpendicular and the obtuse angle. 5. Let A and B be two fixed points, and CD a straight line in the same plane as A, B. Find the position of the point P on the straight line CD, which is such that the sum of the squares on PA, PB is least. 6. Prove that two similar segments of circles which do not coincide can- not be constructed on the same chord, and on the same side of that chord. 7. Let A and B be two points on a circle, ADB, whose centre is C. Let an arc of a circle be described through ACB: let any straight line APQ be drawn cutting the arc ACB at P, and ADB at Q. Then prove that PB = PQ. 8. If AB be the diameter of a circle, CD a fixed straight line perpen- dicular to AB, then, if AQP be any straight line through A, cutting the circle at Q and CD at P, prove that the rectangle contained by AQ and AP is constant. 9. Show how to inscribe, in a given circle, a triangle equiangular to a given one. 10. A polygon having all its sides equal is inscribed in a circle, prove that all its angles are equal. 342 APPENDIX. EDUCATION DEPARTMENT. EXAMINATION FOR ADMISSION INTO TRAINING COLLEGES. Books I. II. — Midsummer, 1889. [All generally understood abbreviations for words may be used, hut no symbols of operations {such as — , + , x ) are admissible. Capital letters, not numbers, must be used in the diagrams.'\ 1. 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, these sides being opposite to equal angles in each, then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other. A, B, C are three given points. Through A draw a line such that the perpendiculars upon it from B, C may be equal. 2. Define parallel straight lines. To draw a straight line through a given point parallel to a given straight line. A is a point without the angle CBD. Draw from A a straight line, AEF, cutting BC in E, BD in F, so that AE may be equal to EF. 3. If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by these two sides is a right angle. PQRS is a rectangle, T any point, show by Bk. I. Prop. 47, that the squares on TP, TE. are together equal to the squares on TQ, TS. 4. What is the objection to the use of algebraical processes in demon- strating the propositions of the Second Book? In every triangle, the square on the side subtending an acute angle, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line inter- cepted between the perpendicular let fall on it from the opposite angle, and the acute angle. EXAMINATION OF STUDENTS IN TRAINING COLLEGES. 1st Year. Books I., II., III. — Christmas, 1889. [Capital letters {excluding A,B,C,D,E,F), not numbers, must be used in the diagrams. All generally understood abbreviations and symbols for words may be used, but not symbols of operations, such as - , +, x , PQ^, PQ, RS.^ 1. Define plane superficies, plane angle, semicircle, gnomon, angle of a seg- ment, sector of a circle. If a triangle PQE. be turned over about its side PQ, show by Prop, 4, Bk. I., that the line joining the two positions of R is perpendicular toPQ. EXAMINATION PAPERS. 343 2. To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. Erora a given point draw a straight line, making equal angles with two given straight lines. (I. 26 may be employed.) 3. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. PQR is an equilateral triangle ; H, K are points in QR, PR, such that QH=RK; QK, PH intersect at S. Show that the angle PSQ is equal to the sum of two angles of the equilateral triangle. 4. Equal triangles upon the same base and upon the same side of it are between the same parallels. Bisect a given triangle by a straight line drawn from a given point in one of its sides. (I. 37, 38.) 6. Show that Propositions 2 and 3 of the Second Bbok are special cases of Proposition 1. What are the corresponding algebraical formulae for these three propositions? 6. If a straight line be divided into two equal and also into two unequal parts; the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. In PR, a diagonal of the square PQRS, a point T is taken. Show that the triangle whose sides are equal to PT, TR and the diagonal of & square described on QT will be right-angled. 7. If a straight line drawn through the centre of a circle, bisect a straight line in it which does not pass through the centre, it shall cut it at right angles. If a straight line be drawn intersecting two concentric circles, prove that the portions of the straight line, intercepted between the two circles, are equal. 8. If two circles touch one another internally, the straight line which joins their centres, being produced, shall pass through the point of contact. Prove also that if a straight line be drawn through their point of contact, cutting the circumferences, two radii drawn to the points of intersection are parallel. 9. The angle at the centre of a circle is double of the angle at the cir- cumference on the same base, that is, on the same arc. GH and KL are two chords in a circle which intersect when pro- duced in the point without the circle; prove that the difference of the angles subtended at the centre by the arcs GK and HL is double of the angle GOK. (Apply I. 32.) 10. If from any point without a circle there be drawn two straight lines, one of which cuts the circle and the other meets it, and 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 on the line which meets the circle, the line which meets the circle shall touch it. From a point T two tangents are drawn to a circle whose centre is U, and TU meets the chord of contact at S; show that the rectangle contained by US, UT is equal to the square on the radius. 344 APPENDIX. 2nd Year. Books I., II., III., IV. and VI. (Props. 1-17). Christmas, 1889. 1. Construct a triangle whose sides are equal to three given straight lines, any two of which are together greater than the third. Show by diagrams what will happen if you attempt to construct triangles whose sides contain units in the following relations: — (1.) 2, 3, 4; (2.) 2, 3, 5; (3.) 2, 3, 6. 2. If a straight line falling on two other straight lines make the alternate angles equal to one another, or make the interior angles on one side together equal to two right angles, these two straight lines shall be parallel. Show that a four-sided figure, which has all its sides equal and one angle a right angle is a square. 3. 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. Show how to produce a straight line, so that the rectangle contained by the given line and the line made up of the given line and the part produced may be equal to the square of the part produced. 4. In any acute-angled triangle the squares of the sides containing the acute angle are together greater than the square of the side subtending the acute angle by twice the rectangle contained by one of the sides and the straight line intercepted between the acute angle and the perpendicular let fall on it from the opposite angle. The straight line PQ is divided in R, so that the rectangle PQ, QR is equal to the square of PR. Circles with centres Q,R, and radii equal to PR intersect in S. Show that PQS is an isosceles triangle, having each of the angles PQS, PSQ double of QPS. 6. The angle at the centre of a circle is double of the angle at the cir- cumference upon the same base, that is, upon the same part of the circum- ference. Two equal circles intersect in P, Q. Any straight line is drawn through P meeting the circles again in R, S. Show that if QR and QS be joined, QR is equal to QS. 6. If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square on the line which touches it. Two circles intersect in P, Q. If tangents be drawn to the circles from any point R in PQ produced they shall be equal. 7. In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. PQR is a triangle, PS, RT, meeting in V, are drawn perpendicular to the opposite sides. QV is produced to meet PR in W. Show that Q\V is perpendicular to PR. 8. To inscribe a circle in a given triangle. M is the centre of the circle inscribed in the triangle PQR; and N is the centre of the circle which touches PQ and PR produced and QR. Show that MN is the diameter of a circle which passes through Q and R. EXAMINATION PAPERS. 345 9. To inscribe an equilateral and equiangular hexagon in a given circle. Compare the areas of regular hexagons inscribed in and described about a given circle. 10. If a straight line be drawn parallel to one of the sides of a triangle it shall cut the other sides, or these produced, proportionally; and con- versely, if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle. MNP, MNQ are equal triangles. MQ and NP intersect in R. Through R a straight line is drawn parallel to MN meeting MP in T and NQ in V. Show that TK is equal to RV. 11. Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional; and conversely, triangles which have one angle of the one equal to one angle in the other, and their sides about the equal angles reciprocally pro- portional, are equal to one another. Prove that the equilateral triangle described on the hypotenuse of a right-angled triangle is equal to the sum of the equilateral triangles described on the other two sides. (VI. 16 may be used.) SCOTCH EDUCATION DEPARTMENT. LEAVING CERTIFICATE, 1889. GEOMETRY— Second (oe Lower) Grade. [Candidates are not expected to attempt more than about three-fourths of this paper. But any omissions, ivhether of reasoning, explanation, or calcu- lation, tvill be treated as errors. All ordinary contractions may be used. Additional marks tvill be given for neatness and good style.] 1. Define a straight line, a right angle, a square, and a parallelogram. The angles which one straight line makes with another straight line on one side of it are either two right angles or are together equal to two right angles, 2. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of every triangle are equal to two right angles. ABCD is a square. On CD an equilateral triangle CDE is de- scribed, so that E lies within the square, AE is joined. Find what part of a right angle the angle EAB is. 3. Parallelograms on the same base and between the same parallels are equal. ABC is a triangle, and AB is bisected in D. From D, DE is drawn parallel to BC, meeting AC in E. Prove that AE = EC, 4. If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part produced, together with the square on half the line bisected, is equal to the square on the line made up of the half and the part produced. Show that the preceding proposition is equivalent to the following: (310) Y 346 APPENDIX. The rectangle contained by the sum and difference of two straight lines is equal to the difference of their squares. 5. Draw a straight line from an external point to touch a given circle. Show that your construction enables two tangents to be drawn, and that these tangents are equal in length. 6. The angle at the centre of a circle is double of the angle at the circum- ference subtended by the same arc. From a point P outside a circle, two straight lines PQR, PST are drawn, cutting the circle. Show that the difference between the angles which RT and QS subtend at the centre of the circle is equal to twice the angle QPS. 7. If two chords of a circle intersect, the rectangle contained by the segments of the one shall be equal to the rectangle contained by the seg- ments of the other, AB is a diameter of a circle, and C is a point in AB produced. , Through C, CD is drawn perpendicular to AB. If through any point P in CD a straight line PBQ be drawn meeting the circle in Q, the rectangle PB, BQ is constant. 8. What is meant by a locus? A and B are two fixed points, and the area of the triangle ABC is constant; find the locus of C. Higher Grade and Honours. 1. If a straight line AB is bisected in C and produced to D, prove that the rectangle contained by AD and DB together with the square on AC is equal to the square on CD. ABC are three given points on a straight line. Find a point D in the line produced such that rect. AD, AB + sq. CD — sq. AC. 2. Prove that in every circle angles at the circumference which stand on the same arc are equal. Prove that if the angles ABC and ADC are equal and B and D are on the same side of AC a circle will pass through the four points ABCD. AB is bisected in O and P is any point in OB; OR is drawn per- pendicular to AB and equal to OP; AR is joined. From R is drawn RS perpendicular to AR towards AB and equal to AR. Prove that SP is perpendicular to AB. 3. Describe a circle which shall touch one side of a triangle and the other two sides produced. What is meant by the locus of a point ? If BC is fixed and A moves on the circumference of a fixed circle passing through B and C, find the locus of the centre of the circle escribed to ABC and opposite to A. 4. When is A to B in the duplicate ratio of C to D ? Two similar parallelograms OABC and Oabc are similarly placed so that the angles AOC and aOc coincide, cb and AB, produced if necessary, meet in D. Prove Oabc : OADc : : OADc : OABC ; and thence show that similar parallelograms are to one another in the duplicate ratio of their homologous sides. 6. Prove that in equal circles angles at the centres are proportional to the arcs on which they stand. EXAMINATION PAPERS. 347 6. When is a straight line harmonically divided? Show that a straight line which is bisected may be looked on as harmonically divided. Prove that any diagonal of a quadrilateral is harmonically divided by the corners of the quadrilateral through which it passes, and the points where it meets the other two diagonals. 7. From a given point outside a plane draw a perpendicular to the plane. O is a point outside a plane. OA is perpendicular to it. BC is a straight line in the plane and AD is perpendicular to it. Prove that OD is also perpendicular to BC. 8. If transversals through the angular points A, B, C of a triangle are concurrent, and intersect the opposite sides in D, E, F respectively, then BD . CE . AF = DC . EA . FB. ABC is a triangle and any straight line CF is drawn meeting AB in F. The angles BFC, AFC are bisected by FD, FE meeting the opposite sides in D, E. Show that AD, BE, CF are concurrent. OXFORD LOCAL EXAMINATIONS. June, 1889. JUNIORS.— Books I. -VI. [Euclid's axioms loill he required, and no proof of any proposition ivill he admitted which assumes the proof of anything not proved in preceding propositions of Euclid. '\ 1. Define a right angle, a rhombus, a parallelogram. 2. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases or third sides equal; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely those to which the equal sides are opposite. If the diagonals of a quadrilateral bisect one another at right angles the quadrilateral is a rhombus or a square. 3. The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are also themselves equal and parallel. 4. Show that every rhomboid is a parallelogram, 5. In any right-angled triangle, the square which is described on the side subtending the right angle is equal to the squares described on the sides which contain the right angle. 6. If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the two parts. 7. If ABC be a triangle obtuse-angled at C, and AD be drawn perpen- dicular to BC produced, prove that the square on AC is less than the squares on AB, BC by twice the rectangle DB, BC. 8. If two circles touch one another internally, the straight line joining their centres being produced passes through the point of contact. 348 APPENDIX. 9. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles in the alternate seg- ments of the circle. AB, AC are tangents to a circle, B, C being the points of contact; BD is a chord through B parallel to AC : show that the arcs CB, CD are equal. 10. Describe a circle about a given triangle. 11. Equal parallelograms which have an angle of the one equal to an angle of the other have their sides about the equal angles reciprocally pro- portional; and, conversely, parallelograms which have one angle of the one equal to one angle of the other and their sides about the equal angles re- ciprocally proportional are equal to one another. Construct a rhombus equal to a given parallelogram and equiangular with it. 12. In equal circles, angles, whether at the centre or at the circumfer- ences, have the same ratio which the arcs on which they stand have to one another. SENIORS.— Books I.-VI. and XL (1-21). 1. 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. 2. If ABC is a triangle such that the square on BC is equal to the squares on AB and AC together, the angle BAG will be a right angle. If the square on BC (on the side remote from A) is BDEC, show that the perpendicular from D on AC (produced if necessary) is equal to AB and AC together. 3. Describe a square which shall be equal to a given rectilineal figure. Of all rectangles having a given area the square is that the sum of the lengths of whose sides is least. 4. If two circles touch one another externally, the straight line which joins their centres will pass through the point of contact. 5. Define a segment of a circle, and the angle in a segment. On the same chord and on the same side of it there cannot be two similar segments of circles, not coinciding with one another. 6. Inscribe a square in a given circle. If a parallelogram can have a circle inscribed in it, it must be equi- lateral. 7. If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base wiU have the same ratio which the other sides of the triangle have to one another. 8. If four straight lines are proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means. On a given base describe an isosceles triangle equal in area to a given rectangle. 9. If two straight lines meeting one another are parallel to two others that meet one another, and are not in the same plane with the first two, the first two and the other two will contain equal angles. EXAMINATION PAPERS. 349 OXFORD UNIVERSITY RESPONSIONS. Books I., II. 1. Define — superficies, centre of a circle, parallel straight lines, plane angle, rhomboid. 2. On the same base, and on the same side of it, there cannot be two tri- angles having their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity. 3. Make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third. 4. 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, namely, either the sides adjacent to the equal angles, or sides which are opposite to equal angles in each, then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other, 5. To a given straight line apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. 6. In any right-angled triangle, the square which is described on the side subtending the right angle is eqxial to the squares described on the sides which contain the right angle. •7. If a straight line be divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle con- tained by the whole and that part, together with the square on the other part. 8. If a straight line "be bisected, and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected and of the square on the line made up of the half and the part produced. 9. Divide a given straight line into two parts, so that the rectangle con- tained by the whole and one of the parts may be equal to the square on the other part. CAMBRIDGE LOCAL EXAMINATIONS. 1889. JUNIORS.— Books I., II., III., IV., VI. [The only abbreviation admitted for 'Hhe square on AB" is " sq. on AB," and for 'Uhe rectanyle contained by AB and CD" "rect. AB, CD.'* All generally understood abbreviations or symbols for words may be used, but not symbols of operations such as - , +, x .] A 1. Define a superficies, a circle, and parallel straight lines. Give one of Euclid's postulates. 350 APPENDIX. A 2. If two angles of a triangle be equal to one another, the sides also which are opposite to the equal angles shall be equal to one another. ABC is a triangle in which the sides AB, AC are equal to each other: equilateral triangles ADB, AEC are described on AB, AC, outside the triangle ABC: BE and CD intersect in 0; prove that CD and OE are equal. A 3. 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, namely sides which are opposite to equal angles in each; then shall the other sides be equal, each to each ; and also the third angle of the one equal to the third angle of the other. AB, AC are two given straight lines. Show how to draw through B a line BPQ cutting AC in P and such that, if produced to Q so that PQ is equal to PA, the angle QCA may be equal to the angle QBA. A 4. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles. A point is taken within a triangle ABC such that the angles AOB, AOC are equal to the exterior angles of the triangle at C and B, prove that the angle BOC is equal to the exterior angle at A. A 5. From a point in one of the lines containing a given angle draw a line which, with the lines containing the angle, shall include a given area. A 6. In any right-angled triangle, the square which is described on the side subtending the right angle is equal to the squares described on the sides containing the right angle. If squares be described on the sides of any triangle ABC as in this proposition, prove that the perpendicular from A on BC divides the square on BC into two parts which differ from the squares on AB and AC by equal areas. A 7. If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line and the several parts of the divided line. A 8. In every triangle, the square on the side subtending an acute angle is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line inter- cepted between the perpendicular let fall on it from the opposite angle and the acute angle. The perpendicular from A meets the base of the triangle ABC in D, and E is the middle point of BC, prove that the difference of the squares on AB and AC is equal to twice the rectangle contained by BC and DE. B 1. Draw a straight line from a given point, without the circumference of a given circle, which shall touch the circle. From a point without a circle draw a line such that the part of it included within the circle may be of a given length less than the diameter of the circle. B 2. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle. EXAMINATION PAPERS. 351 A straight line touches a circle at the point P, and QR is a chord of a second circle parallel to this tangent. PQ, PR cut the first circle in S, T, and the second circle in U, V ; prove that ST and UV are parallel. B 3. Inscribe a circle in a given triangle. B 4. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular to one another, and shall have those angles equal which are opposite to the homologous sides. If a point be taken within a parallelogram such that the line joining it to one of the angular points subtends equal angles at the two adjacent angular points, prove that the lines joining it to any angular point will subtend equal angles at the two angular points adjacent to that angular point. B 6. If any segment of a circle dep bribed on the side BC of a triangle ABC cut BA, CA, produced if necessary, in P and Q, prove that PQ is always parallel to a fixed straight line. SEI^IORS.— Books I., II., III., IV., VI. and XL (1-21). 1. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles. Squares ABDE and ACFG- are described on the sides AB, AC of a triangle ABC external to the triangle: prove that the lines CE and BG are at right angles. 2. If a parallelogram and a triangle be on the same base and between the same parallels, the parallelogram is double of the triangle. HF and EG are the parallelograms about the diagonal AC of a parallelogram ABCD. Show that the sum of the areas of the tri- angles AEG and AHF is equal to the sum of the areas of CHF and CEG. 3. If a straight line be divided into two equal, and also into two un- equal parts, the squares on tlie two unequal parts are together double of the square on half the line and of the square on the line between the points of section. The triangle ABC is equilateral and AD is drawn meeting the base BC at right angles in D. In CB a part CE is taken equal to AD. Prove that the square on ED is equal to the rectangle contained by BE and BC. 4. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle. A line AD is drawn bisecting the angle A of a triangle ABC and meeting the side BC in D. Find a point E in BC produced either way such that the square on ED may be equal to the rectangle contained by EB and EC. 5. Describe a circle about a given triangle. A triangle ABC is inscribed in a circle and B', C are the middle points of the sides AC and AB. The perpendiculars from B and C on the opposite sides meet at P and PB', PC meet the circle again in E and F respectively; prove that EF is equal and parallel to BC. 352 APPENDIX. 6. The sides about the equal angles of triangles which are equiangular to one another are proportionals; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios. Construct a triangle which shall have one angular point at a given point, the other angular points on two fixed straight lines respectively, and its sides proportional to those of a given triangle. 7. If a solid angle be contained by three plane angles, any two of them are together greater than the third. CAMBRIDGE HIGHER LOCAL EXAMINATIONS. June, 1889.— Books I.-IV., VI. and XL (1-21). 1. If a straight line fall on two parallel straight lines, it makes the alter- nate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side. Prove that the straight lines which bisect two opposite angles of a parallelogram either coincide or are parallel to one another. 2. In any right-angled triangle, the square which is described on the side subtending the right angle is equal to the sum of the squares described on the sides which contain the right angle. If ABC be a right-angled triangle having A for the right angle, and squares be described on the lines BC, CA, AB on the sides opposite to the angles A, B, C, then will the diagonal of the square on AB passing through B be parallel to the diagonal of the square on AC passing through C. 3. If a straight line be divided into any two parts, the squares on the whole line and on one of the parts are together equal to twice the rectangle contained by the whole and that part together with the square on the other part. A straight line AB is divided in C and D so that AC is equal to DB: prove that the squares on AB, CD are together double of the sum of the squares on AC and CB. 4. Draw a tangent to a circle from a given point without it. If three circles touch, two and two, the tangents at the points of con- tact meet at a point and are equal, or are parallel. 5. Inscribe a circle in a given triangle. 6. The sides about the equal angles of triangles which are equiangular to one another are proportionals ; and those sides which are opposite to the equal angles are homologous. Two circles ABC, ABD intersect in A and B. Through A a line CAD is drawn cutting the circles in C and D respectively, and the tangent at A to the circle ABD cuts the circle ABC in E. Prove that the chord DA is to the chord CE in the ratio of the radii of the circles. 7. If two straight lines be cut by parallel planes, they are cut in the same ratio. EXAMINATION PAPERS. 353 CAMBRIDGE UNIVERSITY PREVIOUS EXAMINATION. June, 1889.— Books L, IL, III. and VI. (Props. 1-19). 1. Define a triangle, a circle, and a square. What is the difference between a postulate and an axiom? 2. Any two sides of a triangle are together greater than the third side. 3. Construct a triangle, the sides of which shall be respectively equal to three given straight lines, any two of these lines being greater than the third. 4. Describe a parallelogram equal to a given rectilineal figure, and having An angle equal to a given rectilineal angle. 5. If a straight line be divided into any two parts, the squares of the whole line and of one of the parts are equal to twice the rectangle contained by the whole line and that part together with the square of the other part. 6. If a straight line be divided into two equal, and also into two unequal parts, the squares of the two unequal parts are together double of the squares of half the line, and of the line between the points of section. 7. If one circle touch another internally, the straight line which joins their centres, being produced, shall pass through the point of contact. 8. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle. 9. If the sides of two triangles about each of their angles be proportionals, the triangles shall be equiangular. 10. The bisector of the exterior angle A of a triangle ABC meets the side BC produced in D. Prove that the perpendiculars drawn from D to the sides AB, AC produced are equal to each other. 11. In a right-angled triangle prove that the line drawn from the right angle to the middle point of the base is equal to half the base. 12. Construct an isosceles triangle in which the vertical angle shall be equal to four times each angle at the base. CAMBRIDGE MATHEMATICAL TRIPOS. 1888. 1. Parallelograms on the same base, and between the same parallels, are equal to one another. A straight line DE is drawn to cut the base BC of a triangle ABC and is terminated by the sides AB, AC produced if necessary ; prove that, if the quadrilateral BDCE be of constant area, the middle point of DE lies on one of two fixed straight lines. 2. Prove that, if from any point without a circle two straight lines be drawn, one of which cuts the circle and the other touches it, the rectangle contained by the whole line which cuts the circle and the part of it without the circle shall be equal to the square on the line which touches it. 354 APPENDIX. Squares are described on the sides of a triangle ABC, namely BCDE, CAFG, ABHK, and Aj, Bi, Ci are the intersections of BF and CK, CH and AE, BG and AD, respectively; prove that AAi, BBi, CCi meet in a point. Prove that a similar theorem is true if the intersections of BG and CH, CK and AD, AE and BF, be taken. 3. If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base 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. Prove that, if a point on the internal bisector of the angle of a triangle be joined to the two other vertices and the joining lines be produced to intersect the external bisector of the angle, the straight lines joining the points of intersection to the two vertices meet in a second point on the internal bisector. 4. Prove that the sides about the equal angles of triangles which are equiangular to one another are proportionals ; and those sides which are opposite to the equal angles are homologous, that is, are the antecedents or the consequents of the ratios. Construct a rhombus of which two sides lie along two given parallel straight lines while the other two pass each through a fixed point. Of how many solutions does the problem admit? 6. Prove that the locus of a point which is such that its distances from two fixed points are in a constant ratio is a circle. Prove that there are two points, each of which has the property that its distances from the angular points of a triangle are proportional to the opposite sides, and that the straight line joining them passes through the centre of the circumscribed circle. 6. Prove that, if two straight lines be cut by parallel planes, they are cut in the same ratio. Prove that the straight lines which intersect three given non-inter- secting lines that are parallel to the same plane are also all parallel to a plane. 1889. 1. Parallelograms on the same base and between the same parallels are equal to one another. Show how in the three cases of the proposition to cut up one paral- lelogram so that the parts when properly fitted together will form the other parallelogram ; and show also how to do the same for any two equal triangles. 2. In every triangle the square on the side subtending an acute angle is less than the squares on the sides containing it by twice the rectangle con- tained by either of these sides and the straight line intercepted between the perpendicular let fall on it from the opposite angle and the acute angle. A point P is taken within a triangle ABC such that when perpen- diculars PM, PN are let fall on AB, AC respectively the rectangles CN.AC and BM. AB are equal. Prove that P lies on a fixed straight line. EXAMINATION PAPERS. 355 3. Prove that the feet of the perpendiculars drawn to the sides of a tri- angle from any point on the circumscribing circle are collinear. A crossed quadrilateral whose opposite sides are equal is inscribed in a circle. Prove that the feet of the perpendiculars drawn to the sides from any point on the circumference lie on a circle the locus of whose centre is a straight line. 4. Inscribe a circle in a given triangle. A triangle is formed by the centres of the three circles which pass each through two angular points of a given triangle and cut its in- scribed circle orthogonally. Show that (1) its sides are parallel to the lines joining the points of contact with the sides of the inscribed circle of the given triangle, (2) the centres of the circumscribing circles of the two triangles coincide, (3) the radii of these circles differ by one half of the radius of the inscribed circle of the given triangle. 5. If two triangles are equiangular the sides about their equal angles are proportional. A straight line moves so that it is divided in a constant ratio by the sides of a triangle. Prove that the locus of a point which divides one of the segments in a constant ratio is a straight line. 6. If a straight line stand at right angles to each of two straight lines at their point of intersection, it is at right angles to the plane which passes through them. Two tetrahedra ABCD, a^yS are such that the five edges BC, CA, AB, DA, DB of the first are at right angles to the five edges 5a, 5^, Sy, ^y, 7a of the second. Prove that the remaining edges are also at right angles and show that the perpendiculars from the angular points of the one on the corresponding faces of the other are concurrent. INDEX. Abbreviations, 329. Acute angle, 6. Acute-angled triangle, 10. Adjacent angles, 6. Algebraic definition of duplicate ratio, 219. — proof of II. II, p. 114. Aliquot part, 217. Alternando, 221. Alternative proofs of propositions, 326. Altitude of cone, 284. cylinder, 286. figure, 222. parallelogram, 64. prism, 283. pyramid, 283. triangle, 64. Ambiguous case of triangles, 52. Analysis of a problem, 207. Angle, 7. — acute, 6. — at centre of a circle, 131. — at circumference, 131. — between two lines, 286. — in a segment, 130. — obtuse, 6. — of a segment, 130. — plane, 6. — plane rectilineal, 6. — re-entrant, 60. — right, 6. — solid, 283. — unrestricted in size, 261. Antecedent, 219. Arc, 8. — concave, 130. — convex, 130. — major, 130. — minor, 130. Area of a figure, 8. triangle, to find (i), 63. to find (ii), 199. Axiom, definition of, 17. Axiom 12, Playfair's substitute for, 57. Axioms general and geometrical, 17. — unstated but assumed by Euclid, 29, 36, 44, 65, 70, 79. 90- Axis of a cone, 284. cylinder, 285. — — sphere, 284. — radical of two circles, 183. Base of a cone, 284. prism, 283. pyramid, 283. triangle, 11. Bisect, to, 12. Bisector of an angle, 6. Centre of a circle, 8. sphere, 284. similarity, 271. similitude, 270. Chord, 8. Circle, 8. — concentric, 8. — which cut at right angles, 131. Circumference, 8. Circumscribe, to, 131. Coincide, 17. CoUinear, 289. Commensurable, 97. Common section, 282. — tangents, to draw, 181. Compasses, use of, in Euclid, 17. in practical geometry, 45. Complement of an angle, 6. Complements of parallelograms, 14. Complete quadrilateral, 276. Componendo, 221. Compound ratio, 218. Concave arc, 130. Concentric circles, 8. Conclusion, 23. Concurrent, 86. Cone, 284. Consequent, 219. Contact, external and internal, 129. Continued proportion, 219. Converse of a theorem, 26. axiom XII. 40. Convex arc, 130. — solid angle, 283. Corner, 283. Corollaries to propositions on pages 24, 26, 38, 59, 75, 82, 102, 103, 132, 153, 167, 174, 200, 212, 236, 248, 251, 270, 271, 322. Corollary, definition of, 25. Corresponding points, 270. Cube, 285. Cylinder, 285. Decagon, 15. Degree, 7. Described figure, 131. Diagonal of parallelepiped, 286. — — quadrilateral, 12. third, 276. Diameter of circle, 8. parallelogram, 14. sphere, 284. Dihedral angle, 282. Dimensions, three, 281. Direct common tangent, 181. Distance of a point from a line, 42. Dividendo, 221. INDEX. 357 Dodecagon, 15. Dodecahedron, 285. Duplicate ratio, 218. equal to ratio of squares, 251. Edge of a solid, 283. Equal circles, 129. — in all respects, 22. Equality of triangles, absolute, 52. in area, 64. Equiangular triangle, 24. — triangles to one another, 233. Equidistant, 52. — chords of a circle, 130. Equilateral triangle, 10. Equimultiples, 219. Euler's theorem, 277. Ex aequali, 221. Experimental proofs of I. 47, 77, 3. Exterior angle of a triangle, 10. parallels, 14. External contact of circles, 129. — division of a line, 98. Extreme and mean ratio, 222. Extremes, 219. Face of a solid, 283. Figure, 8. Figures, equal and similar solid, 283. Focus of pencil, 322. Foot of perpendicular to a plane, 282. Frustum of cone, 284. Gauche, 319. Generating line of cone, 284. cylinder, 286. Gnomon, 97. Great circle of a sphere, 315. Harmonic conjugates, 322. — pencil, 322. — proportion, 222. — range, 322. Hexagon, 15. Hexahedron, 286. Homologous, 218. Hypotenuse, lo. Hypothesis, 19. Icosahedron, 285. Inclination, planes of same, 282. — of plane to plane, 282. straight line to plane, 282. Incommensurable, 97. Indirect proof, 27. Inscribed fi^re, 131. Internal division of a line, 98. Inverse of a point, 324. Invertendo, 220. Isosceles triangle, 10. Lemma, 327. lyine, 5. — straight, 5. Loci found, 183, 229, 266, 270, 295, 299. Lo^us,, 9. Magnitudes, 17. Mean proportional, 218. Means, 219. Medial section, 113. Median, 87. Multiple, 217. Nine-points circle, 185. Normal to a plane, 282. a tangent, 131. Oblique cone, cylinder, or prism, 284. Oblong, 12, Obtuse angle, 6. Obtuse-angled triangle, 10. Octagon, 15. Octahedron, 285. Opposite angles, 6. Oppositely situated, 271. Orthocentre, 88. Parallel straight lines, 14. — planes, 282. Parallelepiped, 286. Parallelogram, 14. Part, 217. Particular enunciation, 19. Pencil, 322. Pentagon, 15. Perimeter, 8. Perpendicular, line to line, 6. — line to plane, 281. — plane to plane, 281. Plane angle, 6. — geometry, 281. — surface, 5. Point, 5. — projection of, 98. — of section, 98. Polar, 324. Pole, 324. Polygon, 10. Polygons, regular, not inscribable in a circle by Euclid's methods, 194. inscribable, 213. Polyhedral angle, 283. Polyhedron, 286. Postulate, definition of, 17. Prism, 283. Problem, definition of, 19. Projection of point on line, 98. line on line, 98. jjoint on plane, 282. line on plane, 282. Proportion, 218. — continued, 219. — geometric, 222. — harmonic, 222. — reciprocal, 222. Proportional, mean, 218. Ptolemy's theorem, 265. Pyramid, 283. Pythagoras, theorem of, 77. Q.E.D., Q.E.F., meaning of, 18. Quadrant, 131. Quadrilateral, lo. — complete, 276. Quantity, 217. Quantuplicity, 217. 358 Radical axis, 183. — plane, 315. Radius of a circle, 8. sphere, 284. Range, harmonic, 322. Ratio, 217. — compound, 218. — duplicate, 218. — extreme and mean, 222. — of greater inequality, 219. — of less inequality, 219. — reciprocal of a, 219. — triplicate, 218. Ratios, equal, 218. Ray, 322. Rectangle, 14. — contained by two lines, 97. — under two lines, 97. Rectangular parallelepiped, 286. Rectilineal figures, similar, 222. Reductio ad absurdum, 27. Re-entrant, 60. Regular figure, 14. — solids, five only, 316. Revolution, solids of, 287. Rhomboid, 12. Rhombus, 12. Right angle, 5. — cone, cylinder, or prism, 284. — ;line, 6. Right-angled cone, 284. triangle, 10. Scalene triangle, 10. Secant, 129. Section, common, 282. — medial, 113. — of a line, points of, 98. Sections of Book I., 52, 62, 79. Sector, 131. Segment of a circle, 8. Segments of a line, 98. — similar, 131. Self-conjugate triangle, 324. Semicircle, 8. Similar cones and cylinders, 285. — figures, 222. — segments, 131. Similarity, centre of, 271. Similarly situated, 222. Similitude, centre of, 270. Simson, propositions inserted by, 221, 228 262, ^06. Simson's corollary to I. 11, 288. — line, 179. — theorem, 179. Small circle of a sphere, 315. Solid, 281. — angle, 283. Space, 281. Species, triangle of given, 324. Sphere, 284. Square, 12. — definition faulty, 75. Straight line, 5. parallel to plane, 286. Sub-multiple, 217. Subtend, 10. Superficies, 5. Superposition, 23. Supplementary, 7. Surface, 5. Symbols, i. 1- and -, use of, 117, 219. geometrical meaning of, 114. Tangencies, the, 275. Tangent, 129. — as limit of secant, 143. Term, 6. Terms of a ratio, 218, Tetrahedron, 283. — regular, 285. Theorem, definition of, 19 Third diagonal, 276. Three dimensions, 281. Transversal, 320. Transverse common tangents, 181. Trapezium, 12. Trapezoid, 14. Triangle, 10. — of given species, 324. — self-conjugate, 324. Trihedral angle, 283. Triplicate ratio, 218. Trisect, 12. Trisection of angles, 61. a straight line, 80, 81, 229, 237. Vertex of angle, 6. — — cone, 284. pencil, 322. pyramid, 283. triangle, 11. Volume of a solid, 286. BRIEF LIST OF EDUCATIONAL WORKS PUBLISHED BY BLACKIE & SON, Limited. Classics, . CAESAR'S GALLIC WAR. Bks. L, IL, v., and VI. 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