UNIVERSITY OF CALIFORNIA 
 AT LQS ANGELES
 
 THE CONTENTS 
 
 OF THE 
 
 FIFTH AND SIXTH BOOKS 
 OF EUCLID 
 
 (WITH A NOTE ON IRRATIONAL NUMBERS)
 
 CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, 
 
 C. F. CLAY, MANAGER. 
 Hoil&on; FETTER LANE, E.G. 
 100, PRINCES STREET. 
 
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 Berlin: A. ASHER AND CO. 
 
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 fiombag anU Calcutta: MACMILLAN AND CO., LTD. 
 
 [All Rights reserved}
 
 THE CONTENTS 
 
 OF THE 
 
 FIFTH AND SIXTH BOOKS 
 OF EUCLID 
 
 (WITH A NOTE ON IRRATIONAL NUMBERS) 
 
 ARRANGED AND EXPLAINED 
 
 BY 
 
 M. J. M. HILL, M.A., D.Sc., RR.S. 
 
 ASTOB PBOFESSOB OF MATHEMATICS IN THE UNIVERSITY OF LONDON 
 
 LATE EXAMINER IN THE UNIVERSITY OF LONDON 
 
 AND FOR THE CIVIL SERVICE OF INDIA 
 
 SECOND EDITION 
 
 CAMBRIDGE: 
 
 AT THE UNIVERSITY PRESS 
 
 1908
 
 First Edition, 1900. 
 Second Edition, 1908.
 
 Engineering & 
 
 Mathematical 
 
 Sciences 
 
 Library 
 
 c. 
 
 \<\D 
 
 THE object of this work is to remove the chief difficulties felt by those who 
 desire to understand the Sixth Book of Euclid. With the exception of the 
 note at the end of the book on irrational numbers, which is intended for teachers 
 and advanced students only, it contains nothing beyond the capacity of those who 
 have mastered the first four Books, and has been prepared for their use. It is the 
 result of an experience of teaching the subject extending over more than twenty 
 years. 
 
 The Sixth Book depends to a very large extent on the Fifth, but this Fifth 
 Book is so difficult that in actual teaching it is usually entirely omitted with the 
 exception of the Fifth Definition, which is retained not for the purpose of proving 
 all the properties of ratio required in the Sixth Book, but only for demonstrating 
 two important propositions, viz., the 1st and 33rd. 
 
 The other properties of ratio required in the Sixth Book are usually assumed, 
 
 or so-called algebraic demonstrations are supplied. The employment side by side 
 
 Q of these two methods of dealing with ratio confuses the learner, because, not being 
 
 2 equivalent, they do not constitute, when used in this way, a firm basis for the train 
 
 *g of reasoning which he is attempting to follow. A better method is sometimes 
 
 ; attempted. This is to insist on the mastering of the Fifth Book, expressed in 
 
 j modern form as in the Syllabus of the Association for the Improvement of 
 
 Geometrical Teaching, before commencing the Sixth Book. 
 
 But it is far too difficult for all but the best pupils, and even they do not grasp 
 ** the train of reasoning as a whole, though they readily admit the truth of the 
 propositions singly as consequences of the fundamental definitions, which are 
 
 (I) The fifth definition, which is the test for the sameness of two ratios. 
 
 (II) The seventh definition, which is the test for distinguishing the greater 
 of two unequal ratios from the smaller. 
 
 5 * The alterations which have been introduced into this second edition are so extensive that the 
 J/5 preface to the first edition would not be intelligible without a copy of the text. That preface has 
 ^ therefore not been reprinted. So much of it as is relevant to the second edition is here incorporated. 
 
 O- H. K. 
 
 445234
 
 vi PREFACE 
 
 *(III) The tenth definition, which defines " Duplicate Ratio." 
 
 *(IV) The definition marked A by Simson, which defines the process for 
 
 compounding ratios. 
 
 In order to make things clear, it is necessary to explain what it is that makes 
 
 Euclid's Fifth Book so very difficult. 
 
 There is first the difficulty arising out of Euclid's notation for magnitudes 
 and numbers. This has been entirely removed in most modern editions by using 
 an algebraic notation and need not therefore be further considered. 
 
 There is next the difficulty arising out of Euclid's use of the word "ratio," 
 and the idea represented by it. 
 
 His definition of ratio (see Note 4) furnishes no satisfactory answer to the 
 question, "What is a ratio?" and it is of such a nature that no indication is 
 afforded of the answer to the still more important question, " How is a ratio to be 
 measured ? " As Euclid makes no use of the definition in his argument, it is useless 
 to examine it further, but it is worth while to try to get at his view of ratio. He 
 asserts indirectly that a ratio is a magnitude, because in the seventh definition he 
 states the conditions which must be satisfied in order that one ratio may be greater 
 than another. Now the word "greater" can only be applied to a magnitude. 
 Hence Euclid must have considered a ratio to be a magnitude f. To this conclusion 
 it may be objected that if Euclid thought that a ratio was a magnitude he would 
 not so constantly have spoken of the sameness of two ratios, but of their equality. 
 One can only surmise that, whenever it was possible, he desired to leave open all 
 questions as to the nature of ratio, and to present all his propositions as logical 
 deductions from his fundamental definitions. Yet the question as to the nature 
 of ratio is one which forces itself on the careful reader, and is a source of the 
 greatest perplexity, culminating when he reaches the llth and 13th Propositions. 
 
 The llth Proposition may be stated thus: 
 
 If A : B is the same as C : D, 
 
 and if C : D is the same as E : F, 
 
 then A : B is the same as E : F. 
 
 Now if a ratio is a magnitude, this only expresses that if X = Y, and if F = J, 
 then X = Z. 
 
 As this result follows from Euclid's First Axiom it is difficult to see the need 
 for a proof. 
 
 * These are not required until the 6th Book is reached. 
 
 I Some writers maintain that the word "greater" as applied to ratio, is not used in the same sense 
 as when it is applied to magnitudes. This seems to make matters far more difficult.
 
 PREFACE vii 
 
 This only becomes apparent when the reader realises that Euclid's procedure 
 may be described thus: 
 
 Let A, B, G, D be four magnitudes satisfying the conditions of the Fifth 
 Definition, and let C, D, E, F be four magnitudes also satisfying the same 
 conditions, then it is to be proved that A, B, E, F also satisfy the conditions 
 of that definition (see Art. 175). 
 
 Remarks of a somewhat similar nature apply to the 13th Proposition. 
 
 The next difficulty to be considered is the indirectness of Euclid's line of 
 argument, arising from the fact that he uses the Seventh Definition where the 
 Fifth alone need be employed. His Fifth Definition states the conditions which 
 must be satisfied in order that two ratios may be the same (or if ratios are 
 magnitudes, that they may be equal). 
 
 If this definition is a good and sound one, it is evident that it ought to be possible 
 to deduce from it all the properties of equal ratios. This is in fact the case. It is 
 wholly unnecessary to employ the Seventh Definition, which refers to unequal 
 ratios, to prove any of the properties of equal ratios. Its use only renders the 
 proofs of the Propositions indirect and artificial and consequently difficult. Not 
 only does no inconvenience result from avoiding its use, but it is possible to get 
 rid of the latter part of the 8th Proposition, and of the whole of the 10th and 13th 
 Propositions, which deal with unequal ratios, and of the 14th, 20th and 21st 
 Propositions of the Fifth Book, which are particular cases of the 16th, 22nd and 
 23rd Propositions respectively. 
 
 The remaining Propositions are demonstrated by means of the Fifth Definition 
 alone ; and all, with three exceptions, fall under one or other of two well recognized 
 types, which correspond to the two forms of the conditions for the equality of two 
 ratios (Arts. 46, 48). 
 
 The first form of the conditions is Euclid's Test for Equal Ratios as stated 
 in the Fifth Definition of the Fifth Book. It contains three classes of alternatives, 
 one of which appears only when the magnitudes of the ratios are commensurable. 
 Sometimes it is possible to examine all three classes of alternatives in the same 
 way. On the other hand, in the extremely important Propositions Euc. V. 16, 22, 
 23, the examination of the cases in which the ratios are commensurable has to be 
 conducted upon different lines to those which are applicable when they are not 
 commensurable. This it is quite possible to do, but the line of argument is 
 artificial and therefore difficult for a beginner, as will be seen by consulting 
 Notes 6, 9 and 11 at the end of the book. The proofs of Euc. V. 16, 22, 23 as 
 completed by these Notes depend on the use of Prop. 63 (Euc. V. 4), but the way 
 in which that proposition has to be used does not suggest itself naturally. 
 
 62
 
 VJii PREFACE 
 
 It is on this account that the second form of the conditions for the equality 
 of two ratios (Art. 48) has been introduced into this book. So far as the Author 
 knows it was first published by Stolz (see Arts. 47, 48). 
 
 Reference has been made above to three Propositions which do not fall under 
 either of the above recognized types. These are Prop. 62 (Euc. V. 24), Prop. 65 
 (Euc. V. 19), Prop. 66 (Euc. V. 25). The proofs here given are Euclid's. They 
 are very much shorter than any direct deduction of the propositions from either 
 form of the conditions for the equality of two ratios. At the same time their 
 artificial character stands out in striking contrast to the directness of the proofs 
 of the other propositions. 
 
 A still greater difficulty than any of the preceding arises from the fact that 
 Euclid furnishes no explanation of the steps by which he reached his fundamental 
 definitions. 
 
 To write down a definition, and then draw conclusions from it, is a process 
 which is useful in Advanced Mathematics; but it is wholly unsuitable for 
 elementary teaching. It seems not unlikely that Euclid reached his fundamental 
 definitions as conclusions to elaborate trains of reasoning, but that finding great 
 difficulty in expressing this reasoning in words owing to the absence of an algebraic 
 notation, he preferred to write down his definitions as the basis of his argument, 
 and to present the propositions as logical deductions from his definitions. 
 
 Apparently he has left no trace of the steps by which he reached his fundamental 
 definitions ; and one of the chief objects of this book is to reconstruct a path which 
 can be followed by beginners from ideas of a simpler order to those on which his 
 work is based. 
 
 The most important of his definitions is the Fifth, on reaching which the 
 beginner, who has read the first four books of Euclid, experiences a sense of 
 discontinuity. He knows nothing which can lead him directly to it, he has no 
 ideas of a simpler order with which to connect it ; and he is therefore reduced 
 to learning it by rote. His teacher may show him that it contains the definition 
 of Proportion given in treatises on Algebra; but even with this assistance it 
 remains difficult for him to remember its details. He may and frequently does 
 learn to apply it correctly in demonstrating the 1st and 33rd Propositions of the 
 Sixth Book, but the Author's experience both of teaching and examining leads 
 him to the belief that it is not really understood. 
 
 In the first edition of this book, the writer, following a suggestion due to the 
 late Professor De Morgan, derived Euclid's Test for the Sameness (or Equality) of 
 Two Ratios from the Theory of Relative Multiple Scales, and on this theory the 
 whole treatment of the subject was based. This procedure made it possible to
 
 PREFACE ix 
 
 render Euclid's Fifth Book intelligible even to persons not possessed of any special 
 mathematical aptitude, but further study of the subject has led the writer to the 
 conclusion that ideas of a still simpler nature may be taken as a starting point, 
 and accordingly the use of the above-mentioned theory has been replaced by a 
 direct comparison of ratios with rational fractions. 
 
 The arrangement of the argument may be briefly summarised thus : 
 
 The first thing to be done is to consider the ideas connected with the term 
 magnitude. No attempt is made to define a magnitude*; all that is done is to 
 enumerate the characteristic properties of a magnitude. 
 
 It is easy to give illustrations of what is meant by a magnitude, such as a 
 segment of a straight line, an area, a volume, or a weight. In regard to these it 
 is assumed that if any magnitude, which may be called A, exists, then any number 
 of magnitudes exist having exactly the same properties. Each of these is called 
 A, and it is assumed to be possible to add any number of them together to form a 
 new magnitude. Thus if r denote any integer, and r magnitudes each equal to A 
 are added together, the sum is called rA. 
 
 Conversely if any magnitude B exist it is assumed that another magnitude A 
 exists, such that B = rA. It should be carefully noticed that nothing more than 
 the existence of A is assumed. It is not assumed that it is possible to construct 
 the magnitude A when B is given. 
 
 The next step taken is to state the characteristic property of " magnitudes of 
 the same kind." It is easy to give illustrations of what is meant by such magni- 
 tudes, e.g. two segments of straight lines, or two areas, or two volumes, or two 
 weights, and so on. 
 
 But in order to employ such magnitudes in analysis, something further is 
 required. This is expressed as follows : 
 
 Any two magnitudes are said to be of the same kind, whenever it is possible to 
 determine whether any multiple whatever that may be selected of one magnitude 
 is greater than, equal to, or less than any multiple whatever of the other magnitude 
 (Art. 21). In the next place the Axiom of Archimedes is assumed to hold good 
 for any two magnitudes of the same kind (Art. 22). Then the ratio of two 
 commensurable magnitudes is defined in the usual way, viz.: 
 
 The ratio of rA to sA, where r, s denote any two positive integers, is defined to 
 
 r 
 be the rational fraction - (Art. 30). 
 
 * It is not now usual to attempt to define a straight line. All that is done is to state the 
 characteristic property of the straight line that it is determined by two points.
 
 X PREFACE 
 
 It is then proved (Art. 42) that if three magnitudes A, B, C have a common 
 measure, 
 
 and if A = B, then A:C = B:G-, 
 
 but if A > B, then A : G > B : G. 
 
 These are the earlier parts of Propositions 7 and 8 of Euclid's Fifth Book, but 
 up to this stage in the argument they have been proved only if the magnitudes 
 concerned are commensurable. 
 
 If two magnitudes have no common measure, then the definition given for the 
 ratio of commensurable magnitudes does not apply, and the definition of ratio 
 must be extended before it can be applied to incommensurable magnitudes. 
 This requires for its full explanation the introduction into the subject of the 
 irrational number. As this is too difficult for the majority of those for whom this 
 book is written, it is given in a note* at the end of the book, whilst in the text it 
 is simply stated that the irrational number is a " magnitude f of the same kind" 
 (in the sense explained above) as the rational fraction, and a method of arranging 
 all ratios, whether they be ratios of commensurable or incommensurable magnitudes, 
 in order of magnitude, is constructed on the hypothesis that the propositions : 
 
 If A=B, then A:G=B:C, 
 
 and if A > B, then A : G > B : C 
 
 hold good, whenever the magnitudes A, B, C are of the same kind in the sense 
 explained above, whether they have or have not a common measure (Art. 43). 
 
 When irrational numbers have been defined, it is shown that if A, B be 
 magnitudes of the same kind which are incommensurable, then the ratio of A to 
 B is an irrational number. (Arts. 208 210.) 
 
 The hypotheses stated in the last paragraph but one make it possible to compare 
 any ratio with any rational fraction, and to determine whether any two ratios are 
 equal or unequal, and if unequal which is the greater. Thus Euclid's Tests, given 
 in his Fifth and Seventh Definitions, are obtained. 
 
 Euclid however deduces the Propositions just mentioned, which have been 
 taken as hypotheses, from his Fifth and Seventh Definitions. These two definitions 
 present very great difficulties to the beginner. Their meaning is far more difficult 
 to grasp than the propositions quoted above. Accordingly in this second edition 
 the order of the argument has been inverted. 
 
 * Note 15. 
 
 t In the purely arithmetical theory a number is primarily a mark of order, and not a magnitude. 
 This purely arithmetical conception of a number if insisted upon in the text would have made the 
 argument of too abstract a nature to be readily understood by those for whose use it is written.
 
 PREFACE xi 
 
 This is the principal change made in the argument. But it is of so far-reaching 
 a character that it has rendered it necessary to re-write the greater part of the 
 book. 
 
 This method of treatment leads naturally to Dedekind's Theory of the 
 Irrational Number, which, as already mentioned, is explained in a note* at the 
 end of the book, the material of which is drawn largely from Dedekind's work. 
 
 In this edition a section is devoted to the consideration of the ratios of 
 Commensurable Magnitudes, with the object of making it easier to understand 
 the treatment of the ratios of incommensurable magnitudes which follows. 
 
 The technical terms "Compounding of Ratios" and " Duplicate Ratio" present 
 some difficulty to beginners. So far as the Sixth Book of Euclid is concerned, 
 they are only required for the construction of two straight lines whose ratio is 
 equal to the ratio of two areas. The constructions present no difficulty and are 
 given in the text (Art. 36, Props. 40 and 42), whilst the technical terms are 
 relegated to notes in the Appendix (see Notes 13, 14). The effect of these 
 constructions is to reduce the measurement of areas to the measurement of 
 lengths. 
 
 Several alterations have been made in the order of the Propositions. De 
 Morgan pointed out that learners found great difficulty in reading the Fifth Book 
 on account of the abstract character of the reasoning, its application to something 
 concrete not being easily perceived. Accordingly in this work Propositions from 
 the Sixth Book are taken as soon as a sufficient number of Propositions from the 
 Fifth Book have been proved to make it possible to deal logically with those in 
 the Sixth Book. With regard to the enunciations no attempt has been made to 
 adhere to Euclid's words. 
 
 This work contains demonstrations of allf the Propositions in the Fifth Book 
 except Nos. 8, 10, 13 which depend on the Seventh Definition, and which are not 
 required for the understanding of the Sixth Book. 
 
 It contains also the demonstrations of all the Propositions in the Sixth Book 
 except No. 32, which is of no importance. 
 
 * It is hoped that what is there set down will induce some of my readers to study Dedekind's own 
 Tract on Continuity and Irrational Numbers. There is an English Translation by Professor Beman, 
 published by the Open Court Publishing Company, of which the London Agents are Kegan Paul, 
 Trench, Triibner & Co. 
 
 t The demonstration of Euc. V. 3 is involved in the proof of the Proposition here numbered 5. 
 Euc. V. 20 is a particular case of Euc. V. 22, here numbered 37. Euc. V. 21 is a particular case of 
 Euc. V. 23, here numbered 57.
 
 Xll PREFACE 
 
 The Propositions here numbered 7, 8, 13, 14, 26, 39, 60, 61, 64 and 69 are not 
 in Euclid's Text. 
 
 There are given in suitable positions in the book, the definitions of Harmonic 
 Points and Lines, of the Pole and Polar, of Inversion, of the Radical Axis and 
 the Centres of Similitude of Two Circles, and (so far as is possible without 
 explaining the use of the Negative Sign in Geometry) of Cross or Anharmonic 
 Ratio, with the sole object of rendering intelligible the terminology employed in 
 a number of interesting examples in the book. 
 
 The harder examples are marked with an asterisk. 
 
 The Author believes that he has not taken without acknowledgment from 
 other text-books anything which is not common property. 
 
 His special thanks are due to the Cambridge University Press Syndicate, 
 who have made the publication of the book possible ; and to his friend and former 
 pupil, Dr L. N. G. Filon, for valuable suggestions and assistance. 
 
 He will be grateful to his readers for suggestions and corrections.
 
 TABLE OF CONTENTS. 
 SECTION I. 
 
 PROPOSITIONS 1 8. ON MAGNITUDES, THEIR MULTIPLES AND THEIR MEASURES. 
 
 PROP. 1. r(A+B)=rA+rB. 
 
 2. (a + b)R=aR + bR. 
 
 3. IfA>B, tbeur(A-B)=rA-rB. 
 
 4. Ifa>6, then(a-b)R=aR-bR. 
 
 5. r(sA) = (rs)A = (sr)A=s(rA). 
 
 6. (i) If A | B, then rA = rB. (iii) If a | b, then aR | bR. 
 (ii) If rA | rB, then .4 | B. (iv) If atf 1 6^, then a | b. 
 
 7. If the magnitudes A and r4 be each divided into s equal parts, to prove that any oue 
 
 of the parts into which rA is divided will be r times as great as any one of the 
 parts into which A is divided. 
 
 8. If Jf, F, Z be magnitudes of the same kind, and if X be greater than F+Z, then an 
 
 integer t exists such that 
 
 X>tZ>T. 
 
 SECTION II. 
 
 PROPOSITIONS 9 12. COMMENSURABLE MAGNITUDES. 
 
 ART. 31. rA:sA=-. 
 s 
 
 PROP. 9. The areas of parallelograms (or triangles) having the same altitude and standing on 
 
 commensurable bases are proportional to their bases. 
 ART. 35. LEMMA. 
 
 If two intersecting straight lines OX, OF be cut by four parallel straight lines 
 AE, BF, CG, DH ; and if AB=CD, then must EF=GH. 
 
 H. E. C
 
 xiv CONTENTS 
 
 PROP. 10. To divide a finite segment of a straight line into any number of equal parts. 
 
 11. If two straight lines be cut by any number of parallel straight lines, to prove that 
 
 the ratio of any two commensurable segments of one line is equal to the ratio of 
 the corresponding segments of the other line. 
 
 12. In the same circle or in equal circles the ratio of two commensurable angles at the 
 
 centre is equal to the ratio of the arcs on which they stand. 
 
 SECTION III. 
 
 PROPOSITIONS 13, 14. ON RATIO. 
 
 ART. 43. If C= A, then C : B = A : B and conversely. 
 PROP. 13. If A : B = - , then r A | nB and conversely. 
 ART. 46. A:S=C:D, 
 
 if A : .6 = - , according as C:D = - , whatever integers r, s may be. 
 
 PROP. 14. If all values of r t which make rA >sB also make rC>sD, and if all values of 
 r, s which make r A < sB also make rC<sD, then if any values of r, s exist which 
 make rA=s t they also make rC=sD. 
 
 ART. 48. A:B=C:D, 
 
 if A : B < - , according as C : D < - , whatever integers r, s may be. 
 
 SECTION IV. 
 
 PROPOSITIONS 15 26. THE SIMPLER PROPOSITIONS IN THE THEORY OF RATIOS WITH 
 
 GEOMETRICAL APPLICATIONS. 
 
 PROP. 15. A:B = nA:nB. 
 
 16. Parallelograms or triangles having the same altitude are proportional to their 
 
 bases. 
 
 17. If two straight lines be cut by any number of parallel straight lines, then the 
 
 ratio of two segments of one straight line is the same as the ratio of the two 
 corresponding segments of the other straight line. 
 
 18. Given three segments of straight lines, to find a fourth such that the ratio of the 
 
 first and second segments is the same as that of the third and fourth. 
 
 19. To divide a straight line similarly to a given divided straight line.
 
 CONTENTS XV 
 
 PROP. 20. In equal circles or in the same circle, 
 
 (1) angles at the centres are proportional to the arcs on which they stand. 
 
 (2) angles at the circumferences are proportional to the arcs on which they stand. 
 
 (3) sectors are proportional (a) to their arcs, 
 
 (6) to their angles. 
 
 21. If A:B=C:D, 
 then B:A=D:C. 
 
 22. If A = B, thenC:A=C:B. 
 
 23. If C:A = C:B, then A = B. 
 
 24. If A, B, C, D are magnitudes of the same kind, and if 
 
 A :B=C:D, 
 then A : C=B : D. 
 
 25. If two sides of a triangle are divided proportionally, so that the segments terminating 
 
 at the vertex common to the two sides correspond to each other, then the straight 
 line joining the points of division is parallel to the other side. 
 
 26. (i) A given segment of a straight line can be divided internally into segments 
 
 having the ratio of one given line to another in one way only. 
 (ii) A given segment of a straight line can be divided externally into segments 
 having the ratio of one given line to any other not equal to it in one way only. 
 
 SECTION V. 
 
 PROPOSITIONS 27 34. SIMILAR FIGURES. 
 
 PROP. 27. Rectilineal figures which are similar to the same rectilineal figure are similar to 
 one another. 
 
 28. If the three angles of one triangle are respectively equal to the three angles of 
 
 another triangle, then the triangles are similar. 
 
 29. If the sides taken in order of one triangle are proportional to the sides taken 
 
 in order of another triangle, then the triangles are similar. 
 
 30. If two sides of one triangle are proportional to two sides of another triangle and 
 
 the included angles are equal, then the triangles are similar. 
 
 31. If two triangles have one angle of the one equal to one angle of the other, and 
 
 the sides about one other angle in each proportional in such a manner that 
 the sides opposite to the equal angles correspond, then the triangles have their 
 remaining angles either equal or supplementary, and in the former case the 
 triangles are similar. 
 
 32. On a given straight line to describe a rectilineal figure similar and similarly 
 
 situated to a given rectilineal figure. 
 
 33. Two similar rectilineal figures may be divided into the same number of triangles 
 
 such that every triangle in either figure is similar to one triangle in the other 
 figure. 
 
 c2
 
 xvi CONTENTS 
 
 PROP. 34. If a right-angled triangle be divided into two parts by a perpendicular drawn 
 from the vertex of the right angle on to the hypotenuse, then the triangles 
 so formed are similar to each other and to the whole triangle ; the perpendicular 
 is a mean proportional between the segments of the hypotenuse ; and each side 
 is a mean proportional between the adjacent segment of the hypotenuse and the 
 hypotenuse. 
 
 SECTION VI. 
 
 PROPOSITIONS 35, 36. MISCELLANEOUS PROPOSITIONS. 
 
 PROP. 35. To find a mean proportional between two given segments of straight lines. 
 
 36. (i) If the interior or exterior vertical angle of a triangle be bisected by a straight 
 line which also cuts the base, the base is divided internally or externally in the 
 ratio of the sides of the triangle. 
 
 (ii) If the base of a triangle be divided internally or externally in the ratio of the 
 sides, the straight line drawn from the point of division to the vertex bisects 
 the interior or exterior vertical angle. 
 
 SECTION VII. 
 
 PROPOSITION 37. AN IMPORTANT PROPOSITION IN THE THEORY OP RATIO. 
 
 PROP. 37. If A, B, C are magnitudes of the same kind, 
 
 if T, U y V are magnitudes of the same kind, 
 
 and if A : B=T : 17, 
 
 and if B : C= U : V, 
 
 then A : C= T : V. 
 
 SECTION VIII. 
 
 PROPOSITIONS 38 50. ON AREAS. 
 
 PROP. 38. (i) If K:L = M:P, 
 
 then rect. JT.P=rect. L.M. 
 
 (ii) If rect. K.P=rect. L.M, 
 
 the'n K:L=M:P. 
 
 39. The rectangle contained by the diagonals of a quadrilateral cannot be greater than 
 the sum of the rectangles contained by opposite sides. (It may be equal, and 
 then a circle can be described through the vertices of the quadrilateral.)
 
 CONTENTS xvii 
 
 PROP. 40. To construct two straight lines whose ratio shall be equal to the ratio of the areas 
 of two equiangular parallelograms. 
 
 41. The ratio of the areas of two equiangular parallelograms is equal to the ratio of the 
 
 areas of the rectangles contained by their sides. 
 
 42. If three straight lines are in proportion, then the ratio of the area of the square 
 
 described on the first line to the area of the square described on the second line 
 is equal to the ratio of the first line to the third line. 
 
 43. The areas of similar triangles are proportional to the squares described on corre- 
 
 sponding sides. 
 
 44. If four straight lines are proportional, then the squares described on them are 
 
 proportional, and conversely. 
 
 45. If A : B = C : D = E : F, where all the magnitudes are of the same kind, then 
 
 A : B=A+C+E : B + D+F. 
 
 46. The areas of similar rectilineal figures are proportional to the squares described 
 
 on corresponding sides. 
 
 47. (i) If A : B=C : D, 
 
 then figure on A : similar and similarly described figure on B 
 
 = figure on C : similar and similarly described figure on D. 
 
 (ii) If figure on A : similar and similarly described figure on B 
 
 = figure on C : similar and similarly described figure on Z), 
 
 then A : B=C : D. 
 
 48. In any right-angled triangle, any rectilineal figure described on the hypotenuse is 
 
 equal to the sum of the two similar and similarly described figures on the sides. 
 
 49. To describe a rectilineal figure similar to one given rectilineal figure and equal 
 
 in area to another given rectilineal figure. 
 
 50. (i) Equal parallelograms which have one angle of the one equal to one angle of the 
 
 other have the sides about the equal angles reciprocally proportional, 
 (ii) Parallelograms having one angle of the one equal to one angle of the other, and 
 the sides about the equal angles reciprocally proportional, are equal in area. 
 
 COR. (i) 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, 
 (ii) Triangles which have one angle in the one equal to one angle in the other and 
 the sides about the equal angles reciprocally proportional are equal in area. 
 
 SECTION IX. 
 
 PROPOSITIONS 51 56. MISCELLANEOUS GEOMETRICAL PROPOSITIONS. 
 
 PROP. 51. If the vertical angle of a triangle be bisected by a straight line which also cuts 
 the base, the rectangle contained by the sides of the triangle is equal to the 
 rectangle contained by the segments of the base together with the square on 
 the straight line which bisects the angle.
 
 xvni 
 
 CONTENTS 
 
 PROP. 52. If from any vertex of a triangle a perpendicular be drawn to the opposite side, 
 the diameter of the circle circumscribing the triangle is a fourth proportional 
 to the perpendicular and the sides of the triangle which meet at that vertex. 
 
 53. To divide a segment of a straight line internally or externally in extreme and 
 
 mean ratio. 
 
 54. Parallelograms about the diagonal of any parallelogram are similar to the whole 
 
 and to one another. 
 
 55. If two similar parallelograms have a common angle and be similarly situated 
 
 they are about the same diagonal. 
 
 56. If OAB be a given triangle it is required to find a point P on AB or AB produced 
 
 so that if PQ be drawn parallel to OB to cut OA in (J), and if PR be drawn 
 parallel to OA to cut OB in R, then the parallelogram PQOR may have a given 
 area. 
 
 PROPOSITIONS 57 64. 
 
 SECTION X. 
 
 THE REMAINING IMPORTANT THEOREMS IN THE THEORY 
 OP RATIO. 
 
 PROP. 57. If A, B, C be three magnitudes of the same kind, 
 
 if T t U, V be three magnitudes of the same kind, 
 and if A :B=U: V, 
 
 and if B : C= T : U, 
 
 then A : C=T : V. 
 
 58. If A : B = X : F, 
 then A+B : B=X+Y : Y. 
 
 59. If A : B=X : F, 
 then A~B : B=X~Y: Y. 
 
 60. If A : B = X : Y, 
 
 then A~B : A+B=X~ Y : X+ Y. 
 
 61. If A, B, C, D be four harmonic points, A and C being conjugate, and if be the 
 
 middle point of AC, then OC is a mean proportional between OB and OD. 
 
 If A : C=X : Z, 
 
 C=Y:Z, 
 C=X+Y 
 
 62. 
 
 and if 
 then 
 
 63. If 
 then 
 
 64. 
 
 B 
 A + B 
 
 A : B=X : F, 
 rA : sB=rX : sY. 
 
 Z. 
 
 If A', Z, M, P be four straight lines in proportion, if the lengths of L and M be 
 fixed, if the length of K can be made smaller than that of any line however 
 small, to show that the length of P can be made greater than that of any line $, 
 however great Q may be.
 
 CONTENTS 
 
 SECTION XI. 
 
 PROPOSITIONS 65, 66. OTHER PROPOSITIONS IN THE THEORY OF RATIO. 
 
 PKOP. 65. If A, B, C, D are magnitudes of the same kind, and if A : B = C : D, then 
 
 A~C :B~D=A : B. 
 
 66. If A, B, C, D are magnitudes of the same kind, and if A : B=C : D, then the 
 sum of the greatest and least of the four magnitudes is greater than that of 
 the other two. 
 
 PROPOSITIONS IN THE NOTES. 
 
 PROP. 67. If three magnitudes be in proportion, the first has to the third the duplicate ratio 
 of the first to the second. 
 
 68. If two ratios be equal, their duplicate ratios will be equal. 
 
 69. If A : B be any ratio, and C any magnitude, then a magnitude Z of the same kind 
 
 as C exists, such that A : B=C : Z. 
 
 ERRATUM. 
 
 Page 152, line 22. For ' segments of straight lines ' read ' magnitudes of (he same kind.'
 
 LIST OF ABBREVIATIONS. 
 
 + Plus. 
 
 Minus. 
 Equal to. 
 < Less than. 
 
 Greater than. 
 4= Not equal to. 
 
 <t Not less than. 
 
 ^> Not greater than. 
 
 A~B The difference of A and B. 
 
 A : B The ratio of A to B. 
 
 A : B=C : D The ratio of A to B is equal to the ratio of C to D. 
 -X- Compounded with. (Art. 191.) 
 
 <-> Aggregated with. (Art. 194.)
 
 SECTION I. 
 
 PROPOSITIONS 18. 
 ON MAGNITUDES, THEIR MULTIPLES AND THEIR MEASURES. 
 
 Art. 1. Number. 
 
 IN this book except where otherwise stated the word Number will be used 
 as an abbreviation for Positive Whole Number. 
 
 Notation for Number. 
 A Number will always be denoted by a small letter. 
 
 Art. 2. Notation for Magnitude. 
 A Magnitude will be denoted throughout this book by a capital letter*. 
 
 Art. 3. Def. 1. MULTIPLE. 
 
 If A denote any magnitude, then it is usual to denote A + A by 2 A for 
 brevity, and in like manner A + A + A by 3 A for brevity and so on. If A be 
 added successively to itself until there are in the aggregate altogether r A's, the 
 result is written rA for brevity, and called the rth multiple of A. 
 
 It follows from the definition that 
 
 rA +A = (r+l)A. 
 
 If the magnitude B be the same as rA, then B is said to be a multiple of A, 
 and in general 
 
 One magnitude is said to be a multiple of another magnitude, when the 
 former contains the latter an exact number of times. 
 
 It will be in agreement with the above nomenclature, when A is equal to B, 
 to say that A is the first multiple of B; and to call the rth multiple of B and 
 the rth multiple of C the same multiples of B and C. 
 
 * A point will also be denoted by a capital letter, but this will not lead to any difficulty. 
 H. E. 1
 
 2 EUCLID, BOOKS V. AND VI. [4 
 
 Art. 4. It is necessary to prove certain propositions regarding magnitudes 
 and multiples of magnitudes before entering upon the discussion of the relations 
 between magnitudes. 
 
 These propositions depend upon the Commutative and Associative Laws for 
 Addition. 
 
 The Commutative Law is expressed by the formula 
 
 X+Y=Y+X. 
 
 The Associative Law is expressed by the formula 
 
 The effect of these two laws is to render it permissible, when adding up any 
 number of magnitudes, to group them in any way, to add up the members of each 
 group, and then to add up the groups. 
 
 Art. 5. PROPOSITION I.* (Euc. V. 1.) 
 
 ENUNCIATION. To prove that r(A+B) = rA + rB. 
 Since rA means A + A + A + ... to r terms, 
 
 r (A + B) means (A + B) + (A + B) + (A + B) + ... to r terms. 
 
 By means of the commutative and associative laws, the terms in the r groups 
 may be added up in any order. 
 
 Adding up the r A's together, the result is rA, and adding up the r B's together 
 the result is rB. 
 
 Thus r (A + B) = rA + rB. 
 
 Art. 6. EXAMPLE 1. 
 By repeated application of Proposition I. prove that 
 
 r(A + B+ ... + K) = rA+r+ ... + rK. 
 
 Art. 7. PROPOSITION II.* (Euc. V. 2.) 
 ENUNCIATION. To prove that (a + b)R = aR + bR. 
 By definition (a + b) R means R + R + ... until there are (a + b) terms. 
 By the Associative Law these terms may be added up in any way. 
 
 * A more formal proof of this theorem will be found in the Appendix, see Note 2.
 
 10] EUCLID, BOOKS V. AND VI. 3 
 
 Take therefore the first a R's and put them in one group, and put the 
 remaining b R's into another group. 
 
 The sum of the first group is aR. 
 The sum of the second group is bR. 
 
 .-.. (a + b) R = aR + bR. 
 
 Art. 8. EXAMPLES. 
 
 2. Prove that (r + s + t + ...+z) A -rA + sA + tA + ...+ zA. 
 
 3. If A and B are both multiples of G, prove that A + B is a multiple of G. 
 
 Art. 9. PROPOSITION III.* (Euc. V. 5.) 
 
 ENUNCIATION. If A > B, then r (A B)=rA rB. 
 Since A > B, 
 
 let A = B + C. 
 
 .'. rA = r(B+C) 
 
 = rB + rC. [Prop. 1. 
 
 .-. rC=rA -rB. 
 But C=A-B, 
 
 Art. 10. PROPOSITION IV.* (Euc. V. 6.) 
 
 ENUNCIATION. If a > b, prove that (a b)R = aR bR. 
 Since a > b, and each is a positive integer, 
 .'. (a b) is a positive integer which may be called c. 
 
 .. a=b + c. 
 
 = bR + cR. [Prop. 2. 
 
 .-. cR=aR-bR. 
 (a-b)R = a,R-bR. 
 
 * See Note 1. 
 
 12
 
 4 EUCLID, BOOKS V. AND VI. [11 
 
 Art. 11. EXAMPLE 4. 
 If A and B are multiples of G, then the difference of A and B is a multiple of G. 
 
 Art. 12. PROPOSITION V.* 
 
 ENUNCIATION. To prove that 
 
 r (sA) = rs (A) = sr (A) = s(rA). 
 
 Let a rectangle be drawn and divided into compartments standing in r columns 
 and s rows. 
 
 Place the magnitude A in each compartment. 
 
 Then the sum of the magnitudes in any row is rA, and the sum of those in any 
 column is sA. 
 
 Since there are s rows, the sum of all the magnitudes is s(rA). 
 Since there are r columns, the sum of all the magnitudes is r (sA). 
 
 If the number of the magnitudes be counted, it is rs, or it may also be expressed 
 as sr. 
 
 Hence the sum can be written in either of the forms rs (A) or sr (A). 
 
 But by the Commutative and Associative Laws the sum of the magnitudes 
 is the same in whatever way the magnitudes are added together. 
 
 /. r (sA) = rs (A) = sr (A) = s(rA). 
 
 Art. 13. EXAMPLE 5. 
 
 If A and B are multiples of G, then the sum and difference of rA and sB are 
 
 multiples of G. 
 
 Art. 14. PROPOSITION VI (i). 
 ENUNCIATION. If A > B, then rA >rB; 
 
 If A = B, then rA = rB; 
 
 If A <B, then rA < rB. 
 If A>B, 
 
 * See Note 1.
 
 16] EUCLID, BOOKS V. AND VI. 5 
 
 /. rA=r(B + C) 
 
 = rB+rC. [Prop. 1. 
 
 .'. rA > rB. 
 
 If A=B, 
 
 then rA = rB. 
 
 If A < B, 
 
 then B>A. 
 
 Hence rB > rA , by what is proved above. 
 
 .-. rA<rB. 
 
 Art. 15. PROPOSITION VI (ii). 
 
 ENUNCIATION. // r A > rB, then A > B; 
 If rA=rB, then A=B; 
 If rA < rB, then A<B. 
 
 If rA > rB, 
 
 suppose if possible that A is not greater than B. 
 
 Then either A = B, 
 
 or A < B. 
 
 But by the first part of this proposition, 
 if A=B, then rA=rB; 
 
 and if A < B, then r A < rB. 
 
 Both these results are contradictory to the hypothesis that rA > rB. 
 
 Hence A must be greater than B. 
 
 The second and third cases can be proved in like manner. 
 
 Art. 16. PROPOSITION VI (iii). 
 
 ENUNCIATION. If a > b, then aR > bR; 
 
 If a = b, then aR = bR ; 
 
 If a <b, then aR < bR. 
 If a > b, 
 
 let a = 6 + c,
 
 EUCLID, BOOKS V. AND VI. [16 
 
 .-. aR = (b + c)R 
 
 cR; [Prop. 2. 
 
 .-. aR>bR. 
 
 If a = b, 
 
 then ctR = bR. 
 
 If a < b, 
 
 then b > a, 
 
 .'. bR>aR, by what was proved above. 
 .-. aR<bR. 
 
 Art. 17. PROPOSITION VI (iv). 
 ENUNCIATION. If aR > bR, then a > b ; 
 If aR = bR, then a = b ; 
 If aR < bR, then a<b. 
 If aR > bR, 
 
 suppose if possible that a is not greater than b. 
 
 Then either a = b, 
 
 or a < b. 
 
 But by part (iii) of this proposition, 
 
 if a = b, then aR = bR ; 
 and if a < b, then aR < bR. 
 
 Both these results are contradictory to the hypothesis that 
 
 aR > bR. 
 
 Hence a must be greater than b. 
 The second and third cases can be proved in like manner. 
 
 Art. 18. EXAMPLES. 
 
 6. (i) If aU>bV, 
 
 bT>cU, 
 
 prove that aT>cV. 
 
 (ii) If nrU>tV, 
 
 and tT>nsU, 
 
 prove that rT>sV.
 
 20] EUCLID, BOOKS V. AND VI. 7 
 
 7.* If rA>sJB, 
 
 and rC<8D, 
 
 prove that no integers r', s' can exist such that 
 
 r'A<s'B, 
 and r'C>s'D. 
 
 Art. 19. If B = rA, then B was called the rth multiple of A. 
 
 The magnitude B is said to be measured by A ; and A is called a measure or 
 a partf of B. 
 
 When A exists, it is assumed to be always possible to construct a magnitude 
 equal to rA. 
 
 But the reverse operation is not always possible. 
 
 When B is a segment of a straight line, a construction can be given, see Prop. 10, 
 for finding the segment of a straight line A, such that B = rA. 
 
 But if B is an arc of a circle, there exists no known general construction for an 
 arc A, such that B = rA for all values of the integer r. 
 
 Nevertheless it will be assumed that if B be any magnitude, and r any positive 
 integer, then a magnitude A exists, whether it can be constructed or not, such that 
 
 B = rA. 
 This same relation between A and B is denoted by 
 
 B 
 A== r' 
 
 or A = - B. 
 
 r 
 
 Art. 20. PROPOSITION VII. 
 
 ENUNCIATION. If the magnitudes A and rA be each divided into s equal 
 parts, prove that any one of the parts into which rA is divided will be r times 
 as great as any one of the parts into which A is divided. 
 
 Suppose that each of the parts into which A is divided is B. 
 Then A = sB. 
 
 .*. rA = rsB = s (rB). [Prop. 5. 
 
 Hence each of the s equal parts into which rA is divided is rB, which is 
 r times as great as each of the s equal parts into which A is divided. 
 
 t The word part is here used in a technical sense. It does not mean any portion of B.
 
 EUCLID, BOOKS V. AND VI. [20 
 
 If then -- denote the sth part of A, 
 
 s 
 
 will denote the sth part of rA, 
 s 
 
 and this proposition may be expressed thus: 
 
 rA fA'" 
 
 rA (A\ 
 
 - = rB = r - . 
 s \s) 
 
 Art. 21. MAGNITUDES OF THE SAME KIND. 
 
 It is usual to speak of two lengths as magnitudes of the same kind, or two 
 areas, or two volumes, or two weights. 
 
 The characteristic of two such magnitudes is this : 
 
 It is supposed to be possible always to find out whether any multiple of the 
 one is greater than, or is equal to, or is less than any multiple of the other. 
 
 Let the magnitudes be A and B. 
 
 Take any multiple of A say rA, and any multiple of B say sB. 
 
 Then all that is meant by saying that A and B are of the same kind is this : 
 
 It is assumed to be possible always to determine whether rA is greater than 
 sB, or equal to sB, or less than sB. 
 
 In particular it is assumed to be possible to determine whether A is greater 
 than B, or equal to B, or less than B. 
 
 The same thing may be put in a slightly different way thus : 
 
 D 
 
 Suppose that B is divided into r equal parts, each of which is denoted by ; 
 
 that s of these parts are taken, giving s ( ] , then it is supposed to be possible 
 
 /7?\ 
 
 always to determine whether A is greater than, equal to, or less than s ( ) . 
 
 If A>.(f). 
 
 then A > - [Prop. 7. 
 
 r 
 
 .'. rA >sB.
 
 23] EUCLID, BOOKS V. AND VI. 
 
 ' <" 
 
 sB 
 
 , 
 
 r 
 
 .'. rA=sB. 
 
 then A = 
 
 r 
 
 
 sB 
 
 then A < . 
 
 r 
 
 /. r A < sB. 
 
 Art. 22. AXIOM. 
 
 If . I and />' are two magnitudes of the same kind, it is always possible 
 to find a multiple of either which will exceed the other. 
 
 This is usually known as the Axiom of Archimedes. But Euclid uses it in 
 the Fifth Book, see Euc. V. 8, and it is also implied in the fourth definition of the 
 Fifth Book. 
 
 Art. 23. PROPOSITION VIII. 
 
 If X, Y, Z be three magnitudes of the same kind and if 
 
 X>Y+Z, 
 to prove that an integer t exists such that 
 
 X>tZ> Y. 
 
 Let sZ be the greatest multiple of Z which does not exceed Y. 
 Then either (i) Y = sZ 
 
 or (ii) sZ<Y<(s 
 
 In case (i) Y=sZ, 
 
 but X>Y+Z, 
 
 .-. X>(s + I)Z> Y. 
 Hence (s + 1) is the integer required. 
 H. E.
 
 10 
 
 In case (ii) 
 
 but 
 
 EUCLID, BOOKS V. AND VI. 
 F<(*+ l)Z, 
 
 X>Y+Z, 
 Y>sZ, 
 . . X>sZ+Z, 
 
 .: X>(s+l)Z>Y. 
 Hence (s + 1) is the integer required. 
 
 [23 
 
 Art. 24. EXAMPLE 8. 
 
 If X, Y, Z be three magnitudes of the same kind, and if X and Y be 
 unequal, prove that it is always possible to find integers n and t, such that 
 tZ lies between nX and n Y. 
 
 Art. 25. An illustration of the preceding example is given in 
 Fig. 1 in the case where X, Y, Z are segments of straight lines. 
 
 Here OA and OB are the same multiples of X and F, such that 
 AB is greater than Z, and 00 is a multiple of Z, which is greater 
 than OA, but less than OB. 
 
 It should be noticed that 2Z lies between 3X and 3F, 
 that 3Z lies between 4-X and 4F, 
 
 but this is ascertained only after the figure has been drawn, whilst 
 the fact that a multiple of Z, which in this case is 4>Z, lies between 
 bX and oFis determinate from the consideration that 5 (F X) > Z. 
 
 Art. 26. EXAMPLES. 
 
 9. If X, F, Z be three magnitudes of the same kind, and if no multiple 
 of Z can be found which is intermediate in magnitude between the same 
 multiples of X and Y, then X and Y must be equal. 
 
 10. If X=4, Y=5, Z=6, find from a figure the least value of n for 
 which a single multiple of Z is intermediate in magnitude between nX 
 and nY. 
 
 Find also the least value of for which two multiples of Z are inter- 
 mediate in magnitude between nX and nY. 
 
 5Y 
 4Z 
 
 4Y- 
 
 A 
 
 321 
 
 _5X 
 
 3Y- 
 
 -4X 
 
 2Z_ 
 2Y_ 
 
 _3X 
 
 _2X 
 
 -IX 
 
 Y X Z 
 
 Fig. 1.
 
 26] EUCLID, BOOKS V. AND VI. 11 
 
 11.* (i) If rA>sB, 
 
 and rC = sD, 
 
 prove that integers n, t exist such that 
 
 nrA > tB, 
 
 and nrC<tD. 
 
 (ii) If rA = sB, 
 
 and rC<sD, 
 
 prove that integers n, t exist such that 
 
 nrA > t, 
 
 nrC<tD. 
 
 22
 
 SECTION II. 
 
 PROPOSITIONS 912. 
 COMMENSURABLE MAGNITUDES. 
 
 Art. 27. Def. 2. MEASURE. 
 
 If a magnitude A contains another magnitude B an exact number of 
 times, B is said to be a measure of A. 
 
 Art. 28. Def. 3. COMMON MEASURE. 
 
 If the magnitudes A and B each contain another magnitude G an exact 
 number of times, then G is said to be a common measure of A and B. 
 
 Art. 29. Def. 4. COMMENSURABLE MAGNITUDES. 
 
 If two magnitudes have a common measure they are said to be com- 
 mensurable. 
 
 Art. 30. ON COMMENSURABLE MAGNITUDES. 
 If there be two magnitudes B and C, each of which is a multiple of A, viz. : 
 
 C = sA, 
 
 then the ratio of B to C is defined to be the rational fraction - . 
 
 s 
 
 Consequently the ratio of C to B is the rational fraction - . 
 
 f 
 
 Thus the value of the ratio of B to C is independent of the magnitude of their 
 common measure A, but it does depend on the order in which the magnitudes 
 B and C are taken.
 
 33] EUCLID, BOOKS V. AND VI. 13 
 
 Art. 31. NOTATION FOR RATIO. 
 
 If two magnitudes of the same kind be called A and B, then the ratio of A to 
 B is written A : B, and A is called the antecedent or first term of the ratio, 
 whilst B is called the consequent or second term of the ratio. 
 
 If B = rA, C=sA, 
 
 then 
 
 T 
 
 .'. rA :sA -. 
 
 s 
 
 Putting A equal to unity, it follows that 
 
 Consequently rA :sA=r:s. 
 
 In this book, the fact that one ratio A : B is equal to another ratio C : D will 
 be expressed thus : 
 
 A:B = C:D, 
 
 and not as it is written in most modern editions of Euclid : 
 
 A:B::C:D, 
 
 which is read : 
 
 the ratio of A to B is the same as the ratio of C to D, 
 or more briefly, A is to B as C to D. 
 
 Art. 32. Def. 5. THE RATIO OF EQUALITY. 
 
 If in the preceding Article, both r and s be put equal to 1, then it follows that 
 
 A:A-l. 
 Each of two equal magnitudes is said to have to the other the ratio of equality. 
 
 Art. 33. Def. 0. PROPORTION. 
 
 If there are four magnitudes such that the ratio of the first magnitude 
 to the second is the same as that of the third magnitude to the fourth, then 
 the four magnitudes are said to be proportionals, or in proportion. 
 
 If A, B, C, D are four magnitudes, such that 
 
 A:B = C:D, 
 then A, B, C, D are proportionals.
 
 14 EUCLID, BOOKS V. AND VI. [33 
 
 A and D are called the extremes of the proportion. 
 
 B and C are called the means of the proportion. 
 
 D is called the fourth proportional to A, B and C. 
 
 The antecedents A and G of the two equal ratios are said to be corresponding * 
 terms of the ratios ; so also are the consequents B and D. 
 
 The case in which the means of the proportion are equal to one another requires 
 special notice. 
 
 If X:Y=Y:Z, 
 
 then the three magnitudes X, Y, Z are said to be in proportion ; Y is said to 
 be a mean proportional between X and Z, and Z is said to be a third proportional 
 to X and F. 
 
 Art. 34. PROPOSITION IX. 
 
 ENUNCIATION. Two parallelograms, situated between the same parallels, have 
 commensurable bases, to prove that the ratio of the area of the first parallelogram 
 to the area of the second parallelogram is equal to the ratio of the base of the first 
 parallelogram to the base of the second parallelogram. 
 
 DPQRC HSTG 
 
 
 AKLMB ENOF 
 
 Fig. 2. 
 
 Let the parallelograms ABGD, EFGH have their bases AB, -ZF com mensurable. 
 
 Let AK be a common measure of AB, EF. 
 
 Suppose that AB = r (AK), 
 
 and that EF=s(AK). 
 
 Let AB, EF be divided as in the figure into parts each equal to AK, and 
 through the points of division let straight lines be drawn parallel to the sides 
 AD, CB of the parallelogram A BCD ; and to the sides EH, FG of the parallelogram 
 EFGH, so that each parallelogram is divided up into equal parallelograms. 
 
 Then since the bases of all these parallelograms are equal, and they are 
 situated between the same parallels, they are equal in area. 
 
 Since AB contains r lengths each equal to AK, therefore the parallelogram 
 ABCD contains r parallelograms each equal to AKPD, 
 
 .-. ABCD = r (AKPD). 
 
 * Euclid uses the term ' homologous.'
 
 36] 
 
 EUCLID, BOOKS V. AND VI. 
 
 15 
 
 Also EF = s (EN) = s (AK). 
 
 Thus EF contains s lengths each equal to AK, therefore the parallelogram EFGH 
 contains s parallelograms each equal to AKPD, 
 
 .'. EFGH = s(AKPD). 
 
 Since AB=r(AK), 
 
 EF=s(AK), 
 r 
 
 AB:EF= 
 
 Since 
 
 and 
 
 = r(AKPD), 
 EFGH = s (AKPD), 
 
 .-. ABCD:EFGH = ~ 
 
 s 
 
 .-. ABCD:EFGH=AB:EF. 
 
 In the same way it can be shown that 
 
 If two triangles have the same altitude, and have commensurable bases, the ratio 
 of the area of the first triangle to the area of the second triangle is equal to the ratio 
 of the base of the first triangle to the base of the second triangle. 
 
 Art. 35. A particular example of this proposition is the following. If a side 
 of a square be divided into ten equal parts, and perpendiculars be drawn to the 
 side through" the points of division, the square will be divided into ten equal 
 rectangles ; and the area of each rectangle will be one-tenth of the area of the 
 square. 
 
 If the side of the square be divided into a hundred equal parts, and perpen- 
 diculars be drawn to the side through the points of division, the square will be 
 divided into a hundred equal rectangles, and the area of each rectangle will be 
 one-hundredth of the area of the square ; and so on. 
 
 Art. 36. The preceding proposition is sufficient to render it possible to 
 calculate approximately the area of any rectilineal figure. 
 
 Take the rectilineal figure ABODE. 
 (Fig. 3.) 
 
 Join BD. 
 
 Draw CF parallel to BD to meet AB 
 produced at F. 
 
 Then &BDF=&BDC, since they are 
 on the same base BD and between the 
 same parallels BD, FC. 
 
 Add to each the figure ABDE. 
 
 .-. ABDE +&BDF= ABDE + &BDO. 
 .-. AFDE = ABODE. 
 
 Fig. 3.
 
 16 
 
 EUCLID, BOOKS V. AND VI. 
 
 [36 
 
 Now join FE. (Fig. 4.) 
 
 A F O 
 
 Fig. 4. 
 
 Draw DG parallel to EF to meet AF produced at G. 
 
 Then &FED = &FEG, since they are on the same base FE and between the 
 same parallels FE and DG. 
 
 Add to each the &AFE. 
 
 .-. &AFE + A FED =&AFE+& FEG. 
 
 Fig. 5. 
 
 Through A (Fig. 5), draw a perpendicular AX to ^(r, and through E draw 
 parallel to AG to meet AX at #. 
 Join G#. 
 
 Then &AEG = &AHG, for they are on the same base AG and between the 
 same parallels. 
 
 Hence ABODE = AFDE
 
 36] 
 
 EUCLID, BOOKS V. AND VI. 
 
 17 
 
 Hence a right-angled triangle AHG has been constructed having the same area 
 as the given rectilineal figure ABODE. 
 H 
 
 Fig. 6. 
 
 Now let AK (Fig. 6) be taken along AH equal to the unit of length, say 
 one inch. 
 
 Join KG. 
 
 Draw HL parallel to KG to meet AG produced in L. 
 
 Join KL. 
 
 Then &KHG = &.KGL, since they are on the same base KG and between the 
 same parallels KG, HL. 
 
 Add to each &AKG. 
 
 .'. &AKL = ABODE. 
 
 M 
 
 Fig. 7. 
 
 Now bisect KL at M. (Fig. 7.) 
 
 Through M draw a perpendicular to AL, to meet KL at P, and produce it to 
 Q, until PQ = MP. 
 
 Now since M is the mid-point of AL, and MP is parallel to J.#, therefore P is 
 the mid-point of KL, and .MP is half of AK. 
 
 Then AfQ = IMP AK. 
 
 Thus MQ is equal and parallel to AK. 
 
 .. AMQK is a parallelogram, 
 
 H. E. 3
 
 18 
 
 EUCLID, BOOKS V. AND VI. 
 
 [36 
 
 and since KAM is a right angle, 
 
 /. AMQK is a rectangle. 
 Also MP = PQ, 
 
 PL = PK, 
 
 MPL = QPK, 
 .*. the triangles MPL, QPK are congruent, 
 
 Add to each AMPK, 
 
 .'. AMPK + & MPL = AMPK + &QPK, 
 
 Hence a rectangle AMQK has been constructed, whose area is equal to that of 
 ABODE, and which has one side AK equal to the unit of length. 
 
 If now there be measured off consecutive units of length along AM as long as 
 this is possible, and then tenths of units as long as this is possible, and then 
 hundredths of units as long as this is possible, and so on, and if through the points 
 of division parallels be drawn to AK, then it follows from Art. 35 that AMQK 
 contains as many units and fractions of units of area as AM contains units and 
 fractions of units of length. 
 
 Art. 37. LEMMA. 
 
 If two intersecting straight lines OX, Y be cut by four parallel straight lines 
 AE, BF, CO, DH; and if AB = CD, then must EF = OH. 
 
 Draw EJ parallel to OX to meet BF at /, and draw OK parallel to OX to 
 meet DH at K. 
 
 Since AEJB is a parallelogram, 
 
 ET /" A ft 
 
 . . JllJ = A.JJ. 
 
 Since CGKD is a parallelogram, 
 
 but AB = CD, 
 
 .'. EJ=GK. 
 
 Now consider the triangles EJF, GKH. 
 In these triangles 
 EJ=GK. 
 
 EFJ = GHK, v BF is parallel to DH. 
 FEJ= HGK, v EJ is parallel to GK. 
 Hence the triangles are congruent. 
 
 /. EF=GH.
 
 40] 
 
 EUCLID, BOOKS V. AND VI. 
 
 19 
 
 Art. 38. PROPOSITION X. (Euc. VI. 9.) 
 
 ENUNCIATION. To divide a finite segment of 
 a straight line into any number of equal parts. 
 
 It is required to divide the segment OA 
 into r equal parts. 
 
 Through draw any straight line OX. 
 
 On it set off any length OB, and then 
 measure off consecutive lengths each equal to 
 OB, until a point is reached such that 
 OG = r(OB). 
 
 Join AG, and through the points of division 
 of OG draw parallels to AG. 
 
 These parallels will divide OA into r equal 
 parts by the lemma of Art. 37. 
 
 M 
 
 Fig. 9. 
 
 Art. 39. Def. 7. 
 
 CORRESPONDING POINTS AND SEGMENTS ON 
 TWO STRAIGHT LINES. 
 
 When two straight lines are cut by a single system of parallel straight 
 lines, it is convenient to call the two points, in which one of the parallel 
 straight lines cuts the two straight lines, corresponding points; and to call 
 the segments of the two straight lines between any pair of the parallel 
 straight lines corresponding segments. 
 
 Note. If the two straight lines intersect, the point of intersection on one 
 straight line will correspond to itself on the other straight line. 
 
 Art. 40. PROPOSITION XI. 
 
 ENUNCIATION. // two straight lines be cut by any number of parallel straight 
 lines, to prove that the ratio of any two commensurable segments of one line is equal 
 to the ratio of the corresponding segments of the other line. 
 
 Let OX, OF be the two straight lines cut by the parallels AE, BF, CG, DH, 
 and suppose that the segments AB, CD have a common measure AK. 
 
 Let AB = r(AK), 
 
 GD = s(AK). 
 
 3-2
 
 20 
 
 EUCLID, BOOKS V. AND VI. 
 O 
 
 [40 
 
 Fig. 10. 
 
 Then AB may be divided into r equal parts AK, KL, LB; and CD may be 
 divided into s equal parts CM, MN, NU, UD. 
 
 Through the points of division K, L, M, N, /"draw parallels to AE to cut OF 
 at P, Q, R, S, T respectively. 
 
 Then it follows from Art. 37 that 
 
 and since AB is divided into r pieces, each equal to AK, therefore EF is divided 
 into r pieces each equal to EP. This will be stated shortly thus* : 
 
 Since AB = r(AK), 
 
 .'. EF = r(EP), 
 and since CD = s (CM) = s (AK), 
 
 CD = r(AK):s(AK) 
 __r 
 ~s' 
 EF:GH = r(EP):s(EP) 
 
 _ r 
 ~~s' 
 
 .'. AB:CD = EF:GH. 
 
 Similar abbreviations in the argument are used below in this and some subsequent propositions.
 
 41] 
 
 EUCLID, BOOKS V. AND VI. 
 
 Art. 41. PROPOSITION XII. 
 
 21 
 
 ENUNCIATION. In the same circle or in equal circles the ratio of two com- 
 mensurable angles at the centre is equal to the ratio of the arcs on which they stand. 
 
 Fig. 11. 
 
 Let AOB, GPD be angles at the centre of two equal circles having the angle 
 AOE as a common measure. 
 
 Let AOB = r (AOE). 
 
 Let GPD = s(AOE). 
 
 Then the angle AOB is divisible into the r angles AOE, EOF, FOG, GOH, 
 HOB each equal to AOE; and the angle CPD is divisible into the 8 angles 
 CPK, KPL, LPM, MPN, NPQ, QPR, RPD each equal to AOE. 
 
 Now in equal circles, equal angles at the centre stand on equal arcs. 
 /. the arcs EF, FG, GH, HB are each equal to the arc AE; and since 
 
 AOB = r (AOE), 
 .'. arc AE = r (arc AE). 
 Similarly y CPD = s(CPK) = s (AOE), 
 
 .'. arc CD = s (arc CK) = s (arc AE). 
 Now AOB:CPD = r(AdE):s(AOE) 
 
 arc A B : arc CD = r (arc AE) : s (arc AE) 
 
 _ r 
 s' 
 
 .-. AOB : CPD = arc AB : arc CD.
 
 SECTION III. 
 
 
 
 PROPOSITIONS 13, 14. 
 ON RATIO. 
 
 Art. 42. Let there be two magnitudes of the same kind, A and B. 
 
 Suppose that A is an exact multiple of G, say aG ; and that B is also an exact 
 multiple of G, say bG. 
 
 Then the ratio of A to B has been defined to be the rational fraction y . 
 
 o 
 
 Take now another multiple of G, say cG, and let G = cG. 
 
 The ratio of G to B is f . 
 o 
 
 Now if C>4, 
 
 c# > aG, 
 
 .'. oa, 
 c a 
 
 '&>&' 
 
 /. C:B>A:B. 
 
 Hence if C > A, then C:B>A:B. 
 Next if 0=4, 
 
 cG = aG, 
 
 .'. c = a, 
 
 c _a 
 '" b~b' 
 
 .'. C:B = A:B. 
 .'. if C = A, then G:B = A:B.
 
 42] EUCLID, BOOKS V. AND VI. 23 
 
 Next if C<A, 
 
 cG < aG, 
 c<a, 
 
 c a 
 &<&' 
 
 .'. C:B<A:B. 
 
 Hence if C < A, then C:B<A:B. 
 There are therefore the three propositions 
 
 (1) If C>A, then C:B>A:B. 
 
 (2) If C = A, then C:B = A:B. 
 
 (3) If C<A, then C:B<A:B. 
 
 The meaning of these is easily grasped if they are written out with fewer 
 symbols. 
 
 Thus the second says that equal magnitudes have the same ratio to a third 
 magnitude. 
 
 The first says that if C be greater than A, then C has a greater ratio to B 
 than A has. 
 
 The third is not essentially distinct from the first. 
 
 The converse propositions to the above will next be proved. These are 
 
 (4) If C:B>A-.B, then C> A. 
 
 (5) If C:B = A:B, then C = A. 
 
 (6) If C:B<A:B, then C < A. 
 Take (4). 
 
 Suppose that C:B>A:B, 
 
 then one of the following alternatives must hold, 
 
 C>A, or C=A, or C<A. 
 
 Now if C = A, then by (2) C : B = A : B which is contrary to the hypothesis. 
 And if C < A, then by (3) C : B < A : B which is contrary to the hypothesis. 
 Hence C is not equal to A, nor less than A, and therefore G is greater than A. 
 Propositions (5) and (6) depend on (1), (2), (3) in the same way. 
 Consequently (4), (5) and (6) are logical deductions from (1), (2) and (3).
 
 24 EUCLID, BOOKS V. AND VI. [43 
 
 Art. 43. The proofs of (1), (2), (3) of the preceding article depend on the fact 
 that A, B, G are integral multiples of the same magnitude G. 
 
 It will be necessary however in what follows to deal with the ratios of 
 magnitudes which have no common measure. 
 
 Now the definition given for the ratio of magnitudes having a common measure 
 is not applicable to magnitudes which have no common measure. 
 
 Two alternatives therefore present themselves. 
 
 Either two magnitudes, which have no common measure, cannot be said to 
 have a ratio; 
 
 or the definition of ratio must be extended so as to apply to the case of two 
 magnitudes which have no common measure. 
 
 The latter alternative will be selected. 
 
 In the first place it is clear that the ratio of magnitudes having no common 
 measure is not expressible as a rational fraction. 
 
 Before it can be explained what is meant by the ratio of two magnitudes 
 which have no common measure, it is necessary to extend the idea of number, 
 by defining the irrational number. It will then appear that the ratio of two 
 magnitudes which have no common measure is an irrational number. This subject 
 is treated more at length in the Appendix, but here it will be enough to say that 
 the irrational number is a magnitude of the same kind (in the technical sense in 
 which these words are used in Art. 21) as the rational fraction. Hence it is 
 supposed to be possible to determine whether any irrational number is less than 
 or greater than any given rational number, whatever that given rational number 
 may be. Whenever this is possible the irrational number is considered to be 
 known. 
 
 In order to construct a theory of the ratios of magnitudes which have no 
 common measure, it will be assumed that the propositions 
 
 if C>A, then C:B>A:B; 
 if C = A, then C:B = A:B; 
 and if G < A, then G:B<A:B 
 
 hold good whether A, B, C have a common measure or not. 
 From these as explained in Art. 42, it follows 
 
 that if C: B > A : B, then G > A ; 
 if C:B = A :B, then C=A- 
 and if C : B < A : B, then C < A. 
 From the above propositions the following will be deduced.
 
 44] EUCLID, BOOKS V. AND VI. 
 
 Art. 44. PROPOSITION XIII. 
 
 To prove that 
 
 n 
 
 (1) // A : B > - -, then rA > nB. 
 
 T 
 
 (2) // A:B = -, then rA=nB. 
 
 T 
 
 (3) // A : B < - , then rA < nB. 
 And conversely, (4) If rA > nB, then A:B>-. 
 
 T 
 
 (5) // rA=nB, then A:B = -. 
 
 (6) // rA<nB, then A:B<-. 
 
 r 
 
 (These propositions are very important and will be used repeatedly in what 
 follows.) 
 
 To prove (1). It is assumed * that a magnitude Q exists such that B = rQ. 
 Further by definition nQ : rQ = - . 
 
 tit 
 
 Now, by hypothesis, A : B > - , 
 
 r 
 
 .'. A:rQ>nQ:rQ, 
 
 .'. A>nQ, [Art. 43. 
 
 /. rA > rnQ, 
 .-. rA >n(rQ), 
 .'. rA> nB. 
 To prove (2). Here, by hypothesis, 
 
 A:B = ^, 
 r 
 
 .-. A:B = nQ:rQ. 
 But B = rQ, 
 
 .'. A:rQ = nQ : rQ, 
 
 .'. A = nQ, [Art. 43. 
 
 .'. r A = r (nQ) = n (rQ) = nB. 
 
 * It is not necessary to be able to construct Q. All that is assumed is that Q exists. 
 H. E. 4
 
 26 EUCLID, BOOKS V. AND VI. [44 
 
 To prove (3). Here, by hypothesis, 
 
 n n 
 
 A :B< -, 
 
 r 
 
 .'. A:B<nQ:rQ, 
 .-. A:rQ<nQ:rQ, 
 
 .-. A<nQ, [Art. 43. 
 
 .'. r A < r(nQ), 
 .'. rA <n(rQ), 
 .'. r A < nB. 
 To prove (4). Here, by hypothesis, 
 
 rA > nB. 
 Take B = rQ, 
 
 .-. rA>n (rQ), 
 .'. rA >r(nQ), 
 .'. A>nQ', 
 .-. A:rQ>nQ: rQ, [Art. 43. 
 
 r W 
 
 .'. A :B>-. 
 r 
 
 To prove (5). Here, by hypothesis, 
 
 rA = nB, 
 
 .'. A:B = nQ:rQ = . [Art. 43. 
 
 To prove (6). Here, by hypothesis, 
 
 r A < nB. 
 .'. rA< n(rQ), 
 .'. rA<r(nQ), 
 
 /. A < nQ, 
 
 .'. A:rQ<nQ:rQ, [Art. 43. 
 
 .'. A:B<nQ:rQ; 
 
 r 
 
 Art. 45. It is now possible to find approximate values for the ratios of 
 quantities which have no common measure. 
 Let A and B have no common measure.
 
 46] EUCLID, BOOKS V. AND VI. 27 
 
 Assume that a quantity Q exists such that B = rQ. 
 
 Now A is not a multiple of Q, for if it were A and B would have a common 
 measure, Q. 
 
 Consequently some integer s exists such that sQ is less than A, but (*+ 1)Q 
 is greater than A. 
 
 Since A > sQ, 
 
 .-. A :B>sQ: B, [Art. 43. 
 
 .-. A:B>sQ:rQ, 
 
 /. A:B>-. 
 r 
 
 Since A < (s + 1) Q, 
 
 .'. A:B<(s+l)Q:B, [Art. 43. 
 
 .*. A:B<(s + l)Q:rQ; 
 
 D * + l 
 
 .'. A:B< . 
 
 r 
 
 Hence the ratio of A to B lies between 
 
 s , s + l 
 
 - and . 
 
 r r 
 
 The difference of these two fractions is - , which can be made as small as 
 
 r 
 
 we please, by sufficiently increasing r. This is possible, on the hypothesis made 
 that, B being given, and any integer whatever r taken, a quantity Q exists 
 
 such that 
 
 B = rQ. 
 
 In this way an approximate measure of the ratio of any two magnitudes 
 of the same kind can be obtained. 
 
 But it is not enough to decide the important question whether two given 
 ratios are or are not equal. 
 
 If each can only be measured approximately, it still remains possible for 
 their difference to be less than the degree of accuracy of measurement. 
 
 Art. 46. ON EQUAL RATIOS. 
 
 Two ratios are equal if no rational fraction lies between them. 
 Now if no rational fraction lies between A : B and C : D, then if any rational 
 
 S S 
 
 fraction whatever - be selected, and it be found that A : B is greater than - , 
 r r 
 
 o o 
 
 then C : D must also be greater than - ; if however A : B is equal to - , then 
 
 42
 
 28 EUCLID, BOOKS V. AND VI. [46 
 
 <j o 
 
 C : D must also be equal to - ; but if A : B is less than - , then C : D must 
 
 g 
 
 also be less than - . 
 r 
 
 These conditions must be satisfied for every value of the integers r, s. 
 The conditions, written more shortly are 
 
 If A : B > - , then must C:D>-: 
 r r 
 
 If A : B = - , then must C : D = - ; 
 r r 
 
 If A : B < - , then must C:D<-, 
 r r 
 
 whatever values the integers r, s may have. 
 
 By Prop. 13 these conditions are equivalent to the following. 
 If r, s be any integers whatever, and 
 
 if all values of r, s which make rA > sB, also make rC > sD (1); 
 
 if all values of r, s which make rA = sB, also make rC = sD (2); 
 
 if all values of r, s which make rA < sB, also make rC < sD (3), 
 
 then A: B = C :D. 
 
 These are the conditions given by Euclid in the Fifth Definition of the Fifth 
 Book for the equality of the ratios A : B and C : D. 
 
 The three sets of conditions given above may be called collectively the Test 
 for the Equality of two Ratios*. 
 
 Art. 47. They may be replaced by equivalent conditions in various waysf. 
 
 One of these is as follows: 
 
 If r, s be any integers whatever, and 
 
 if rC > sD, then must rA > sB ; (4) 
 
 i(rC = sD, then mustrvl=s.5; (5) 
 
 if rC<sD, then must r A < sB. (6) 
 
 To show that this is so, let it be supposed that (1), (2) and (3) hold. 
 
 If possible let some values of r, s exist, such that 
 
 rC > sD, but rA $ sB. 
 Then either r A = sB, or rA < sB. 
 
 * See Note 3. f See Prop. 14 and Examples 12 and 13 in Art. 50.
 
 48] EUCLID, BOOKS V. AND VI. 29 
 
 If rA = sB, then by (2) rC = sD, 
 and if rA < sB, then by (3) rC < sD, 
 
 both of which are contrary to the hypothesis that rC > sD. 
 Hence rA > sB, and .'. (4) holds. 
 In like manner (5) and (6) can be proved. 
 
 The most important result in regard to the conditions (1), (2) and (3) is this, 
 that (2) is included in (1) and (3)*. This is proved in Proposition 14. 
 
 Art. 48. PROPOSITION XIV. 
 
 To prove that if all values of r, s which make 
 
 rA > sB, also make rC>sD ........................... (1), 
 
 and if all values of r, s which make 
 
 rA<sB, also make rC<sD ........................... (3), 
 
 then if any values of r, s exist which make rA = sB, they also make rC = sD. 
 
 Suppose that when r = r 1} s = s 1} 
 
 rA = sB, 
 
 but 
 
 i.e. r^A = Sj5, 
 
 but either rfi> s^D 
 
 or r l C<s 1 D. 
 
 Suppose first r,C > ^D. 
 
 Then i\G s t D is a magnitude of the same kind C or D. 
 
 Hence by Archimedes' Axiom an integer n exists, such that 
 
 but r l A=s 1 B, 
 
 Now putting r = nr lt s = ns l + 1 in (3) it follows that as 
 
 < (ns l 
 
 * The first explicit mention of this result known to the author is in Stolz's Vorlesungen iiber Allge- 
 meine Arithmetik, Part i. p. 87, published in 1885. 
 
 t The proof may be completed by taking n (riC - siD)>C.
 
 30 EUCLID, BOOKS V. AND VI. [48 
 
 it is necessary to have 
 
 (ns 1 + 1) D, 
 which is contrary to what has been proved above that 
 
 nr l C>(ns l + l)D. 
 Hence r^C is not greater than s t D. 
 
 Take next the case 
 
 r l C<s l D. 
 
 Then s l D r^C is a magnitude of the same kind as G or D. 
 Hence by Archimedes' Axiom an integer n exists such that 
 
 n(s l D-r l C)>C*, 
 
 i.e. (r, + !)</< ns^. 
 
 But r^A =!#, 
 
 /. nr^A = ns l B, 
 .'. (nr l + 1) A > ns l B. 
 Now putting r = nr^ + 1, s = n! in (1), it follows that as 
 
 (nr-! + 1) A > nSiB, 
 
 then must (nr-j + 1) C > ns 1 D, 
 
 which is contrary to what has been shown above that 
 
 Hence r^C is not less than SjZ>. 
 
 It was proved before that 
 
 r,(7 is not greater than s l D. 
 
 Consequently rfi s l D. 
 
 It follows therefore that there is no need to consider the set of conditions (2). 
 In fact this set of conditions is never satisfied unless the magnitudes concerned 
 are commensurable. 
 
 For if r A = sB, and if a magnitude Q be taken such that B = rQ, then 
 
 rA =8(rQ) = r(sQ), 
 
 /. A=sQ. 
 Thus A and B have a common measure Q. 
 
 Although it is not necessary to consider the set of conditions (2), this will 
 nevertheless be done in many of the proofs which follow, because in the comparison 
 
 * The proof may be completed by taking n ($iD - r l C)>D.
 
 49] EUCLID, BOOKS V. AND VI. 31 
 
 of any ratio with any rational fraction there are always three alternatives to be 
 considered, and the consideration of the alternative corresponding to the condition 
 (2) is sometimes instructive, and does not involve any additional difficulty. 
 
 It follows therefore that in order to prove that A : B = C : D, it is sufficient 
 and necessary to shoiu that 
 
 if A:B>-, then C:D>-, 
 
 r r 
 
 s s 
 
 and if A : B < - , then C:D <-, 
 
 w *' * ' 
 
 whatever integers r, s may be. 
 
 Art. 49. UNEQUAL RATIOS. 
 
 In connection with Euclid's Test for Equal Ratios, it is interesting to consider 
 his test for distinguishing the greater of two unequal ratios from the smaller. 
 
 It is as follows: 
 
 If rA be greater than sB, 
 
 but rC be not greater than sD, 
 
 then A:B>C:D. 
 
 Now if rA > sB, 
 
 then A : B > - ; 
 
 r 
 
 and if r 
 
 then C:D$-. 
 
 r 
 
 
 
 Hence A : B is greater than - , 
 
 o 
 
 but C : D is not greater than - . 
 
 Consequently A : B is greater than C : D. 
 
 Notice that the conditions are equivalent to either 
 
 rA>sB, rC=sD, 
 
 or rA > sB, rC < sD, 
 
 each of which is inconsistent with the conditions laid down in the Test for Equal 
 Ratios.
 
 32 EUCLID, BOOKS V. AND VI. [50 
 
 Art. 50. EXAMPLES. 
 
 12.* If all values of r, s which make rA > sB, also make rC > sD, and if all values 
 of r, s which make rC > sD, also make rA > sB, prove that A : B - C : D. 
 
 13.* If all values of r, s which make rA < sB, also make rC < sD, and if all values 
 of r, s which make rC < sD, also make rA < sB, prove that A : B = C : D. 
 
 14. If for a single value of the integer r, say r lf and a single value of the integer s, 
 say 8 1) it is true that 
 
 r^A = !# 
 
 and r^G sfl, 
 
 then prove that any values of the integers r, s which make 
 
 (1) .rA>sB, also make rC>sD, 
 
 (2) rA = sB, also make rC = sD, 
 
 (3) rA < sS, also make rC < sD. 
 
 15. If rA<sB, 
 if rC>sE, 
 if tA > uB, 
 if tC < uF; 
 and if further A:B = C:D, 
 prove that E<D<F.
 
 SECTION IV. 
 
 PROPOSITIONS 1526. 
 
 THE SIMPLER PROPOSITIONS IN THE THEORY OF RATIOS 
 WITH GEOMETRICAL APPLICATIONS. 
 
 Art. 51. PROPOSITION XV. (Euc. V. 15.) 
 To prove that A : B = nA : nB. 
 
 
 
 Take any rational fraction - . 
 
 If A:B>-, 
 
 r 
 
 then rA > sB, [Prop. 13. 
 
 .-. n (rA) > n (sB) ; 
 i.e. r (nA) > s (nB) ; 
 
 .'. nA : nB > - . [Prop. 13. 
 
 Butif A:B<-> 
 
 r 
 
 then rA < sB, [Prop. 13. 
 
 /. n(rA)< n(sB), 
 i.e. r (nA ) < s (nB), 
 
 .'. nA:nB<-. [Prop. 13. 
 
 Hence it has been shown 
 
 A t 
 
 if A : B > - , then nA : nB > - , 
 r r 
 
 n O 
 
 but if A:B<-, then nA:nB<-. 
 
 r r 
 
 H. E.
 
 34 EUCLID, BOOKS V. AND VI. [51 
 
 Hence by Art. 48 
 
 A:B = nA:nB*. 
 
 The case of the above Proposition in which n = 2 will often be required, 
 
 i.e. A:B = 2A:2B. 
 
 Since n represents any whole number whatever, it may have an infinite number 
 of values. 
 
 Hence nA and nB represent an infinite number of pairs of magnitudes, e.g. 
 
 2A and 2J5, 3J. and SB, such that the ratio of any pair is the same as 
 
 that of A, B. 
 
 Hence there are an infinite number of pairs of magnitudes which have the 
 same ratio. 
 
 Hence if a ratio be given, the magnitudes of which it is the ratio are 
 not given. 
 
 Thus two magnitudes of the same kind determine a definite ratio ; but if a 
 ratio only be given, the magnitudes of which it is the ratio are not determined. 
 
 Art. 52. PROPOSITION XVI. (Euc. VI. 1.) 
 
 ENUNCIATION. The ratio of the areas of two parallelograms (or triangles} 
 which have the same altitude is equal to the ratio of the lengths of their bases. 
 
 Any two parallelograms which have the same altitude may be placed so as to 
 lie between the same parallels. 
 
 P W X Y ZC H B L G 
 
 Fig. 12. 
 Let A BCD, EFGH be two parallelograms lying between the same parallels. 
 
 * If A and B are numbers, then denoting numbers by small letters it follows that 
 
 a : 6 = na : nb.
 
 52] EUCLID, BOOKS V. AND VI. 35 
 
 It is required to prove that 
 
 AB:EF= ABCD : EFGH. 
 
 o 
 
 Take any rational fraction - . 
 
 r 
 
 The ratio AB : EF will be compared with this fraction. 
 
 Cut EF into r equal parts EP, PQ, QF. 
 
 Set off s consecutive lengths AS, ST, TU, UV each equal to EP along AB. 
 
 Then AV = s(EP). 
 
 Then three alternatives are possible : 
 
 (i) AB>AV; 
 
 (ii) AB = AV\ 
 
 (iii) AB<AV. 
 
 Through P, Q draw parallels to EH, FG ; and through 8, T, U, V draw 
 parallels to AD. 
 
 Then the parallelograms EPRH, PQLR, QFGL are equal, and the parallelo- 
 gram EFGH is equal to r times the parallelogram EPRH. 
 
 In like manner, 
 
 A VZD = s (ASWD) = s (EPRH). 
 
 Now take case (i), (Fig. 12), AB>AV. 
 
 .-. AB:EF>AV:EF [Art. 43. 
 
 >s(EP):r(EP) 
 
 but since AB > AV, 
 
 .'. ABCD>AVZD; 
 
 .-. ABCD: EFGH > AVZD : EFGH [Art. 43. 
 
 > s (EPRH) :r (EPRH) 
 
 s 
 
 >-. 
 r 
 
 It has therefore been shown that, 
 
 if AB:EF>-, 
 
 r 
 
 then ABCD: EFGH >-. 
 
 r 
 
 52
 
 36 
 
 EUCLID, BOOKS V. AND VI. 
 AB = AV. 
 
 [52 
 
 Next take case (ii), 
 In this case F falls on B, 
 and the parallelograms AVZD, A BCD are the same, 
 
 V AB = AV, 
 .-. AB:EF=AV:EF 
 
 = s(EP):r(EP) 
 
 [Art. 43. 
 
 Also 
 
 ABCD : EFOH = A VZD : EFOB 
 
 = s(EPRH):r(EPRH) 
 
 [Art. 43. 
 
 It has therefore been shown that, 
 
 if 
 
 then 
 
 AB:EF=-, 
 r 
 
 ABCD:EFGH = -. 
 r 
 
 Lastly take the case (iii), (Fig. 13), AB < AV. 
 In this case the figure has the following form. 
 D W X Y KG M 2 
 
 B J 
 
 Here 
 
 Fig. 13. 
 
 AB<AV, 
 AB:EF<AV:EF 
 
 <s(EP):r(EP) 
 
 s 
 < 
 
 r 
 
 Q 
 
 [Art. 43.
 
 53] EUCLID, BOOKS V. AND VI. 37 
 
 But since AB <AV t 
 
 .-. ABCD<AVZD; 
 
 . \ ABCD : EFGH < A VZD : EFGH [Art. 43. 
 
 <s(EPRH):r(EPRH) 
 
 s 
 
 < - . 
 r 
 
 It has therefore been shown that 
 
 if AB : EF < - , then ABCD : EFGH < - : 
 r r 
 
 and it was previously shown that 
 
 if AB:EF=-, then ABCD : EFGH =-, 
 r r 
 
 but if AB : EF > - , then ABCD : EFGH > - . 
 
 r r 
 
 Putting these results together it follows by Art. 46 that 
 AB:EF= ABCD : EFGH. 
 
 Art. 53. EXAMPLES. 
 
 16. Given two rectilineal areas, and a straight line, find another straight line such 
 that the ratio of the areas is the same as that of the lines. 
 
 17. Given two straight lines, and a rectilineal area, find another rectilineal area 
 such that the ratio of the lines is the same as that of the areas. 
 
 18. Given three rectilineal areas, find a fourth such that the ratio of the first to 
 the second area is the same as that of the third to the fourth. 
 
 19. Prove that the ratio of the areas of two triangles on equal bases is the same as 
 the ratio of their altitudes. 
 
 20. If ABC be a triangle, and any point in its plane, and if AO cut EC at D, 
 prove that 
 
 BD :DC=&AOB : &AOC. 
 
 445234
 
 38 EUCLID, BOOKS V. AND VI. [54 
 
 Art. 54. PROPOSITION XVII. (Containing the first part of Euc. VI. 2.) 
 
 ENUNCIATION. If two straight lines be cut by any number of parallel straight 
 lines, to prove that the ratio of any two segments of one line is equal to that of the 
 corresponding segments of the other line. 
 
 Let the intersecting lines OX, OF be cut by the parallel straight lines AB, 
 CD, EF, GH. 
 
 Then the segment AC corresponds to the segment ED, 
 and the segment EG corresponds to the segment FH. 
 
 It is required to prove that AC:EG = BD: FH. 
 
 Fig. 14. 
 
 Take any rational fraction - . 
 
 r 
 
 Divide EG into r equal parts, EK, KL, LG. 
 
 From A set off s consecutive lengths, each equal to EK, along A C. 
 
 Let AQ = s(EK\ 
 
 Three cases are possible : 
 
 (i) AQ<AC, 
 
 (ii) AQ = AC, 
 
 (iii) AQ>AC.
 
 54] EUCLID, BOOKS V. AND VI. 39 
 
 (i) Through the points of division K, L, W, Q draw parallels to AB. 
 Then by Art. 37 FM = MN=NH = BP = PR, 
 
 In figure 14 AC > AQ, 
 
 .-. AC:EG>AQ: EG [Art. 43. 
 
 >s(EK):r(EK) 
 
 s 
 r' 
 
 Again from the figure, 
 
 since AG>AQ, 
 
 and AB, QR, CD are parallel, 
 
 .-. BD > BR, 
 
 /. BD:FH>BR: FH [Art. 43. 
 
 >s(BP):r(BP) 
 
 s 
 r' 
 
 It is therefore proved that 
 
 if AC:EG>-, 
 
 then BD:FH>-. 
 
 r 
 
 (ii) In this case Q falls on C, 
 and therefore R falls on D, 
 
 AG = s (EK), EG = r (EK\ 
 
 BD = s (BP\ FH = r (BP); 
 
 .-. AC:EG = - 
 r 
 
 and BD:FH = -. 
 
 r 
 
 It is therefore proved that if AC : EG = - , then BD:FH = ~.
 
 40 EUCLID, BOOKS V. AND VI. 
 
 (iii) In this case (Fig. 15) AC < AQ, 
 
 .-. AC:EG<AQ:EG 
 
 <s(EK):r(EK} 
 
 [54 
 [Art. 43. 
 
 In this case 
 
 and 
 
 AC<AQ, 
 AB, QR, CD are parallel, 
 
 /. BD<BR, 
 .-. BD:FH<BR:FH 
 
 <s(BP):r(BP) 
 
 < -. 
 r 
 
 Fig. 15. 
 
 
 
 It is therefore proved that if AC: EG < - , 
 
 \ 
 
 then 
 
 BD:FH<-. 
 r 
 
 [Art. 43. 
 
 Putting the three cases together, it follows by Art. 46 that 
 
 AG:EG=BD:FH. 
 
 Art. 55. COROLLARY. 
 
 As a particular case of the preceding, it follows that if ABC be a triangle, 
 and if the sides AB, AC be cut by any straight .line parallel to BC, then the 
 sides AB, AC are divided proportionally.
 
 56] EUCLID, BOOKS V. AND VI. 
 
 Let DE, parallel to BC, cut AB at D and AC at E. 
 There are three varieties of figure. 
 
 B 
 
 \ 
 
 Fig. 16. 
 The point A corresponds to the point A 
 
 " 
 
 7) ff 
 
 >y a " 
 
 The segment AB segment AC 
 
 AD AE 
 
 D 7"l f^V 
 
 nu u/Sr. 
 
 From the above, by means of Proposition 17, the following results may be 
 deduced, the first being the one most frequently required. 
 
 AD:DB = AE:EC. 
 DB:AD=EG:AE. 
 AB:AD=AC:AE. 
 AD:AB=AE:AC. 
 A D . j)T) AC 1 ' C^T? 
 
 These six results are not independent. 
 
 Any one of them being given, the rest follow as consequences by means of the 
 properties of ratio as will be seen hereafter. 
 
 Art. 56. PROPOSITION XVIII. (Euc. VI. 12.) 
 
 ENUNCIATION. To find a fourth proportional to three given straight lines'*. 
 Let AB, CD, EF be three straight lines. 
 
 * My attention was called to the construction here given by the late Mr Budden. 
 H. E. 6
 
 42 EUCLID, BOOKS V. AND VI. [56 
 
 It is required to find a straight line OL, such that AB:CD = EF : OL. 
 Then OL is called the fourth proportional to AB, CD, EF. 
 
 Y 
 
 Fig. 17. 
 
 Let OX, OF be two straight lines intersecting at 0; measure off on OX in 
 opposite directions OG = AB, OH= CD. 
 On OF measure off OK=EF. 
 
 Join GK, and draw HL parallel to GK, meeting OF at L. 
 Then the intersecting lines OX, OF are cut by the parallels GK, HL. 
 The points G, 0, H correspond to K, 0, L respectively. 
 The segments GO, OH correspond to KO, OL respectively. 
 
 /; GO:OH = KO: OL, [Prop. 17. 
 
 .-'. AB:CD = EF:OL. 
 
 Art. 57. EXAMPLE 21. 
 
 Given three straight lines AB, CD, EF, it is required to construct another 
 straight line P such that 
 
 P :AB = CD-.EF, 
 
 or AB : P = CD : EF, 
 
 or AB:CD= P : EF. 
 
 Art. 58. DEFINITION 8. 
 
 If in the figure of proposition 18 
 
 CD = EF,
 
 61] EUCLID, BOOKS V. AND VI. 43 
 
 then OL is such that 
 
 AB:CD = CD:OL, 
 
 and OL is called the third proportional to AB and CD. 
 
 Art. 59. EXAMPLE 22. 
 
 In the triangle ABC, a straight line DE is drawn parallel to EC cutting AB at 
 D and AC at E. DF ia drawn parallel to BE, cutting AC at F. 
 
 Prove that AF is a third proportional to AC and AE. 
 
 Art. 60. Def. 9. SIMILARLY DIVIDED STRAIGHT LINES. 
 
 Two straight lines are said to be similarly divided, when the ratio of 
 any two parts of one straight line is the same as that of the two corre- 
 sponding parts of the other straight line. 
 
 Art. 61. PROPOSITION XIX. (Euc. VI. 10.) 
 
 ENUNCIATION. To divide a straight line similarly to a given divided straight 
 line. 
 
 It is required to divide the given straight line AB similarly to the way in 
 which the line CF is divided at D and E. 
 
 Through A draw any straight line AX (not in 
 the same straight line as AB), and on it measure 
 off 
 
 Join BK, and draw GL, HM parallel to BK 
 cutting AB in L, M respectively. 
 
 Then since GL, HM, BK are parallel lines, the 
 segments AG, AL correspond; so do GH, LM; 
 and HK, MB. 
 
 :. AL:LM = AG:GH 
 
 = CD:DE. 
 
 Also LM:MB=GH:HK 
 
 = DE:EF, 
 
 and so on. 
 
 Hence AB is divided similarly to CF. 
 
 L M B 
 
 Fig. 18. 
 
 [Prop. 17. 
 [Prop. 17. 
 
 62
 
 44 EUCLID, BOOKS V. AND VI. [62 
 
 Art. 62. PROPOSITION XX. (Euc. VI. 33.) 
 
 ENUNCIATION. In the same circle or in equal circles 
 
 (i) angles at the centre are proportional to the arcs on which they stand, 
 
 (ii) angles at the circumference are proportional to the arcs on which they 
 stand. 
 
 (iii) angles at the centre are proportional to the sectors bounded by the sides of 
 the angles and the arcs on which they stand. 
 
 If the angles are in the same circle, the figure may be drawn twice over, so 
 that it is sufficient to consider the case where there are two equal circles. 
 
 (i) Let A, B, Figs. 19 24, be the centres of two equal circles. 
 
 Let CA D, EBF be two angles at the centres standing on the arcs CD, EF. 
 
 It is required to prove that 
 
 CAD : EBF = arc CD : arc EF. 
 
 o 
 
 Let the fraction - be compared with the ratio 
 
 CAD : E&F. 
 There are three alternatives : 
 
 (1) CAD:EBF>-; 
 
 (2) CAD:EBF=-; 
 
 r 
 
 (3) CAD:EBF<-. 
 
 r 
 
 Make GAG equal to r(CAD), then since equal angles at the centre of a 
 circle stand on equal arcs it follows that 
 
 the arc CO =r(arc CD). 
 
 Make next EBH equal to s(E$F), then 
 
 the arc EH = s (arc EF). 
 In case (i), see Figs. 19 and 20. 
 
 v CAD:EBF>-, 
 r 
 
 r(CAD)>s( EBF ) ; [Prop. 1 3.
 
 62] 
 
 EUCLID, BOOKS V. AND VI, 
 
 .-. CAG>EBH; 
 .'. arc CO > arc EH* ; 
 
 G 
 
 45 
 
 .-. r(arc CD) > s(arc EF) ; 
 
 /. arc CD :&rcEF>-. 
 r 
 
 [Prop. 13. 
 
 In case (ii), see Figs. 21 and 22. 
 
 CAD:EBF=-; 
 r 
 
 [Prop. 13. 
 
 = *TC EH 
 
 .*. &rcCD:arcEF=- . 
 r 
 
 [Prop. 13. 
 
 This follows immediately from the proposition that in equal circles equal angles at the centres 
 stand on equal arcs.
 
 EUCLID, BOOKS V. AND VI. 
 
 In case (iii), see Figs. 23 and 24. 
 
 CAD:EBF<-; 
 
 T 
 
 .'. r(CAD)<s(EBF)\ 
 
 [62 
 
 .-. arc CO < arc EH* ; 
 .-. r(arc CD) < a (arc EF) 
 
 .'. arc CD : arc EF < - . 
 r 
 
 [Prop. 13. 
 
 [Prop. 13. 
 
 Hence 
 
 if CAD: EBF>- , 
 
 then arc CD : arc EF > - ; 
 
 r 
 
 if CAD :EBF=-, 
 r 
 
 then arc CD : arc EF = - : 
 r 
 
 and if CAD:EBF<-, 
 r 
 
 then arc CD : arc EF < - . 
 r 
 
 .'. CAD : EBF= arc OD : arc EF. 
 
 [Art. 46. 
 
 * This follows immediately from the proposition that in equal circles equal angles at the centres 
 stand on equal arcs.
 
 65] EUCLID, BOOKS V. AND VI. 47 
 
 (ii) An angle at the centre of a circle is double the angle at the circumference 
 standing on the same arc. 
 
 Hence the ratio of two angles at the centre of the same or of equal circles is 
 the same as that of the angles at the circumference on the same arcs. [Art. 51. 
 
 Hence, by case (i), the ratio of two angles at the circumference of the same or 
 of equal circles is the same as that of the arcs on which they stand. 
 
 (iii) The proof of this is derivable from that of (i) by replacing therein each 
 arc by the corresponding sector. 
 
 Art. 63. Def. 10. RECIPROCAL RATIOS. 
 The ratios A : B and B : A are called reciprocal ratios. 
 
 Art. 64. EXAMPLE 23. 
 
 If two reciprocal ratios are equal, prove that each of them is a ratio of equality. 
 
 Art. 65. PROPOSITION XXI. (Corollary to Euc. V. 4.)* 
 
 ENUNCIATION. If two ratios are . equal their reciprocal ratios are equal, 
 i.e. if A:B=C:D, 
 
 to prove that B : A = D : C. 
 
 a 
 
 Compare the ratio B : A with the rational number - . 
 
 By Art. 48 it is necessary to consider only the alternatives 
 
 (1) B:A> S -, 
 
 (2) B:A< 8 -. 
 
 m 
 
 In case (1) B : A > - ; 
 
 r 
 
 .'. rB>sA; [Prop. 13. 
 
 .-. sA < rB\ 
 
 .-. A : B< r -\ [Prop. 13. 
 
 s 
 
 but A : B = C : D ; 
 
 V 
 
 .'. C:D<-\ 
 
 8 
 
 .'. sC<rD\ [Prop. 13. 
 
 * Simson numbers this proposition Euc. V. B.
 
 48 EUCLID, BOOKS V. AND VI. [65 
 
 /. rD>sC; 
 .'. D: G> S -. [Prop. 13. 
 
 In case (2) B : A < - ; 
 
 .-. rB<sA\ [Prop. 13. 
 
 .'. sA 
 
 .'. A :>-; [Prop. 13. 
 
 s 
 
 but A : B = C : D ; 
 
 .-. C:D>-; 
 s 
 
 .'. sG>rD\ [Prop. 13. 
 
 .-. rD<sG- y 
 
 .'. D:C<^. [Prop. 13. 
 
 Hence if B : A > - , then D:C>-; 
 
 r r 
 
 but if B : A < - , then D : C < - . 
 
 r r 
 
 .'. B: A=D:C. [Art. 48. 
 
 Art. 66. PROPOSITION XXII. (Euc. V. 7, 2nd Part.) 
 
 ENUNCIATION. If A, B, C be three magnitudes of the same kind, and if A be 
 equal to B, then 
 
 C:A = C:B. 
 
 If A be equal to B, then by Art. 43 
 
 A :C = B:C; 
 /. C:A=C:B. [Prop. 21. 
 
 Art. 67. PROPOSITION XXIII. (Euc. V. 9, 2nd Part.) 
 
 ENUNCIATION. If A, B, C be three magnitudes of the same kind, and if 
 
 C:A = C:B, 
 then A=B. 
 
 Now if C:A = C:B, 
 
 then A:C = B:C; [Prop. 21. 
 
 .'. A = B. [Art. 43.
 
 69] EUCLID, BOOKS V. AND VI. 49 
 
 Art. 68. EXAMPLE 24. 
 
 (i) If rA : B = sA : C, prove that sB = rC. 
 (ii) If A:rC = B:sC t prove that sA = rB. 
 
 Art. 69. PROPOSITION XXIV. (Euc. V. 16.)* 
 
 ENUNCIATION. If A, B, C, D be four magnitudes of the same kind, and if 
 A:B=C:D, to prove that A:C=B:D. 
 
 o 
 
 Compare the ratio A : C with any rational fraction - . 
 
 By Art. 48 it is necessary to consider only the two alternatives 
 
 (1) A ' ' l '""' ' 
 
 (2) A:C<*-. /. 
 
 o 
 
 In case (1), A : C > - , 
 
 rA > sC. [Prop. 13. 
 
 Hence rA sC is a magnitude of the same kind as A, B, C, D. 
 Comparing it with 5f, then by Archimedes' Axiom (Art. 22) an integer n 
 
 exists such that 
 
 n(rA-sC)>B, 
 
 .'. nrA nsC > B, 
 .-. nrA >nsC+B. 
 Hence a multiple of B, say tB, exists such that 
 
 nrA >tB> nsC; [Prop. 8. 
 
 V nrA >tB, 
 
 .'. A : B > , [Prop. 13. 
 
 nr 
 
 but A:B = C:D, 
 
 .'. C:D>-, 
 nr 
 
 .-. nrC>tD (I), [Prop. 13. 
 
 * See Note 6. t The reader should complete the proof by taking D instead of B. 
 
 H. E. 7
 
 50 EUCLID, BOOKS V. AND VI. [69 
 
 but tB > nsC (II), 
 
 /. snrC>stD from (I), 
 
 and rtB > rnsC from (II), 
 
 .*. rtB>stD, 
 :. rB>sD*, 
 
 .'. B:D>-. [Prop. 13. 
 
 r 
 
 o 
 
 In case (2), A : C < - , 
 
 rA < sC. [Prop. 13. 
 
 Hence sC rA is a magnitude of the same kind as A, B, C, D. 
 
 Comparing it with J5f, then by Archimedes' Axiom an integer n exists such 
 
 that 
 
 n(sC-rA)>B, 
 
 :. nsC >nrA + B, 
 
 Hence (by Prop. 8) a multiple of B, say tB, exists such that 
 
 nsC >tB> nrA. 
 V nrA <tB, 
 
 .-. A:B< , [Prop. 13. 
 
 nr 
 
 but A:B = C:D, 
 
 /. C:J)< , 
 nr 
 
 /. nrC<tD (III), [Prop. 13. 
 
 but nsG > tB (IV), 
 
 /. snrC<stD from (III), 
 rnsC > rtB from (IV), 
 .'. rtB<stD, 
 .*. rB < sD, 
 
 :. B-.D< 8 -. [Prop. 13. 
 
 * This result is an algebraic consequence of the inequalities (I and II). It is obtained by trans- 
 forming them so that the multiple of C (which is the magnitude appearing in both) becomes the same in 
 each. 
 
 t The reader should complete the proof by taking D instead of B.
 
 71] EUCLID, BOOKS V. AND VI. 51 
 
 Hence it has been proved that 
 
 if A:C>-, then B:D>-: 
 
 r r 
 
 andif A:C<-, then B:D<-. 
 
 r r 
 
 .'. A:G = B:D. [Art. 48. 
 
 Art. 70. COROLLARY. (Euc. V. 14.) 
 
 If A, B, C y D are all magnitudes of the same kind, and if 
 
 A:B = C:D, 
 then A = C, according as B = D, and conversely. 
 
 Since A :B = C: D, and the magnitudes are all of the same kind, 
 
 .-. A:C=B:D. [Prop. 24. 
 
 Compare now the ratio B : D with the rational fraction - . 
 
 If B:D>, 
 
 then B > D. [Prop. 13. 
 
 Since B:D = A:C, 
 
 /. A:C>\, 
 
 .'. A>C. [Prop. 13. 
 
 The other cases may be proved in the same way. 
 
 Art. 71. EXAMPLES. 
 
 25. If A:B = G:D, 
 if E : C = F : A, 
 if E : D = F : G, 
 
 and if the magnitudes A, B, C, D, E, F, G are all of the same kind, prove that B = G. 
 
 [The proposition is also true if A, B, F, G are of the same kind, and if C, D, E are 
 of the same kind, as may be proved by using Prop. 37 below.] 
 
 26. If A, B, C, D are four points on a straight line such that B divides AC 
 internally in the same ratio as D divides it externally; prove that C divides BD 
 internally in the same ratio as A divides it externally. 
 
 72
 
 52 
 
 EUCLID, BOOKS V. AND VI. 
 
 [72 
 
 Art. 72. Def. 11. HARMONIC POINTS. 
 
 Four points A, B, C, D on a straight line are said to be four harmonic 
 points if B and D divide AC in the same ratio, one internally and the 
 other externally. 
 
 Then A and C are called conjugate points; as are also B and D. 
 
 Art. 73. PROPOSITION XXV. (Euc. VI. 2, 2nd Part.) 
 
 ENUNCIATION. If two sides of a triangle are divided proportionally so that 
 the segments terminating at the vertex common to the two sides correspond to 
 each other, then the straight line joining the points of division is parallel to the 
 other side. 
 
 Let the points D and E divide the sides AB, 
 AC of the triangle A BC, so that 
 
 AD: DB = AE:EC, 
 then will DE be parallel to BC. 
 
 If DE be not parallel to BC, draw BF parallel 
 to DE cutting AC at F. 
 
 AD:DB = AE: EF, [Prop. 17. 
 /. AE:EC = AE:EF, 
 
 .'. EC=EF, [Art. 43. 
 
 which is impossible. 
 
 Hence DE is parallel to BC. 
 
 Fig. 25. 
 
 Art. 74. EXAMPLE 27. 
 
 If ABCD be a plane quadrilateral, and if E, F,G,Hbe points on AB, BC, CD, DA 
 
 respectively such that 
 
 AE : AB = CF:CB = CG :CD = AH :AD, 
 prove that EFGH is a parallelogram.
 
 75] EUCLID, BOOKS V. AND VI. 53 
 
 Art. 75. PROPOSITION XXVI. 
 
 ENUNCIATION. (1) A given segment of a straight line can be divided 
 internally into segments having the ratio of one given line to another in one 
 way only. 
 
 (2) A given segment of a straight line can be divided externally into 
 segments having the ratio of one given line to any other not equal to it in one 
 way only. 
 
 (1) Let AB be the straight line to be divided internally at some point C, 
 so that 
 
 where K, L are two given segments of straight lines. 
 
 Through A, one of the extremities of AB, draw any straight line AX, and 
 measure off AD equal to K, and DE equal to L in the same direction as AD. 
 
 K- 
 
 L- 
 
 Join BE, and through D draw DC parallel to EB, cutting AB at C. Then 
 C will divide AB, so that AC:CB = K:L. 
 
 Since AB, AX are cut by the parallel lines CD, BE; the segments AC, CB 
 correspond respectively to AD, DE. 
 
 .-. AC : CB = AD : DE [Prop. 17. 
 
 = K:L. 
 
 Hence C is one point which satisfies the required condition. 
 
 If possible let F be some other point which also satisfies the required 
 condition. 
 
 Join FD, and draw BQ parallel to FD cutting AE at G. 
 
 * See Note 7.
 
 54 EUCLID, BOOKS V. AND VI. [75 
 
 Then the segments AF, FB correspond to AD, DO respectively. 
 
 .'. AF:FB = AD:DG, [Prop. 17. 
 
 but by hypothesis AF : FB = K : L. 
 
 .'. AD:DG = K:L 
 
 = AD:DE. 
 
 .: DG = DE, [Art. 43. 
 
 which is impossible. 
 
 Hence C is the only point which satisfies the required condition. 
 
 It is important to notice that if K<L, then AC <CB, and C is nearer to 
 A than to B. 
 
 If K = L, then AC=CB, and is the middle point of AB. 
 
 If K>L, then AC>CB, and C is further from A than from B. 
 
 These three cases correspond to different figures in the second part of the 
 proposition. 
 
 (2) In this case the figures differ from that of the first case in having 
 the length DE (equal to L) measured in the opposite direction to AD; and 
 that is the only difference in the constructions for the cases K < L, K > L. 
 
 Fig. 27. 
 
 If K<L, then E must fall on DA produced through A as in Fig. 27, and 
 C is nearer to A than to B. 
 
 If K> L, then E must fall between D and A, as in Fig. 28, and C is further 
 from A than from B. 
 
 The proofs for the cases K < L and K > L are the same as in Case (1). They 
 need not therefore be repeated.
 
 76] 
 
 EUCLID, BOOKS V. AND VI. 
 
 55 
 
 A 
 
 Fig. 28. 
 
 Art. 76. NOTE ON CASE (2) OF THE PRECEDING ARTICLE. 
 
 If K = L, E coincides with A. 
 Hence BE coincides with BA. 
 Hence the parallel through D to BE is parallel to BA, as in Fig. 29. 
 
 Hence in this case from Euclid's point of view the construction fails, and there 
 is no point corresponding to C. See Note 7. 
 
 K=L 
 
 Fig. 29.
 
 SECTION V. 
 
 PROPOSITIONS 2734. 
 SIMILAR FIGURES. 
 
 Art. 77. Def. 12. 
 
 Similar rectilineal figures are those which satisfy the following two sets 
 of conditions. 
 
 (1) The angles of one of the figures taken in order must be respectively 
 equal to the angles of the other figure taken in order. 
 
 (2) Those sides in the two figures which join the vertices of equal 
 angles being defined as corresponding sides, the ratio of any pair of 
 corresponding sides must be equal to the ratio of every other pair of 
 corresponding sides. 
 
 Let A^B l C l DiE l , AJ$$J)JEi be two similar figures, then the two sets of 
 conditions are as follows : 
 
 (1) 
 
 I = BI,
 
 81] EUCLID, BOOKS V. AND VI. 57 
 
 (As there is in this figure only one angle at each vertex it is sufficient to 
 indicate each angle by the letter standing at its vertex.) 
 
 It is often convenient to indicate the equality of two angles in two similar 
 rectilineal figures by marking the equal angles with the same number, e.g. in the 
 above figures the equal angles at A 1} A. 2 are both marked 1. 
 
 (2) Since A, = A 3 , 
 
 and 5, = B z , 
 
 .'. the side A l B l corresponds to the side A 2 B 2 . 
 
 In like manner B l G l corresponds to B 2 C 2 , and so on. 
 
 Art. 78. Def. 13. 
 
 The ratio of a side of the first figure to the corresponding side of 
 the second figure is called the ratio of similitude of the first figure to the 
 second. 
 
 Art. 79. Note. It is obvious that two congruent figures are similar to one 
 another. For such figures the ratio of similitude is the ratio of equality. 
 
 Art. 80. Similar figures are said to be similarly described on two straight 
 lines, when these two straight lines are corresponding sides of the figures, e.g. : 
 
 The figures A^^C^E^ A 2 B 2 C 2 D 2 E 2 are similarly described on A&, A 2 B 2 ; 
 or on .BiCu B 2 C Z ; and so on. 
 
 (In general language two similar figures are similarly described on two straight 
 lines to which they have the same relation.) 
 
 Art. 81. EXAMPLE 28. 
 
 If B be the middle point of AC, and BX, CY be drawn perpendicular to AC, and 
 if A be joined to any point P on BX, and if on AP on the side remote from B a 
 triangle APQ be described similar to ABP so that the sides AP, PQ of APQ may 
 correspond to the sides AB, BP of ABP, prove that Q is equidistant from A and from 
 the straight line CY. 
 
 H. E. 8
 
 58 
 
 EUCLID, BOOKS V. AND VI. 
 
 [82 
 
 Art. 82. PROPOSITION XXVII. (Euc. VI. 21.) 
 
 ENUNCIATION. Rectilineal figures which are similar to the same rectilineal 
 figure are similar to one another. 
 
 Let the figure A BCD be similar to A^CJ)^ and also to A$J)^ it is 
 required to prove that A-^E-fj-J)^ and AJB^CJ)^ are similar to each other. 
 
 A D 
 
 [2 3J 
 
 B C 
 
 4 
 
 2 3 
 
 'i 
 
 Fig. 31. 
 
 Since A BCD is similar to A^C^ 
 
 .'. A=A 1} B = B lt C=C 1} = A (!) 
 
 Since A BCD is similar to AJ$ 2 CJ) Z 
 
 . j _ D D r* f* T\ D fQ\ 
 
 . . .il -tl -j, -D -*-'2> ^ ~~ '-^2> -^ -*-^2 ' \*^/ 
 
 From (1) and (3) it follows that 
 
 ^i == A 2) B! = B 2 , Q = C 2 , D! = D 2 (5). 
 
 From (2) AB: A.B^BC : B&, 
 
 .-. AB : BC = -4^ : BjC^ [Prop. 24. 
 
 Similarly from (4) AB : BC = A 2 B 2 : B Z C* 
 
 A T) . T) fi A T) . T) Ci 
 
 . . -/XJ.DJ . XJjl^i ./Ig-Og ^'2^2 
 
 . . A 1 B 1 :A 2 B 2 = B& : B 2 C 2 . [Prop. 24. 
 
 In like manner it can be shown that 
 
 . . A & : A 2 B 2 = B& : B 2 C 2 = C l A : CJ) t = A^i : D*A* (6). 
 
 Now (5) and (6) are the two sets of conditions which must be satisfied in 
 order that A 1 B 1 C 1 D 1 and A 2 B 2 C 2 D 2 may be similar (see Art. 77). 
 
 Hence A 1 B 1 C 1 D 1 and A^B^C 2 D Z are similar.
 
 85] EUCLID, BOOKS V. AND VI. 59 
 
 Art. 83. ON SIMILAR TRIANGLES. 
 
 The different cases, in which two triangles are similar, correspond to some 
 extent to the cases in which two triangles are congruent. 
 
 For this reason the cases in which two triangles are congruent will first be 
 enumerated. 
 
 Art. 84. Two triangles are congruent if 
 
 (1) The three sides of one triangle are respectively equal to the three sides 
 of the other triangle. 
 
 (2 a)* Two sides and the included angle of one triangle are respectively 
 equal to two sides and the included angle of the other triangle. 
 
 (3a)f Two angles and the adjacent side of one triangle are respectively 
 equal to two angles and the adjacent side in the other triangle. 
 
 (36)f One side, the opposite angle, and one other angle of one triangle are 
 respectively equal to one side, the opposite angle, and one other angle in the 
 other triangle. 
 
 Besides the above cases there should be noted the following, in which three 
 elements (sides or angles) of one triangle are respectively equal to the three 
 corresponding elements of the other triangle, viz. those in which 
 
 (26)* One angle, the opposite side, and one other side of one triangle are 
 respectively equal to one angle, the opposite side and one other side of the 
 other triangle. 
 
 In this case the angles opposite the other pair of equal sides are either equal 
 or supplementary, and in the former alternative the triangles are congruent. 
 
 (This case is usually known as the Ambiguous Case.) 
 
 (4) Three angles of one triangle are respectively equal to three angles of the 
 other triangle. 
 
 This last case is only mentioned in order to complete all the possible cases 
 in which three elements of one triangle are respectively equal to the three 
 corresponding elements of another triangle. In it the triangles are not generally 
 congruent, but are always similar (see Prop. 28). 
 
 Art. 85. To case (1) above corresponds the proposition that if the sides of 
 one triangle taken in order are proportional to the sides of another triangle 
 taken in order, then the triangles are similar. 
 
 * The numbers attached to the cases (2 a) and (2 I) both contain the same number 2 because 
 in each there r.re two sides and one angle given equal. 
 
 t The numbers attached to the cases (3 a) and (3 b) both contain the same number 3 because 
 in each there are two angles and one side given equal. 
 
 82
 
 60 EUCLID, BOOKS V. AND VI. [85 
 
 To case (2 a) corresponds the proposition that if two sides of a triangle are 
 proportional to two sides of another triangle, and the included angles are equal, 
 then the triangles are similar. 
 
 To case (2 b) corresponds the proposition that if two triangles have one angle 
 of the one equal to one angle of the other, and the sides about one other angle 
 proportional in such a manner that the sides opposite the equal angles correspond, 
 then the triangles have their remaining angles either equal or supplementary, 
 and in the former case the triangles are similar. 
 
 To cases (3 a), (36) and (4), in all of which the three angles of the one 
 triangle are respectively equal to the three angles of the other triangle, corre- 
 sponds the single proposition that if the angles of one triangle are respectively 
 equal to the angles of another triangle, then the triangles are similar. 
 
 Hence there are four cases of similar triangles to be dealt with. 
 
 It should be noticed that the first and last amount to the proposition that, in 
 the case of triangles, if either of the two sets of conditions for the similarity of 
 rectilineal figures be satisfied, then the other set must also be satisfied. 
 
 So that the two sets of conditions for the similarity of rectilineal figures 
 are not independent when the rectilineal figures are triangles. 
 
 Art. 86. In dealing with similar triangles the reader will find it useful 
 to draw the similar triangles separately if 
 they happen to overlap, and to mark equal 
 angles with the same numbers, as in the 
 figure. 
 
 Then those sides which join the equal 
 angles have the same numbers at their ex- 
 tremities, and it is therefore at once evident 
 that they are corresponding sides. 
 
 If in the triangles ABC, DEF, A = D, B=E, and d = P, let A and D be 
 marked 1, let B and E be marked 2, and let C and F be marked 3. 
 
 Write down all the possible pairs of the numbers 1, 2, 3, viz. : 23, 31, 12. 
 
 Now 2 and 3 are at the extremities of BC in one triangle, and at the 
 extremities of EF in the other. 
 
 Hence BC, EF are corresponding sides. 
 
 In like manner the positions of the numbers 3 and 1 indicate that CA, FD 
 are corresponding sides, and the positions of the numbers 1 and 2 indicate that 
 A B and DP] are corresponding sides.
 
 87] EUCLID, BOOKS V. AND VI. 61 
 
 Art. 87. PROPOSITION XXVIII. (Euc. VI. 4.) 
 
 ENUNCIATION. If the three angles of one triangle are respectively equal to 
 the three angles of another triangle, then the triangles are similar. 
 
 Those sides correspond which join the vertices of equal angles. 
 In the triangles ABC, DEF, let 
 
 r F 
 
 |_y ^ J} . 
 
 To prove that the triangles are similar*. G 
 
 From A on AB measure off a length AG equal to 
 
 DE, and then draw GH parallel to BC cutting AC K c E 
 
 at H. 
 
 It will first be shown that the triangles AGH, DEF are congruent. 
 
 Since GH is parallel to BC, 
 
 AGH = ABC = DEF. 
 
 Also GAH = BAC= EDF, 
 
 and AG = DE. 
 
 Hence the triangles AGH, DEF are congruent. 
 
 /. AH=DF, 
 
 GH = EF. 
 Since GH is parallel to BC, 
 
 .'. BA:GA = CA: HA. [Prop. 17. 
 
 .-. BA:DE=CA : DF. 
 fNow draw HK parallel to AB. 
 Then CA:HA = CB: KB. [Prop. 17. 
 
 * Observe that if the triangles are similar, the vertex A of the triangle ABC corresponds to 
 the vertex D of the triangle DEF ; 
 
 and the side AB of the triangle ABC corresponds to the side DE of the triangle DEF. 
 t It has now been shown that the sides which meet at A are proportional to the corresponding sides 
 which meet at D. 
 
 Consequently a similar proof will show that the sides which meet at B are proportional to the 
 corresponding sides which meet at E. 
 
 .: BA : ED = BC : EF. 
 
 .: AB:DE = BC:EF=CA:FD.
 
 62 EUCLID, BOOKS V. AND VI. [87 
 
 Now BGHK is a parallelogram, 
 
 and HA = DF. 
 
 :. CA:DF=CB:EF. 
 
 Hence BA:DE = CA:DF=CB:EF, 
 
 which, taking the letters in order, may be more conveniently written 
 
 AB:DE = BC:EF = GA : FD. 
 
 Now AB, DE join the vertices of equal angles, and are therefore corre- 
 sponding sides. 
 
 In like manner BC corresponds to EF, and CA to FD. 
 
 Butalso A = D\ B = E;C = F. 
 
 Hence the two sets of conditions for the similarity of the triangles ABC, DEF 
 are satisfied. 
 
 Hence the triangles are similar. 
 
 It should be noticed that corresponding sides of the triangles are opposite to 
 equal angles; e.g. AB corresponds to DE, and they are opposite to the equal 
 
 angles C and F respectively. 
 
 Art. 88. COROLLARY TO PROP. 28. 
 
 If a triangle be cut by a straight line parallel to one of the sides, the 
 triangular portion cut off is similar to the whole triangle. 
 
 For with the figure of Prop. 28, OH may be regarded as any straight line 
 parallel to BC, the triangles AOH, ABC are equiangular, and therefore similar 
 by Prop. 28. 
 
 Art. 89. EXAMPLES. 
 
 29. Show how to draw a straight line across two of the sides of a triangle, but 
 not parallel to the third side, which will cut off a triangle similar to the original 
 triangle. 
 
 When will it be impossible to do this ? 
 
 30. If ABC be a triangle inscribed in a circle, and CD a diameter of the circle, 
 and AE a perpendicular from A on the side BC, show that the triangles AEB, 
 AC 'D are similar.
 
 90] EUCLID, BOOKS V. AND VI. 63 
 
 Art. 90. PROPOSITION XXIX.* (Euc. VI. 5.) 
 
 ENUNCIATION. If ike sides taken in order of one triangle are proportional 
 to the sides taken in order of another triangle, prove that the triangles are 
 similar, and that those angles are equal which are opposite to corresponding 
 sides. 
 
 lu the triangles ABC, DEF let it be given that 
 
 AB:DE=BC:EF=CA:FD t (I) 
 
 to prove that the triangles ABC, DEF are similar. 
 
 B C E F 
 
 Fig. 34. 
 
 From A the vertex of the triangle ABC which corresponds to D, measure off 
 on AB, the side corresponding to DE, a length AG equal to DE. 
 
 Draw GH parallel to BC, cutting A C at H. 
 
 It will first be shown that the triangles AGH and DEF are congruent. 
 
 The triangles A GH and ABC have the angles of the one respectively equal 
 to the angles of the other. 
 
 Therefore by Prop. 28 they are similar. 
 
 .-. AB:AG = BC:GH = CA:HA. (II) 
 
 Now DE = AG, 
 
 .'. AB:DE = AB:AG. [Prop. 22. 
 
 Hence each of the three ratios marked (I) is equal to each of the three 
 ratios marked (II). 
 
 /. BC:EF=BC:GH, 
 
 .-. EF=GH. [Prop. 23. 
 
 * See Note 8.
 
 64 EUCLID, BOOKS V. AND VI. [90 
 
 Also CA:FD = CA:HA, 
 
 .-. FD = HA. [Prop. 23. 
 
 Hence in the triangles DBF, AGH, 
 
 DE=AG, 
 EF=GH, 
 
 FD = HA- 
 .'. they are congruent. 
 
 .% EDF = GAH = BAC, 
 
 DBF = AGH = ABC, 
 
 Hence in the triangles ABC, DEF 
 
 AB:DE=BC:EF=CA:FD, 
 
 A=b, B = E, C = F. 
 Hence the triangles are similar. 
 
 The equal angles BAG, EDF are opposite the corresponding sides EC, EF. 
 The equal angles ABC, DEF are opposite the corresponding sides CA, FD. 
 The equal angles BCA, EFD are opposite the corresponding sides AB, DE. 
 
 Art. 91. NOTE ON PROPOSITION 29. 
 
 The proviso that the sides of the triangles are proportional when taken in 
 order is very important. 
 
 It is quite possible for the sides of one triangle to be proportional to the 
 sides of another without the triangles being similar. 
 
 Suppose that in the triangles ABC, DEF, 
 
 BC:CA=EF:DE, 
 
 and BC:AB = FD:DE, 
 
 then it may be proved (see Proposition 57 below) that 
 
 CA:AB = FD:FE. 
 But the triangles are not similar. 
 In the first proportion BC corresponds to EF. 
 In the second proportion BC corresponds to FD. 
 Hence the sides of the two triangles cannot be made to correspond.
 
 92] EUCLID, BOOKS V. AND VI. 65 
 
 Art. 92. PROPOSITION XXX. (Euc. VI. 6.) 
 
 ENUNCIATION. If two sides of one triangle be proportional to two sides of 
 another triangle, and if the included angles be equal, then the triangles are 
 similar ; and those angles are equal which are opposite to corresponding sides. 
 
 In the triangles ABC, DEF let it be given that 
 
 and BAC=EbF, 
 
 it is required to prove that the triangles are similar ; and that the angles BOA, 
 EFD opposite the corresponding sides BA, ED are equal ; and that the angles 
 ABC, DEF opposite the corresponding sides AC, DF are equal. 
 
 D 
 
 From A, the vertex of the triangle ABC which corresponds to D, measure 
 off on AB, the side corresponding to DE, a length AG equal to DE. 
 
 Draw GH parallel to BC cutting AC at H. 
 
 It will first be proved that the triangles A GH, DEF are congruent. 
 
 Since GH is parallel to BC, 
 
 .'. BA:GA = CA:HA, [Prop. 17. 
 
 .-. BA:CA = GA:HA; [Prop. 24. 
 
 but it is given that BA:CA = ED: DF, 
 
 .-. ED:DF=GA:HA. 
 But ED = GA by construction, 
 
 .-. DF=HA. [Prop. 23. 
 
 H. E. 9
 
 66 EUCLID, BOOKS V. AND VI. [92 
 
 Now in the triangles DEF, AGH, 
 
 = AG, 
 DF=AH, 
 
 EDF = BAC = GAH. 
 Hence the triangles DEF, AGH are congruent. 
 
 .-. DEF = AGH = ABC, 
 
 and DFE=AHG = ACB. 
 
 Hence the angles of the triangle DEF are respectively equal to the angles 
 of the triangle ABC. 
 
 Hence by Prop. 28 the triangles DEF, ABC are similar, and those angles 
 are equal which are opposite to corresponding sides. 
 
 Art. 93. EXAMPLES. 
 
 31. Two parallel straight lines are cut by any number of straight lines passing 
 through a fixed point. 
 
 Prove that the intercepts made on the parallel lines by any two of the straight 
 lines through the fixed point have a constant ratio. 
 
 32. If the tangents at A and B to a circle meet at C, and if P be any point 
 on the circle, and if PQ, PR, PS be drawn perpendicular to AC, CB, BA respectively, 
 then prove that the triangles PAS, PBR are similar; and that the triangles PBS, PAQ 
 are similar ; and that PS is a mean proportional between PQ and PR. 
 
 33*. If C be the centre of a circle, F any point outside it, if FA, FB be tangents 
 to the circle at A and B respectively, if FP be any straight line through F cutting 
 the circle at P ; and if through P a straight line be drawn perpendicular to FP 
 cutting CA at Q and CB at R ; then prove that the triangles CFQ, CRF are similar ; 
 and that CF is a mean proportional between CQ and CR. 
 
 34. Let be the centre of a circle, and C a fixed point in its plane. Let CO 
 cut the circle at A and B. Let P be any point on the circle, and through P let 
 a straight line be drawn perpendicular to CP, cutting the tangents at A and B 
 at Q and R respectively, then prove that 
 
 (1) the triangles ACQ, BCR are similar. 
 
 (2) AQ : AC = BC : BR. 
 
 (3) the angle QCR is a right angle.
 
 94] EUCLID, BOOKS V. AND VI. 67 
 
 Art. 94. PROPOSITION XXXI. (Euc. VI. 7.) 
 
 ENUNCIATION. If two triangles have one angle of the one equal to one angle 
 of the other, and the sides about one other angle in each proportional in such a 
 manner that the sides opposite to the equal angles correspond, then the triangles 
 have their remaining angles either equal or supplementary, and in the former 
 case the triangles are similar. 
 
 In the triangles ABC, DEF it is given that 
 
 ABC = DBF, 
 and BA:AC=ED:DF, 
 
 to prove that either 
 
 (1) ACB = DFE, 
 
 and the triangles ABC, DEF are similar; 
 or (2) ACS + DFE = two right angles*. 
 
 A 
 
 B C E T K 
 
 Fig. 36. 
 
 On AB, the side corresponding to DE, take a length AG equal to DE, 
 and draw GH parallel to BC cutting AC at H. 
 The triangles AGH, DEF will first be compared. 
 Since GH is parallel to BC, 
 
 :. BA : GA = CA : HA, [Prop. 17. 
 
 .-. BA : CA = GA : HA. [Prop. 24. 
 
 But BA:CA = ED:DF, 
 
 .-. ED:DF=GA:HA. 
 
 * Notice that the sides about the angles BAG, EDF are proportional in such a manner that 
 the sides AC, DF opposite the equal angles ABC, DEF correspond. 
 
 Notice further that in each triangle two angles have been referred to, viz. ABC, BAG in the 
 triangle ABC and DEF, EDF in the triangle DEF, and therefore the remaining angles are 
 
 ACB, DFE. 
 
 92
 
 68 EUCLID, BOOKS V. AND VI. [94 
 
 But AG = DE, 
 
 .'. DF=HA. [Prop. 23. 
 
 Now in the triaugles DEF, AGH, 
 
 DE = AG, 
 DF=AH, 
 
 DEF = ABC = AGH, 
 
 where it is to be noticed that the equal angles are opposite to equal sides. 
 Now there are necessarily two alternatives, 
 
 either (1) GAH = EDF, 
 
 or (2) GAH is not equal to Ef)F. 
 
 (1) If GAH = EDF, 
 then BAG=EDF, 
 and since ABC = DEF, 
 
 :. BCA=EFD, 
 and in this case the triangles ABC, DEF are similar. [Prop. 28. 
 
 (2) If GAH be not equal to EDF, draw DK, making EDK equal BAG, 
 and cutting EF at K. 
 
 Then in the triangles AGH, EDK, 
 
 GAH =EDK, 
 
 AG =DE. 
 
 Hence the triangles A GH, DEK are congruent. 
 
 = DK, 
 
 But AH = DF, 
 
 .'. DF=DK, 
 .-. DKF=DFK. 
 
 Therefore A CB + DFE = AHG + DFE 
 
 = DFK+DFE 
 = two right angles.
 
 96] EUCLID, BOOKS V. AND VI. 69 
 
 Art. 95. NOTE. 
 
 Euclid's method of stating Proposition 31 amounts to the insertion of 
 additional conditions in the statement here given, the effect of which is to 
 exclude the second alternative in those cases in which the two alternatives 
 are really distinct. 
 
 It is as follows : 
 
 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 be either less or greater than a right angle, or if one of them be a right 
 angle, the triangles are similar and have those angles equal about which the sides 
 are proportionals. 
 
 Hence the additional conditions are that ACS and DFE are both less than 
 a right angle, or both greater than a right angle, or one of them is a right 
 angle. 
 
 If they are both less than a right angle, their sum is less than two right 
 angles. 
 
 If they are both greater than a right angle, their sum is greater than two 
 right angles. 
 
 In neither of these cases can the second alternative hold. 
 
 Hence the first alternative must hold, and the triangles are similar ; the angles 
 between the proportional sides being equal. 
 
 If next one of the two angles AGB, DFE is a right angle, then, whichever 
 alternative hold, the other angle is a right angle, hence the remaining angles are 
 equal, and the triangles are similar. 
 
 This is the case in which the two alternatives are not really distinct. 
 
 Art. 96. EXAMPLE 35. 
 
 If B and C are the centres of two circles, and A the point of intersection of their 
 internal or of their external common tangents, and if APQ be any straight line through 
 A cutting the first circle at P and the second at Q, prove that the angles APB, AQG 
 are either equal or supplementary.
 
 70 
 
 EUCLID, BOOKS V. AND VI. 
 
 [97 
 
 Art. 97. PROPOSITION XXXII. (Euc. VI. 18.) 
 
 ENUNCIATION. On a given straight line to describe a rectilineal figure similar 
 and similarly situated to a given rectilineal figure. 
 
 Let AiB&DiEi be the given rectilineal figure; it is required to describe 
 a similar figure on the given straight line A 2 B 2 , so that A& and A 2 B 2 may be 
 corresponding sides of the figures. 
 
 Fig. 37. 
 
 Join 
 
 A 
 
 At B 2 draw a straight line making with A. 2 B 2 an angle equal to AJZfi^ and at 
 A 2 draw a straight line making with A 2 B 2 an angle equal to B 1 A 1 G 1 . Let these 
 straight lines meet at C 2 . 
 
 At (7 2 draw a straight line making with A 2 G Z an angle equal to A l C l D lt and at 
 A., draw a straight line making with A. 2 C 2 an angle equal to Cj^A- Let these 
 straight lines meet at D z . 
 
 At A draw a straight line making with D 2 A 2 an angle equal to AiD l E l> and at 
 A 2 draw a straight line making with A 2 D 2 an angle equal to D 1 A 1 E 1 . Let these 
 straight lines meet at E 2 . 
 
 It will be proved that A^CJ)^ and A 2 B 2 C 2 D 2 E 2 are similar figures, and that 
 A 1 B l and A 2 B 2 are corresponding sides. 
 
 In the three pairs of triangles in Figure 37, viz. : -4iJ5 1 C' 1 and A 2 B 2 G 2 , A 1 C 1 D 1 
 and A 2 C 2 D 2 , A 1 D 1 E 1 and A. 2 D 2 E 2 , let the equal angles be marked with the same 
 numbers. Then in each pair of triangles two angles of the one triangle are 
 respectively equal to two angles in the other triangle. Therefore the remaining 
 angles are equal. Let these be marked with the same number.
 
 98] EUCLID, BOOKS V. AND VI. 71 
 
 Then it is at once apparent that the angles at A l} B lt (7,, D lf E l of the figure 
 i are respectively equal to the angles at A 2 , B 2 , C 2> D 2 , E 2 of the figure 
 t\ so that the first set of conditions (see Art. 77) for the similarity of 
 the two figures is satisfied. 
 
 Next the triangles A- i B l C l and A^B^G^ are equiangular and therefore by 
 Prop. 28 are similar. 
 
 In like manner the triangles A&D^ and A<JJ) 2 are similar; and the triangles 
 A^D^E-i and A^D 3 E t are similar. 
 
 From these pairs of similar triangles follow the relations 
 
 (1), 
 (2), 
 A,D, '.AtD^D&'.DiEt^E^: E Z A Z ..................... (3). 
 
 Hence 
 
 Hence the second set of conditions (see Art. 77) for the similarity of the two 
 figures is also satisfied. 
 
 Hence the two figures A 1 B 1 C 1 D 1 E 1 and A Z B Z CJ) Z E Z are similar figures, and 
 A 1 B l and A Z B 2 are corresponding sides. Therefore the figures are similarly 
 described on A& and A^B 2 (see Art. 80). 
 
 Art. 98. PROPOSITION XXXIII. (Included in Euc. VI. 20.) 
 
 ENUNCIATION. Two similar rectilineal figures may be divided into the same 
 number of triangles such that every triangle in either figure is similar to one 
 triangle in the othei' figure. 
 
 Fig. 38.
 
 72 EUCLTD, BOOKS V. AND VI. [98 
 
 Let AiBiCiDi, A^B^CJ) Z be two similar figures, such that 
 
 D.A.B, = D*A 2 B 2 , 
 
 and ^B, : A Z B, = B& : B& 
 
 Let O l be any point in the plane of A l B l C-J) l) and join Oj-d,, Ojfi,, O,^, OiZ),- 
 
 Through A z draw a straight line making with AJ3. 2 an angle equal to jB^d a Q. 
 
 Through 5 2 draw a straight line making with BA% an angle equal to A^O^. 
 
 Let these two straight lines meet at 2 . 
 
 Join 2 (7 2 , 2 D 2 ' 
 
 It will be proved that the triangles 0^^, O^A^B^ are similar; that the 
 triangles O l B^C l , 2 fi 2 C 2 are similar, and so on. 
 
 A A 
 
 Since 1 A 1 B 1 =0 2 A. t B 2 , 
 
 Hence the triangles A^O^, J. 2 2 5 2 are similar. [Prop. 28. 
 
 /. A.B, : AsB^B^ : B 
 Since A#i : A 2 B 2 = B& 
 
 /. 5,0! : Bfi^BtOs : B&. [Prop. 24. 
 
 Also A&C^A&Ci; 
 
 A&Oi-A&O,, 
 
 .'. L/i-DxGj = C' 2 A> 2 C' 2 . 
 
 Hence by Prop. 30 the triangles Oj5j(7j, OJB. 2 C<i are similar. 
 
 In like manner the triangles O^A and 2 (7 2 D 2 can be proved to be similar ; 
 and also OjA-^i. 2 D 2 ^2 can be proved to be similar. 
 
 So the two similar figures are divided up into the same number of triangles, 
 such that every triangle in either figure is similar to one triangle in the other 
 figure. 
 
 The point O l in the one figure corresponds to the point O 2 'in the other figure. 
 
 Since 0^ is any point in the one figure, it follows that to every point in one of 
 the figures corresponds one and only one point of the other figure.
 
 101] EUCLID, BOOKS V. AND VI. 73 
 
 Art. 99. COROLLARY. 
 
 If in Figure 38 the first figure be placed on the second so that 0^ falls on 2 , 
 0j.4j falls along 2 A Z , 0^ falls along 2 fi 2 , it can be shown that the sides of the 
 first figure will then be parallel to and in the same direction as the corresponding 
 sides of the second, and that the distances from O l or 0^ to a point on either figure 
 along any straight line are in the ratio of similitude of the figures. 
 
 If Oi be placed on O a , 1 A 1 along A 3 0. 2 produced through 2 , and O l B l along 
 jB 2 2 produced through 2 , the sides of the first figure will then be parallel but in 
 the opposite direction to the corresponding sides of the second figure. 
 
 When the two figures have been placed as described in either of the two 
 preceding cases, then the point 2 , with which 0! coincides, is called a centre 
 of similitude of the two figures. 
 
 The term centre of similitude is not however restricted to rectilineal figures. 
 (See Art. 100, Ex. 37 below.) 
 
 Art. 100. EXAMPLES. 
 
 36. If two similar rectilineal figures are placed so that two consecutive sides 
 of one figure are respectively parallel and both in the same direction as, or both in 
 the opposite direction to, the corresponding sides of the other figure, then each side 
 of the one figure will be parallel to the corresponding side of fche other figure, and 
 the straight lines joining corresponding angular points of the two figures are all parallel 
 or meet in a point ; and in the latter case the distances from that point along any 
 straight line to the points where it meets corresponding sides of the figures are in 
 the ratio of similitude of the figures. 
 
 What is the ratio of similitude when the lines joining corresponding angular points 
 are parallel? 
 
 37. If the straight line joining the centres A, B of two circles be divided internally 
 and externally in the ratio of the radii of the circles, (the segment of the line AB 
 terminated at A corresponding to the radius of the circle whose centre is A), then show 
 that the points of division may be regarded as centres of similitude of the circles. 
 
 Art. 101. PROPOSITION XXXIV. (Euc. VI. 8.) 
 
 ENUNCIATION. If a right-angled triangle be divided into two parts by a 
 perpendicular drawn from the vertex of the right angle on to the hypotenuse, then 
 the triangles so formed are similar to each other and to the whole triangle ; the 
 perpendicular is a mean proportional between the segments of the hypotenuse ; and 
 each side is a mean proportional between the adjacent segment of the hypotenuse and 
 the hypotenuse. 
 
 H. E. 10
 
 74 
 
 EUCLID, BOOKS V. AND VI. 
 
 [101 
 
 If ABC be the triangle, and B the vertex of the right angle, and if ED be 
 drawn perpendicular to AC, it is required to prove 
 
 (1) that the triangles ABC, ABD, BDC are similar. 
 
 (2) that BD is a mean proportional between AD and DC. 
 
 (3) that EC is a mean proportional between CD and AC. 
 
 (4) that BA is a mean proportional between AD and AC. 
 
 B B B 
 
 Fig. 39. 
 
 The triangles ABC, ABD will be compared first. 
 
 BAG = BAD 
 
 ABC=ADB = a right angle. 
 .-. ACS = ABD. 
 Hence the triangles are similar (Prop. 28). 
 
 .'. BC:DB = CA:BA=AB: AD. 
 Since CA :BA=BA :AD, 
 
 .'. BA is a mean proportional between AC and AD. 
 
 In like manner it can be shown that the triangles ABC, DBG are similar ; and 
 that EG is a mean proportional between AC and CD. 
 
 Since ABD, GBD are similar to ABC they are similar to one another. 
 
 BAD = DBG 
 ABD = BCD 
 ADB = BDC, 
 
 .-. DB :DC=BA : CB = AD:BD. 
 Hence DB : DC = AD : DB, 
 
 .'. DC : DB = DB : DA. [Prop. 21. 
 
 Hence DB is a mean proportional between DA and DC. 
 
 Art. 102. EXAMPLE 38. 
 
 If in any triangle ABC, BD is drawn to cut AC at D so that BDC is equal to ABC, 
 prove that the triangles ABC, BCD are similar; that BC is a mean proportional between 
 AC and CD, and that AC:AB = BC: BD.
 
 SECTION VI. 
 
 MISCELLANEOUS GEOMETRICAL PROPOSITIONS. Props. 35, 36. 
 
 Art. 103. PROPOSITION XXXV. (Euc. VI. 13.) 
 
 ENUNCIATION. To find a wiean proportional between two given segments of 
 straight lines. 
 
 K 
 
 A 
 Fig. 40. 
 
 Let the given straight lines be K and L. 
 
 Take a straight line AD equal to K, and produce AD to C, so that DC is equal 
 to L. 
 
 On AC as diameter describe a semicircle. 
 
 Through D draw DB perpendicular to AC to cut the semicircle at B. 
 
 Then DB is the mean proportional between K and L required. 
 
 Join AB, BG. 
 
 Then ABC being the angle in a semicircle is a right angle. 
 
 Also BD, being drawn perpendicular to AC from the vertex B of the right 
 angle, is by Prop. 34 part (2) a mean proportional between AD and DG. 
 
 .'. BD is a mean proportional between K and L. 
 
 Art. 104. EXAMPLES. 
 
 39. Solve the problem of the last proposition by means of Proposition 34 (3) or (4). 
 
 40. If two circles touch each other and also touch a given straight line, prove 
 that the part of the straight line between the points of contact is a mean proportional 
 between the diameters of the circles. 
 
 102
 
 76 
 
 EUCLID, BOOKS V. AND VI. 
 
 T104 
 
 41. If through the middle point A of the arc BAG of a circle, a chord be drawn 
 cutting the chord of the arc BC at D and the circle again at E, prove that AB is a mean 
 proportional between AD and AE. 
 
 42. If G be the centre of a circle, a point outside it, OT a tangent from to the 
 circle, TP a perpendicular from T on OC, then prove that the radius of the circle is a 
 mean proportional between CO and CP. 
 
 Art. 105. PROPOSITION XXXVI. (i). (Euc. VI. 3 and A, 1st Part.) 
 
 ENUNCIATION. If the interior or exterior vertical angle of a triangle be bisected 
 by a straight line which also cuts the base, the base is divided internally or externally 
 in the ratio of the sides of the triangle. 
 
 Ac D 
 
 Fi - 41. Fig. 42. 
 
 Let ABC be a triangle. 
 
 Lefc BD bisect the interior angle ABC in Fig. 41, but the exterior angle A'BC 
 between CB and AB produced to A' in Fig. 42. 
 Let BD cut the base at D. 
 To prove that AD : DC = AB : BC. 
 Draw CE parallel to BD cutting AB at E. 
 
 Then BEG = DBA in Fig. 41 (or DBA' in Fig. 42) 
 
 = DBC 
 = BCE, 
 
 .'. BC=BE. 
 
 Since ADC, ABE are cut by parallel lines BD, CE, 
 
 .'. AD : DC=AB: BE, 
 .-. AD :DC = AB:BC. 
 
 [Prop. 17.
 
 106] 
 
 EUCLID, BOOKS V. AND VI. 
 
 77 
 
 Art. 106. PROPOSITION XXXVI. (ii). (Euc. VI. 3 and A, 2nd Part.) 
 
 ENUNCIATION. // the base of a triangle be divided internally or externally in 
 the ratio of the sides of the triangle, the straight line drawn from the point of 
 division to the vertex bisects the interior or exterior vertical angle. 
 
 A' 
 
 Fig. 44. 
 
 Let ABC be a triangle. 
 
 Let D divide the base AC, internally in Fig. 43, externally in Fig. 44, so that 
 
 AD:DC = AB:BC. 
 
 To prove that DB bisects the interior angle ABC in Fig. 43, but the exterior 
 angle between CB and AB produced through B in Fig. 44. 
 
 Join DB, and draw CE parallel to DB cutting AB at E, 
 
 Then AD : DC = AB : BE, [Prop. 17. 
 
 but AD:DC = AB:BC, 
 
 .'. AB:BE = AB:BC, 
 
 .'. BE = BC, [Prop. 23. 
 
 .-. BCE = BEC. 
 
 Now ABD in Fig. 43, or A'BD in Fig. 44= BEG 
 
 = BCE 
 
 /. ABD in Fig. 43, or A'BD in Fig. 44 = DBC. 
 
 Hence DB bisects the interior vertical angle in Fig. 43, but the exterior 
 vertical angle in Fig. 44.
 
 78 EUCLID, BOOKS V. AND VI. [107 
 
 Art. 107. EXAMPLES. 
 
 43. If the internal and external angles at B of the triangle ADC be bisected 
 by straight lines which cut the side AC at D and E respectively, show that A, D, C, E 
 are four harmonic points. 
 
 44. By means of Proposition 36 construct the fourth harmonic to three given 
 points A, B, C on a straight line ; considering separately the cases which arise according 
 as the fourth harmonic is to be conjugate to A or B or C. 
 
 45. If ABC be a triangle inscribed in a circle, PQ a diameter of the circle 
 perpendicular to AC, if CB cut PQ at R, and AB cut PQ at S, prove that 
 
 QS:JKP=QS:SP. 
 
 Hence construct the fourth harmonic to three given points on a straight line. 
 
 46. Divide a given arc of a circle into two parts so that the chords of these parts 
 may be to each other in a given ratio. 
 
 47. A point P moves in a plane so that the ratio of its distances from two fixed 
 points A, B in that plane is always the same. Show that in general the locus of P is 
 a circle, the extremities of one diameter of which are the points dividing AB internally 
 and externally in the given ratio. What is the exceptional case 1 
 
 48. The side BC of a triangle ABC is bisected at D, and the angles ADC, ADB are 
 bisected by the straight lines DE, DF meeting AC, AB at E, F respectively. Prove 
 that EF is parallel to BC. 
 
 49. If the bisector of the angle A of the triangle ABC cut BC at D, and if the 
 bisector of the angle B cut AC at E, and if DE be parallel to AB, prove that the 
 triangle ABC is isosceles. 
 
 50. If ABC be a triangle, if D be the middle point of BC, if any straight line 
 through D cut AB at E, AC at F, and a parallel through A to BC at G ; then prove 
 that E, D, F, G are four harmonic points. 
 
 Hence show that if any point be joined to four harmonic points, they will be cut 
 by any transversal in four harmonic points. 
 
 Art. 108. Def. 14. HARMONIC LINES. 
 
 If four concurrent straight lines be cut by any transversal in four harmonic 
 points they are called four harmonic lines, or are said to form a harmonic 
 pencil.
 
 SECTION VII. 
 
 AN IMPORTANT PROPOSITION IN THE THEORY OF RATIO. 
 
 Art. 109. PROPOSITION XXXVII. 
 
 If A , B, C are three magnitudes of the same kind ; 
 if T, U t V are three magnitudes of the same kind ; 
 if A:B = T:U 
 
 and B:C=U:V, 
 
 then will A:C=T:V. 
 
 o 
 
 Compare A : C with any rational fraction - . 
 
 By Art. 48 it is necessary to consider only the two cases 
 
 (1) A:C>-. 
 
 r 
 
 (2) A:C<-. 
 
 r 
 
 o 
 
 In case (1) A :C>- , 
 
 r 
 
 rA > sC. [Prop. 13. 
 
 Hence rA sC is a magnitude of the same kind as B. 
 Hence by Archimedes' Axiom (Art. 22) an integer n exists, such that 
 
 n (rA - sC) > B, 
 .'. nrA >nsC + B. 
 Hence a multiple of B, say tB, exists, such that [Prop. 8. 
 
 nrA >tB> nsC. 
 Since nrA > tB, 
 
 .: A:B> . [Prop. 13. 
 
 nr
 
 80 EUCLID, BOOKS V. AND VI. [109 
 
 But A:B=T:U, 
 
 .-. T:U>-, 
 nr 
 
 :. nrT>tU. [Prop. 13. 
 
 Since tB > nsC, 
 
 :. B:G>^. [Prop. 13. 
 
 I 
 
 But B:G=U:V, 
 
 /. tU>nsV. [Prop. 13. 
 
 But nrT>tU, 
 
 .'. nrT> nsV, 
 .-. rT>sV, 
 
 /. T: V>- r . [Prop. 13. 
 
 In case (2) A : C < - , 
 
 r 
 
 .-. rA < sC. [Prop 13. 
 
 Hence sC rA is a magnitude of the same kind as B. 
 Hence by Archimedes' Axiom an integer n exists, such that 
 
 n (sC - rA) > B, 
 
 .'. nsC>nrA + B. 
 Hence a multiple of B exists, say tB, such that [Prop. 8. 
 
 nrA <tB < nsC. 
 Since nrA < tB, 
 
 .'. A:B< . [Prop. 13. 
 
 nr 
 
 But A:B = T:U, 
 
 nr 
 
 :. nrT<tU. [Prop. 13. 
 
 Since tB < nsC, 
 
 .-. B:C<^. [Prop. 13. 
 
 t
 
 110] EUCLID, BOOKS V. AND VI. 81 
 
 But B:G=U:V, 
 
 /. tU <nsV. [Prop. 13. 
 
 But nrT<tU, 
 
 .*. nrT< nsV, 
 .'. rV<sV, 
 
 T:V<*-. [Prop. 13. 
 
 It has therefore been proved 
 
 (1) If A:C>-,theuT:V>-. 
 
 r r 
 
 (2) If A:C<-, thenT:F<-. 
 
 r r 
 
 Hence by Art. 48 
 
 A: C=T:V. 
 
 Art. 110. EXAMPLE 51. 
 
 Two circles whose centres are C and C" intercept equal chords AB and A'ff on 
 a straight line cutting both circles. 
 
 The tangents at A and A' meet at T. 
 
 Prove that AT :A'T = AC : A'C'. 
 
 H. F, 11
 
 SECTION VIII. 
 
 AREAS. 
 PROPOSITIONS 3850. 
 
 Art. 111. PROPOSITION XXXVIII (i). (Euc. VI. 16, 1st Part.) 
 
 ENUNCIATION. If four straight lines are proportional, the rectangle contained 
 by the extremes is equal to the rectangle contained by the means. 
 Let K, L, M, P be the straight lines, such that 
 
 K:L = M:P, 
 
 it is required to prove that the rectangle contained by K and P is equal to that 
 contained by L and M. 
 
 L- 
 M- 
 P- 
 
 F EG 
 
 
 
 V B 
 
 
 D H 
 
 Fig. 45. 
 
 Take AB equal to K. 
 
 Produce AB to C so that EG is equal to L. 
 
 Through B draw BD equal to M perpendicular to AB, and produce DB to E 
 so that BE is equal to P. 
 
 Complete the rectangles ABEF, BCGE, BGHD, 
 
 Since rectangles having equal altitudes are proportional to their bases (Prop. 16) 
 ABEF : BCGE = AB : BC = K : L, 
 BD:BE=M:P.
 
 112] EUCLID, BOOKS V. AND VI. 83 
 
 Now K:L = M:P, 
 
 .-. A BEF : BCGE = BCHD : BCGE, 
 
 :. A BEF = BCHD. [Art. 43. 
 
 Now ABEF is the rectangle contained by AB and BE, i.e. by If and P. 
 Whilst BCHD is the rectangle contained by BC and BD, i.e. by L and M. 
 :. the rectangle contained by K and P is equal to that contained by L and M. 
 
 Art. 112. PROPOSITION XXXVIII (ii). (Euc. VI. 16, 2nd Part.) 
 
 ENUNCIATION. Let there be four straight lines, which taken in a definite order 
 are K, L, M, P ; and let it be given that the rectangle contained by the first and 
 fourth, K and P, is equal to the rectangle contained by the second and third, L and 
 M ; to prove that 
 
 K:L=M:P. 
 
 K- 
 L- 
 M- 
 P- 
 
 Fig. 46. 
 
 Make the same construction as in the preceding part of the proposition. 
 
 The rectangle contained by K and P is the rectangle contained by AB and 
 BE and is therefore ABEF. 
 
 The rectangle contained by L and M is the rectangle contained by BC and 
 BD, and is therefore BCHD. 
 
 :. ABEF is given equal to BCHD. 
 
 :. ABEF: BCGE '= BCHD: BCGE, [Art. 43. 
 
 but ABEF:BCGE = AB:BC=K:L, [Prop. 16. 
 
 and BCHD : BCGE = BD : BE = M : P, [Prop. 1 6. 
 
 .-. K:L = M:P. 
 
 112
 
 84 EUCLID, BOOKS V. AND VI. [113 
 
 Art. 113. COROLLARY TO PROPOSITION 38. (Euc. VI. 17.) 
 
 PART 1. If three straight lines are proportionals, the rectangle contained by 
 the extremes is equal to the square on the mean. 
 
 PART 2. Let there be three straight lines, which taken in a definite order are 
 K, L, and P; and let it be given that the rectangle contained by the first and third, 
 K and P, is equal to the square on the second, L, then it will follow that 
 
 K:L = L:P. 
 
 The first part is the particular case of Proposition 38 (i), and the second part 
 the particular case of Proposition 38 (ii), when M = L. 
 
 * 
 
 Art. 114. EXAMPLES. 
 
 52. Let C be the centre of a circle, any point in its plane ; let A and B be the 
 extremities of the diameter through 0; let P and Q be the extremities of any chord 
 through 0. Prove that the circle drawn through C, P and Q cuts 00 in a point D 
 which is the same for all directions of the chord OPQ, and show that 
 
 OA : OD = 00 : OB. 
 
 53. (i) Let A be the centre of a circle, B a point outside it, BD and BE tangents 
 to the circle, C the point in which DE cuts AB, BFG a straight line through B cutting 
 the circle at F and G ; then prove that the rectangle BA . EG is equal to the rectangle 
 BF.BG. Prove that CD bisects the angle FOG, and that if CD cut FG at H, then 
 B, F, H, G are four harmonic points. ^ 
 
 (ii) Let A be the centre of a circle, B a point inside the circle, and let any chord 
 GBF\>e drawn through B, and produced to H so that G, B, F, H are four harmonic 
 points, prove that the locus of H is a straight line which cuts AB at right angles at a 
 point C such that the rectangle BA . BC is equal to the rectangle BF. BG. 
 
 54. It A, B, C, D are four harmonic points and the middle point between the two 
 conjugate points A and C, prove that the rectangle contained by OB and OD is equal to 
 the square on OC. 
 
 Art. 115. Def. 15. POLE AND POLAR. 
 
 If through any point a straight line be drawn cutting a circle at P and 
 Q, and on OPQ a point R be taken so that 0, P, R, Q are four harmonic 
 points, and R being conjugates ; then the locus of R is called the polar line 
 of 0, and is called the pole of the locus of R. 
 
 It is a result of Example 53 that the polar line is a straight line.
 
 118] EUCLID, BOOKS V. AND VI. 85 
 
 Art. 116. EXAMPLES. 
 
 55. If C be the centre of a circle, any point in its plane and T the foot of the 
 perpendicular from C on to the polar line of 0, then prove that the rectangle contained 
 by CO and CT is equal to the square on the radius of the circle. 
 
 56. If A lie on the polar of B with regard to a circle, show that B lies on the polar 
 of A with regard to that circle. 
 
 (Two such points as A, B are said to be conjugate with regard to the circle.) 
 
 57. If two circles cut at right angles prove that the extremities of any diameter of 
 either circle are conjugate points with regard to the other circle. 
 
 58. If A, B be two points, and if from A a perpendicular AP be drawn to the polar 
 line of B with regard to a circle whose centre is C, and if from B a perpendicular BQ be 
 drawn to the polar line of A, prove that 
 
 CA : CB = AP : BQ, 
 and show that the triangles CA P, CBQ are similar. 
 
 59. Let C be the centre of a circle, Fany fixed point in its plane, let CV cut the 
 circumference at A, and let a point P be taken on CV so that the rectangle CV . CP is 
 equal to the square on CA. Let a straight line PY be drawn through P perpendicular 
 to CP, and let PY be cut by any straight line through V in W y and by a perpendicular 
 through C to FTFin X, prove that the rectangle PX . PW will always be equal to the 
 rectangle CP . P V in whatever direction the straight line V W may be drawn. 
 
 Art. 117. Def. 16. INVERSE LOCUS. CENTRE OF INVERSION. 
 
 If from any point a straight line be drawn to cut any curve at P, and 
 on OP a point Q be taken so that the rectangle OP. OQ has a constant area, 
 then the locus of Q is called the inverse of the locus of P with regard to as 
 centre (or origin) of inversion. 
 
 The side of the square whose area is equal to the constant rectangle OP . OQ 
 is called the radius of inversion. 
 
 Also P and Q are said to be inverse points with regard to the circle whose 
 centre is 0, and whose radius is the radius of inversion. 
 
 Art. 118. EXAMPLES. 
 
 60. If the locus of P is a circle, show that the inverse locus is generally a circle, 
 but will be a straight line if the centre of inversion be a point on the circle on which 
 P lies.
 
 86 
 
 EUCLID, BOOKS V. AND VI. 
 
 [118 
 
 61. If the locus of P is a straight line show that the inverse locus is a circle passing 
 through the centre of inversion. 
 
 62. If two circles or a straight line and a circle or two straight lines intersect one 
 another, show that their angle of intersection is equal to the angle of intersection of 
 their inverse loci. 
 
 Art. 119. Def. 17. THE RADICAL AXIS OF TWO CIRCLES. 
 
 The locus of points from which tangents drawn to two circles are of 
 equal length is called the radical axis of the two circles. 
 
 Art. 120. EXAMPLES. 
 
 63. If two circles intersect, show that the straight line joining their points of 
 intersection is their radical axis. 
 
 If they do not intersect, show that the radical axis is perpendicular to the line 
 joining the centres of the circles, and cuts it at a point which is such that double the 
 distance of this point from the point half way between the centres of the circles is 
 a fourth proportional to the distance between the centres of the circles, the sum of 
 their radii and the difference of their radii. 
 
 64. Show that the difference between the squares of the tangents from any point P 
 to two circles is equal to twice the rectangle contained by the perpendicular from P on 
 the radical axis, and the distance between the centres of the circles. 
 
 65. Show how to choose the centre and the radius of inversion so that two given 
 circles may be inverted each into itself. 
 
 Art. 121. PROPOSITION XXXIX. 
 
 ENUNCIATION. The rectangle contained by the diagonals of a quadrilateral 
 cannot be greater than the sum of the rectangles contained by opposite sides. (It 
 may be equal, and in that case a circle can be described through the vertices of 
 the quadrilateral.) 
 
 B 
 
 Fig. 47.
 
 121] EUCLID, BOOKS V. AND VI. 87 
 
 Let ABCD be a quadrilateral. 
 
 It is required to prove that the rectangle AC .BD cannot be greater than the 
 sum of the rectangles AD. BC and AB . CD. 
 
 On EG describe the triangle BCE similar to ABD, so that the side BC of 
 BCE may correspond to the side BD of ABD. 
 
 Then C&E=ABD, 
 
 = BDA, 
 BEG = BAD. 
 
 Also BD:BC=DA:CE=AB: EB. 
 
 From the first and second ratios it follows by Prop. 38 (i) that 
 
 rect. BD.GE = rect. AD .EG. 
 Also from the first and third ratios 
 
 BD:BG = BA: BE, 
 
 but also DBG = DBE + EBG 
 
 = DBE + DBA 
 
 = ABE. 
 Hence by Prop. 30 the triangles DBG, ABE are similar. 
 
 E -C 
 
 Fig. 49. 
 The side BD corresponds to BA, 
 
 the side BC corresponds to BE, 
 and the side CD corresponds to AE. 
 
 .-. BD:BA = DC:AE=CB: EB. 
 From the first and second ratios by Prop. 38 (i) 
 
 rect. BD.AE= rect. AB . CD. 
 Now it has been shown that 
 
 rect. BD.CE = rect. AD . BC, 
 /. rect. BD.AE + rect. BD . GE = rect. AE . CD + rect. AD . BC.
 
 88 EUCLID, BOOKS V. AND VI. [121 
 
 Now AC cannot be greater than AE+ EG. 
 
 (The case in which AC is equal to AE + EG will be considered below.) 
 
 .-. rect. BD . AC cannot be greater than rect. BD . AE + rect. BD . CE. 
 
 .'. rect. BD . AC cannot be greater than rect. AB . GD + rect. AD . BC. 
 
 If AC = AE + EC, 
 
 then E lies on AC, 
 and rect. BD.AG = rect. AB.CD + rect. AD . BC. 
 
 Now BCE = BDA, 
 
 whilst in this case BCE = BCA. 
 
 :. BCA=BDA, 
 and therefore the circle circumscribing ABD passes through C. 
 
 Hence a circle can be described about the vertices of the quadrilateral A BCD. 
 
 Art. 122. It is interesting to examine what happens when D and C are 
 points on the straight line AB. 
 
 In this case the straight line can be regarded as a circle of infinite radius. 
 
 Hence taking the points on the line in the order A, B, C, D the lines 
 corresponding to the diagonals are AC, BD; whilst AB, CD correspond to one 
 pair of opposite sides; and BC, AD to the other pair. 
 
 Hence rect. AC.BD = rect. AB.CD+ rect. BC . AD. 
 
 This result is easily verified. 
 
 Art. 123. PROPOSITION XL. (Euc. VI. 23.) 
 
 ENUNCIATION. To construct two straight lines whose ratio is equal to the ratio 
 of the areas of two equiangular parallelograms. 
 
 Since the parallelograms are equiangular they may be placed so as to have 
 a common angle, and their sides along the same straight lines, but in opposite 
 directions. 
 
 M C 
 
 Fig. 50.
 
 124] EUCLID, BOOKS V. AND VI. 89 
 
 When this has been done let ABGD be one parallelogram, let AEFG be the 
 other. 
 
 Produce MD and FE to meet at H. 
 
 Join HA. 
 
 Produce HA to meet FG at K. 
 
 Draw KLM parallel to AD to meet AB, DC at L, M respectively. 
 
 Then AEFG is equal to ALMD, since they are complements about the 
 diameter of a parallelogram. 
 
 /. ABCD : AEFG = ABGD : ALMD [Prop. 22. 
 
 = AB:AL. [Prop. 16. 
 
 The two lines required are therefore AB and AL. 
 
 In this way the problem of finding the ratio of two areas is reduced to the 
 simpler one of finding the ratio of two lengths. 
 
 The close connection of the construction employed in this proposition with that 
 given by Euclid in the 44th proposition of the First Book should be noted. 
 
 Art. 124. It is worth while to notice the relation of the line AL to the sides 
 of the parallelograms ABGD, AEFG. 
 
 AB-.AE is the ratio of one side of the first parallelogram to one side of the 
 second parallelogram. 
 
 AD:AG is the ratio of the other side of the first parallelogram to the other 
 side of the second parallelogram. 
 
 But since the triangles ADH, AGK are equiangular, they are similar, and 
 therefore 
 
 = DH:GK=HA:KA. 
 
 Now DH = AE, 
 
 and GK = AL. 
 
 Hence AD : A G = AE : AL. 
 
 Hence the two ratios of the sides of the parallelograms are AB:AE, and 
 AD:AG which is equal to AE-.AL. 
 
 And the ratio of the areas of the parallelograms is AB: AL. 
 H. E. 12
 
 90 
 
 EUCLID, BOOKS V. AND VI. 
 
 [125 
 
 Art. 125. PROPOSITION XLI. 
 
 ENUNCIATION. The ratio of the areas of two equiangular parallelograms is 
 equal to the ratio of the areas of the rectangles contained by their sides. 
 
 This follows immediately from the preceding proposition by observing that the 
 ratio AB:AL depends only on the lengths of the sides, and not at all on the 
 common angle of the parallelograms. 
 
 For AL is determined by the proportion AD :AG = AE:AL. 
 
 Hence if a rectangle be constructed whose sides are equal to AB, AD and 
 another rectangle constructed whose sides are equal to AE, AG', the ratio of the 
 first rectangle to the second will as before be equal to AB : AL, and therefore equal 
 to the ratio of the parallelogram A BCD to the parallelogram AEFG. 
 
 Art. 126. PROPOSITION XLII. 
 
 ENUNCIATION. // three straight lines be in proportion, then the ratio of the area 
 of the square described on the first line to the area of the square described on the 
 second line is equal to the ratio of the first line to the third line. 
 
 Fig. 51. 
 
 Let A BCD, AEFG be two squares placed so that they have a common vertex 
 and their sides on the same straight lines but in opposite directions. 
 
 Constructing the figure with the same letters as in Prop. 40, then 
 
 area ABCD : area AEFG = AB :AL, 
 where as before AD : A G = AE : AL.
 
 127] EUCLID, BOOKS V. AND VI. 91 
 
 But AD=AB, 
 
 and AG = AE, 
 
 .-. AB:AE = AE:AL. 
 Hence the square described on AB : the square described on A E 
 
 = AB:AL, 
 where AL is determined by the proportion 
 
 AB-.AE=AE:AL. 
 
 This result will be repeatedly required in what follows, and should be carefully 
 remembered. 
 
 Art. 127. PROPOSITION XLIII. (Euc. VI. 19.) 
 
 ENUNCIATION. The ratio of the areas of two similar triangles is equal to the 
 ratio of the areas of the squares described on corresponding sides*. 
 
 Fig. 52. 
 
 The triangles being similar may be placed so as to have a common vertex, and 
 the corresponding sides which meet at this vertex along the same straight lines 
 and in the same direction. 
 
 When this is done, let the triangles be ABC and ADE. 
 
 Since ABC ' = ADE, 
 
 .-. EG is parallel to DE. 
 Join DC. 
 
 Draw BF parallel to DC to cut AE at F. 
 Join DF. 
 
 * On account of the importance of this proposition an independent proof is given. See Note 10. 
 
 122
 
 92 EUCLID, BOOKS V. AND VI. [127 
 
 Since BF and DC are parallel, 
 
 Add to each &ABF, 
 
 .-. &ADE : &ABC=&ADE : &ADF [Prop. 22. 
 
 = AE:AF. [Prop. 16. 
 
 But AE:AC = AD:AB 
 
 since the triangles ADE, ABC are similar. 
 
 And AD:AB = AC:AF 
 
 since DC is parallel to BF. 
 
 .-. AE : AC = AC : AF. 
 
 Hence by Prop. 42 
 
 square described on AE : square described on AC 
 =AE:AF 
 
 Now AE and AC are corresponding sides of the triangles ADE, ABC. 
 
 Consequently the ratio of the areas of two similar triangles is equal to the ratio 
 of the areas of the squares described on corresponding sides. 
 
 Art. 128. EXAMPLE 66. 
 
 If ABC be a triangle, and if BE, CF be drawn perpendicular to the sides AC, 
 AB respectively; prove that the triangle ABE is to the triangle ACF as the square 
 on A B is to the square on AC. 
 
 Art. 129. PROPOSITION XLIV. (i). 
 
 ENUNCIATION. If four straight lines are proportional, then the squares described 
 on them are proportional. 
 
 ABODE F Q H 
 
 Fig. 53.
 
 131] EUCLID, BOOKS V. AND VI. 93 
 
 Let A B, CD, EF, GH be four straight lines such that 
 
 AB:CD = EF: GH. (1) 
 
 It is required to prove that 
 
 sq. on AB : sq. on CD = sq. on EF : sq. on GH. 
 Construct straight lines X and Y such that 
 
 AB:CD = CD:X, (2) 
 
 EF:GH=GH:Y. (3) 
 
 It follows from (1), (2), (3) that 
 
 CD:X = GH:Y. (4) 
 
 From (1) and (4) by Prop. 37 it follows that 
 
 AB : X = EF : Y, 
 
 but AB : X = sq. on A B : sq. on CD [Prop. 42. 
 
 and EF:Y = sq. on EF : pq. on GH, [Prop. 42. 
 
 .'. sq. on AB : sq. on CD = sq. on EF : sq. on GH. 
 
 Art. 130. PROPOSITION XLIV. (ii). 
 
 ENUNCIATION. If four squares are in proportion, their sides will be in pro- 
 portion. 
 
 Let AB, CD, EF, GH be four lines such that 
 
 sq. on AB : sq. on CD = sq. on EF : sq. on GH. 
 
 Take a line X such that 
 
 AB:CD = EF:X. 
 
 Then it has been shown that 
 
 sq. on AB : sq. on CD= sq. on EF : sq. on X, [Prop. 44 (i). 
 .*. sq. on EF : sq. on GH = sq. on EF : sq. on X, 
 
 .'. sq. on GH = sq. on X, [Prop. 23. 
 
 /. GH=X. 
 .-. AB:CD = EF: GH. 
 
 Art. 131. PROPOSITION XLV. (Euc. V. 12.) 
 
 ENUNCIATION. // there be any number of equal ratios in which the magnitudes 
 are all of the same kind, then the ratio of any antecedent to its consequent is equal to 
 the ratio of the sum of the antecedents to the sum of the consequents ; i.e. if 
 
 A:B = C :D = E:F, 
 then A: B = A+C + E:B + D + F.
 
 94 EUCLID, BOOKS V. AND VI. [131 
 
 o 
 
 Compare A : B with the ratio - . 
 There are three possible alternatives, 
 
 (1) A:B>'-. 
 
 (2) A:B = 8 -, 
 
 (3) A:B< 8 -. 
 
 Of these it is not necessary to consider the second. [Art. 48. 
 
 o 
 
 In case (1) A : B >-, 
 
 r 
 
 /. rA > sB. [Prop. 13. 
 
 Since A : B = C : D = E : F, 
 
 /. C:D>-, 
 r 
 
 .'. rC > sD, [Prop. 13. 
 
 and E:F>-, 
 
 r 
 
 /. rE>sF, [Prop. 13. 
 
 but also rA > sB, 
 
 .'. rA + rC + rE > sB + sD + sF, 
 
 ~ r - [Prop. 13. 
 
 Hence if A : B > - , 
 
 r 
 
 then A + C+E: B + D + F>-. 
 
 r 
 
 *In like manner it can be shown that in case (3) if 
 
 s 
 r' 
 
 then A + C + E: B + D + F<-. 
 
 r 
 .'. A : B = A + C+E:B + D + F. [Art. 48. 
 
 * The demonstration of case (3) can be deduced from that of case (1) by replacing the sign > by 
 the sign < throughout.
 
 133] EUCLID, BOOKS V. AND VI. 95 
 
 Art. 132. EXAMPLE 67. 
 
 The perimeters of similar triangles (or similar rectilineal figures) are to one another 
 in the ratio of corresponding sides. 
 
 Art. 133. PROPOSITION XLVI. (Euc. VI. 20.) 
 
 ENUNCIATION. The ratio of the areas of two similar rectilineal figures is equal 
 to the ratio of the areas of the squares described on corresponding sides. 
 
 Let A 1 B l CiD 1 and A. 2 B^C^D^ be similar figures, and A l B l , A Z B 2 corresponding 
 sides. 
 
 To prove that 
 
 z = square on A^ : square on A Z B 2 . 
 
 D 
 
 Fig. 54. 
 
 Taking the figure of Proposition 33 it was proved that the triangles O l A 1 B l , 
 O^AzBz were similar, as were also O^B^ and 2 J?j(7 2 , 1 C 1 D 1 and 2 (7 8 A. OiA-^i 
 and Z D^A 2 . 
 
 Hence by Prop. 43 
 
 ^0 1 A 1 B 1 : A0 2 -4 2 5 2 = square on A l B l : square on A z B y> 
 A 1 B 1 C 1 : A 2J B 2 C z = square on B l C^ : square on B z (7 2 , 
 A Oi G 1 D l : A 2 (7 2 -D 2 = square on GI D v : square on (7 2 D 2 i 
 &0 1 D 1 A l : A 2 D 2 ^ 2 = square on -D^i : square on D Z A^. 
 
 But since the figures are similar 
 
 A^ : A A =B 1 C l : B 2 C 2 = C, D, : C,D 2 = A^i : A^ 2 . 
 
 Hence by Prop. 44 the ratios of the squares described on these lines are equal. 
 .'. A CMA : ACUA - A0AC7, : A0 2J B 2 3 = A
 
 96 EUCLID, BOOKS V. AND VI. [133 
 
 Hence by Prop. 45 
 
 = figure A 1 B 1 G 1 D 1 : figure A y B 3 OyD a . 
 
 .'. figure A i B 1 C 1 D l : figure A 2 B 3 C 2 D, t = square on A 1 B 1 : square on A t B 2 . 
 
 Art. 134. PROPOSITION XLVII. (i). (Euc. VI. 22, 1st Part.) 
 
 ENUNCIATION. Let there be four straight lines A, B t C, D which are in 
 proportion. 
 
 Let two similar rectilinear figures be similarly described on A and B. 
 Let two similar rectilinear figures be similarly described on C and D. 
 It is required to prove that 
 
 the area of the figure on A : the area of the figure on B 
 = the area of the figure on C : the area of the figure on D. 
 
 Let the areas of the similar figures similarly described on A and B be U 
 and V respectively. 
 
 V 
 
 A B C D 
 
 Fig. 55. 
 
 Then U:V= square on A : square on B. [Prop. 46. 
 
 Let the areas of the similar figures similarly described on C and D be W 
 and X respectively. 
 
 Then W : X = square on C : square on D. [Prop. 46. 
 
 Now A:B = C:D, 
 
 .'. the square on A : the square on B 
 = the square on C : the square on D. [Prop. 44 (i).
 
 136] EUCLID, BOOKS V. AND VI. 97 
 
 Art. 135. PROPOSITION XL VII. (ii). (Euc. VI. 22, 2nd Part.) 
 ENUNCIATION. Let there be four straight lines A, B, C, D. 
 Let two similar figures be similarly described on A and B. 
 Let two similar figures be similarly described on C and D. 
 
 Let it be given that 
 
 the area of the figure on A: the area of the figure on B 
 
 = the area of the figure on C : the area of the figure on D. 
 
 It is required to prove that 
 
 A:B=G:D. 
 
 Fig. 56. 
 
 Let the areas of the similar figures on A and B be U and V respectively. 
 Then U : V = the square on A : the square on B. [Prop. 46. 
 
 Let the areas of the similar figures on G and D be W and X respectively. 
 Then W : X = the square on C : the square on D. [Prop. 46. 
 
 It is given that U : V = W : X, 
 
 .'. the square on A : the square on B 
 = the square on C : the square on D. 
 
 .-. A:B=C:D. [Prop. 44 (ii). 
 
 Art. 136. PROPOSITION XLVIII. (Euc. VI. 31.)* 
 
 ENUNCIATION. In any right-angled triangle, any rectilineal figure described on 
 the hypotenuse is equal to the sum of the two similar and similarly described figures 
 on the sides. 
 
 Let ABC be a triangle right-angled at C. 
 
 On AB let any rectilineal figure Z be described. 
 
 * I am indebted to Mr H. M. Taylor, the author of the Pitt Press Euclid, and to the Syndicate of 
 the Pitt Press for their kind permission to use this proof, which is substantially the same as that given 
 in the Pitt Press Euclid. 
 
 H. E. 13
 
 98 
 
 EUCLID, BOOKS V. AND VI. 
 
 [136 
 
 On BC let a rectilineal figure X be described similar to Z so that the side 
 BC of X corresponds to the side AB of Z; and on AC let a rectilineal figure Y 
 be described similar to Z so that the side AC of Y corresponds to the side 
 AB of Z. 
 
 It is required to prove that 
 
 Z = X + Y. 
 
 Since Z and X are similar figures, and AB, BC are corresponding sides, 
 therefore by Proposition 46 
 
 Z : X = square on AB : square on BC. 
 In like manner 
 
 Z: Y= square on AB : square on AC. 
 
 Now Z, X, Y and the squares on AB, BC, 
 CA are all magnitudes of the same kind, viz. 
 areas. 
 
 .'. by Prop. 24 
 
 Z : square on AB = X : square on BC, Fig. 57. 
 
 and Z : square on AB = 
 .'. X : square on BC = 
 
 X : square on BC, 
 Y: square on AC, 
 Y: square on AC. 
 X : square on BC = X + Y: square on BC + square on AC 
 
 = X + Y : square on AB. 
 . Z: square on AB = X + Y: square on AB. 
 .'. Z = X+Y. 
 
 Art. 137. EXAMPLE 68. 
 
 [Prop. 45. 
 
 [Art. 43. 
 
 In an acute-angled triangle similar figures are similarly described on the sides, shew 
 that the sum of the areas of any two of them is greater than the area of the third. 
 
 Art. 138. PROPOSITION XLIX. (Euc. VI. 25.) 
 
 ENUNCIATION. To describe a rectilineal figure similar to one given rectilineal 
 figure and equal in area to another given rectilineal figure. 
 
 (In ordinary language to describe a figure having the shape of one given figure 
 
 and the size of another.) 
 
 Let it be required to describe a figure similar to the figure ABCDE and 
 equal to the figure FGHK.
 
 139] 
 
 EUCLID, BOOKS V. AND VI. 
 
 99 
 
 F G 
 
 Fig. 58. 
 
 On AB describe a rectangle ABLM equal to ABODE. 
 
 On BL describe a rectangle BLNO equal to FGHK. 
 
 Take PQ a mean proportional between AB and BO. [Prop. 35. 
 
 On PQ describe a figure PQRST similar to ABODE, so that PQ may 
 correspond to AB. [Prop. 32. 
 
 It will be shewn that PQRST is the figure required. 
 Since AB:PQ = PQ: BO, 
 
 .'. square on AB : square on PQ = AB : BO. [Prop. 42. 
 
 Now AB:BO = ABLM : BLNO [Prop. 16. 
 
 = ABODE: FGHK. 
 
 Also ABODE : PQRST = square on AB : square on PQ. [Prop. 46. 
 
 .-. ABODE: FGHK = ABODE: PQRST. 
 
 .'. FGHK = PQRST. [Prop. 23. 
 
 Hence PQRST is equal to FGHK and similar to ABODE. 
 It is therefore the figure required. 
 
 Art. 139. Def. 18. FIGURES WITH SIDES RECIPROCALLY 
 
 PROPORTIONAL. 
 
 A figure is said to have the two sides about one angle reciprocally 
 proportional to the two sides about an angle of another figure when these 
 four sides are proportional in the following manner: 
 
 a side of the first figure : a side of the second figure 
 = the other side of the second figure : the other side of the first figure. 
 
 132
 
 100 EUCLID, BOOKS V. AND VI. [140 
 
 Art. 140. PROPOSITION L. (i). (Euc. VI. 14, 1st Part.) 
 
 ENUNCIATION. Parallelograms having equal areas and having one angle of 
 the one equal to one angle of the other have the sides about the equal angles 
 reciprocally proportional. 
 
 The ratio of the areas of two equiangular parallelograms is equal to the ratio 
 of the areas of the rectangles contained by their sides. [Prop. 41. 
 
 In this proposition it is given that the areas of the parallelograms are equal. 
 Hence the rectangles contained by their sides are equal. 
 Hence by Prop. 38 (ii) 
 
 a side of the first rectangle : a side of the second rectangle 
 = the other side of the second rectangle : the other side of the first rectangle. 
 /. a side of the first parallelogram : a side of the second parallelogram 
 
 = the other side of the second parallelogram : the other side of the first 
 parallelogram. 
 
 Hence the sides about the equal angles are reciprocally proportional. 
 
 Art. 141. PROPOSITION L. (ii). (Euc. VI. 14, 2nd Part.) 
 
 ENUNCIATION. Parallelograms having one angle of the one equal to one angle 
 of the other, and the sides about the equal angles reciprocally proportional are equal 
 in area. 
 
 Since the sides about the equal angles are reciprocally proportional, therefore 
 by Prop. 38 (i) the rectangle contained by the sides of one parallelogram is equal 
 to the rectangle contained by the sides of the other. 
 
 Therefore by Prop. 41 the parallelograms have equal areas. 
 
 Art. 142. COROLLARY TO PROP. 50. (Euc. VI. 15.) 
 
 ENUNCIATION (i). Two triangles having equal areas and having one angle of 
 the one equal to one angle of the other have their sides about the equal angles 
 reciprocally proportional. 
 
 (ii) Two triangles which have one angle in the one equal to one angle in the 
 other and the sides about the equal angles reciprocally proportional are equal 
 in area. 
 
 These propositions may be deduced from Proposition 50 by constructing the 
 parallelograms of which the triangles are the halves.
 
 143] EUCLID, BOOKS V. AND VI. 101 
 
 Art. 143. EXAMPLES. 
 
 69. Triangles which have one angle in the one supplementary to one angle in the 
 other and their sides about the supplementary angles reciprocally proportional are equal 
 in area. 
 
 70. Triangles having equal areas and having one angle of the one supplementary 
 to one angle of the other, have their sides about the supplementary angles reciprocally 
 proportional. 
 
 71. If P be any point on the side AC of the triangle ABC, and if PQ be drawn 
 parallel to BC to cut AB at Q, then if a straight line through P cut BA produced 
 through A at R and BC at S so as to make the triangles ABC, BRS equal, prove that 
 QR will be a third proportional to QA and QB. 
 
 72. The triangles ABC, DEF are similar, and on DE the side corresponding to 
 AB a point K is taken so that DK is a third proportional to DE and AB, prove that 
 the triangles ABC, DKF are equal in area. 
 
 (This is the proposition on which Euclid's proof that the areas of similar triangles 
 are to one another as the squares described on corresponding sides is based.) 
 
 73. If a straight line DE be drawn parallel to the base BC of the triangle ABC 
 cutting AB at D and AC at E, and if AF be drawn perpendicular to DE, prove that the 
 rectangle AF . BC is double of the triangle AEB.
 
 SECTION IX. 
 
 MISCELLANEOUS GEOMETRICAL PROPOSITIONS. 
 PROPOSITIONS 5156. 
 
 Art. 144. PROPOSITION LI. (Euc. VI. B.) 
 
 ENUNCIATION. If the vertical angle of a triangle be bisected by a straight line 
 which also cuts the base, the rectangle contained by the sides of the triangle is equal to 
 the rectangle contained by the segments of the base together with the square on the 
 straight line which bisects the angle. 
 
 Let ABC be the triangle. 
 Bisect BAG by AD cutting the base BC at D. 
 It is required to prove that 
 rect. AB . AC= rect. BD . DC + square on AD. 
 
 Describe a circle round the triangle ABC, and 
 let AD cut the circle at E. 
 
 Join CE. 
 
 In the triangles ABD, A EC 
 
 BAD = EAC 
 
 ABD = AEC, since they stand on the same arc AC. 
 
 [Prop. 28. 
 
 [Prop. 38 (i). 
 
 Hence the triangles are similar. 
 
 .-. BD:EC = DA:CA = AB:AE. 
 From the second and third ratios 
 
 rect. AB . AC = rect. AD . AE 
 
 = square on AD + rect. AD . DE 
 = square on AD + rect. BD . DC.
 
 146] EUCLID, BOOKS V. AND VI. 103 
 
 Art. 145. PROPOSITION LII. (Euc. VI. C.) 
 
 ENUNCIATION. If from any vertex of a triangle a perpendicular be drawn to 
 the opposite side, the diameter of the circle circumscribing the triangle is a fourth 
 proportional to the perpendicular and the sides of the triangle which meet at that 
 vertex. 
 
 Let ABC be a triangle. 
 
 Let BD be perpendicular to AC. 
 
 Let a circle be circumscribed about ABC. 
 
 Let BE be the diameter through B. 
 
 Join CE. 
 
 Fig. 60. 
 
 In the triangles ABD, EBC 
 
 BEG = BAD, for they stand on the same arc BC. 
 
 A A 
 
 BCE = BDA, for each is a right angle. 
 /. CBE=ABD. 
 Hence by Prop. 28 the triangles are similar. 
 
 /. DB:CB = BA:BE=AD:EC. 
 
 From the equality of the first and second ratios it follows that the diameter 
 BE is a fourth proportional to the perpendicular BD and the sides BC, BA. 
 
 Art. 146. EXAMPLE 74. 
 
 If D is any point on the side BC of a triangle ABC, then the diameters of the circles 
 circumscribing the triangles ABD and ACD are proportional to the sides A, AC.
 
 104 
 
 EUCLID, BOOKS V. AND VI. 
 
 [147 
 
 Art. 147. PROPOSITION LIII. (Euc. VI. 30.) 
 
 ENUNCIATION. To divide internally or externally a finite straight line in 
 extreme and mean ratio; i.e. so that the whole line is to one segment as that 
 segment is to the other segment. 
 
 Let AB be the straight line, it is required to find a point C on it so that 
 
 AB:AG = AC: CB. 
 On AB describe the square ABDE. 
 Bisect AE at F. 
 Join FB. 
 
 C' 
 
 E K 
 
 H' G 
 
 Fig. 61. 
 
 On EFA measure FO = FG' = FB. 
 On A G describe the square A CHG. 
 On AG' describe the square AC'H'G'. 
 The points C, C' fall on AB, and are the required points. 
 It is proved in Prop. 11 of the Second Book of Euclid that 
 the square on AG = rect. AB . BC. 
 
 :. AB:AC = AG:BC. [Cor. to Prop. 38. 
 
 To prove the same property for the point G'. 
 Since AE is bisected at F and produced to 0', 
 
 /. square on FG' = square on FA + rect. EG' .AG'. 
 
 /. square on FB = square on FA + rect. EG'H'K'.
 
 149] EUCLID, BOOKS V. AND VI. 105 
 
 /. square on FA + square on AB = square on FA +rect. EG'H'K'. 
 .-. square on AB = rect. EG'H'K'. 
 
 :. AEDB = EG'H'K'. 
 
 :. AEDB + AEK'C' = EG'H'K' + AEK'C'. 
 .-. BDK'C' = AC'H'G'. 
 
 .'. rect. BA . BC' = square on AC'. 
 
 .'. AB:AC' = AC': O'B. [Cor. to Prop. 38. 
 
 Art. 148. EXAMPLE 75. 
 
 If ABC be a triangle right-angled at A, and AD be drawn perpendicular to the 
 hypotenuse cutting it at /), and if D divide BC in extreme and mean ratio, then prove 
 that the sides of the triangle ABC are in proportion. 
 
 Art. 149. PROPOSITION LIV. (Euc. VI. 24.) 
 
 ENUNCIATION. Parallelograms about the diagonal of any parallelogram are 
 similar to the whole and to one another. 
 
 Let A BCD be a parallelogram. 
 
 Let AEFH, FKCG be parallelograms about the diagonal AC of the 
 parallelogram A BCD. 
 
 f* t. 
 
 It is required to prove that they are 
 similar to ABCD and to one another. 
 
 Since EF is parallel to BC, 
 
 the triangles AEF, ABC are similar. 
 
 [Cor. to Prop. 28. 
 
 .-. AE:AB = EF:BC = FA:CA. 
 Since FH is parallel to CD, 
 
 the triangles AFH, ACD are similar. Fi 62 
 
 [Cor. to Prop. 28. 
 
 /. FA : CA = AH : AD = HF: DC. 
 
 Further HAE = DAB, 
 
 AEF = ABC, 
 EFH = BCD, 
 FHA = CDA. 
 
 H. E. 14
 
 106 EUCLID, BOOKS V. AND VI. [149 
 
 Hence the two sets of conditions for the similarity of AEFH, A BCD are 
 satisfied (Art. 77). 
 
 In like manner FKCG is similar to A BCD. 
 
 .-. AEFH, FKGG are similar. [Prop. 27. 
 
 Art. 150. PROPOSITION LV. (Euc. VI. 26.) 
 
 ENUNCIATION. If two similar parallelograms have a common angle and be 
 similarly situated they are about the same diagonal. 
 
 Let ABCD, AEFH be two similar parallelograms having the same angle A. 
 Let them be similarly situated, and let AB, A E B 
 
 AE be corresponding sides. f~ T"~ 
 
 It is required to prove that the diagonals / / 
 
 AC, AF coincide in direction. 
 
 Since the parallelograms are similar 
 
 ABC=AEF, 
 AB:AE = BC:EF. 
 
 .'. AB:BC=AE:EF. [Prop. 24. Fig. 63. 
 
 Hence the triangles ABC, AEF are similar. [Prop. 30. 
 
 In these triangles BC, EF are corresponding sides. 
 Hence the angles opposite them are equal. 
 
 /. BAC=EAF. 
 
 Hence AC coincides in direction with AF. 
 
 Hence the parallelograms ABCD, AEFH are about the same diagonal. 
 
 Art. 151. EXAMPLE 76. 
 
 Let the straight line AB be produced through A to P and through B to Q, so that 
 AP is equal to BQ. On BQ, BP let similar parallelograms be similarly described, 
 viz. BQRS and BPTU. Prove that the parallelogram whose adjacent sides are QA, QR 
 is equal in area to that whose adjacent sides are PA, PT.
 
 152] EUCLID, BOOKS V. AND VI. 107 
 
 Art. 152. PROPOSITION LVI. (Euc. VI. 27, 28, 29.) 
 
 ENUNCIATION. If OAB be a given triangle it is required to find a point P 
 on AB or AB produced so that if PQ be drawn parallel to OB to cut OA 
 in Q, and if PR be drawn parallel to OA to cut OB in R, then the parallelogram 
 PQOR may have a given area. 
 
 There are two kinds of cases. 
 
 Case I. Suppose the point P to have been found and to lie between A 
 and F, the middle point of AB. 
 
 Let E be the middle point of OA. 
 
 Complete the parallelogram EA VF. 
 
 Let QP cut FV in T. 
 
 Let PR cut EF in S and A V in U. 
 
 Then OQPR = OESR + EQPS 
 = EAUS+PUVT 
 = EAVF-SPTF o~~ E Q ^A 
 
 = &OAF-SPTF. Fig - 64 - 
 
 Hence if P be between A and F the area OQPR is less than the triangle 
 OAF. (This is equivalent to the result of Euc. VI. 27.) 
 
 Hence SPTF= &OAF- OQPR 
 
 = % (given triangle OAB) (a given area). 
 Hence the parallelogram SPTF has a known area. 
 It is also known to be similar to the known parallelogram EA VF. 
 
 Hence it can be constructed by Prop. 49, and if it be placed so that the side 
 corresponding to FV falls along FV, and the side corresponding to FE falls along 
 FE, then its diagonal will fall on FA by Prop. 55. 
 
 Hence the position of P is known. 
 (This is equivalent to the result of Euc. VI. 28.) 
 
 In order that the construction for P may be possible it is necessary that the 
 given area should not exceed half the given triangle OAB. 
 
 The above construction applies only to the case where P lies between A and F, 
 the middle point of AB. 
 
 142
 
 108 EUCLID, BOOKS V. AND VI. [152 
 
 If P be one position of the required point, let a point P' be taken on FB 
 so that PF= P'F, and let Q f , R', S', T' be the points corresponding to Q, R, S, T. 
 
 Then the parallelograms 8PTF, S'P'T'F are equal. 
 Hence the parallelograms OQPR, OQ'P'R' are equal. 
 Hence P is another position of the required point. 
 
 Case II. Let P be on BA produced through A, and let the same construction 
 be made. 
 
 Then OQPR = OESR + EQPS 
 
 = SPTF -EAVF 
 
 = SPTF -&OAF. 
 .'. SPTF=&OAF + OQPR 
 
 = (given triangle OAB)+ (a given area). 
 Hence the parallelogram SPTF has a known area. 
 It is also known to be similar to the known parallelogram EA VF. 
 
 Hence it can be constructed by Prop. 49, and if it be placed so that the side 
 corresponding to FV falls along FV, and the side corresponding to FE falls along 
 FE, then its diagonal will fall along FA by Prop. 55. 
 
 Hence the position of P is known. 
 
 (This is equivalent to the result of Euc. VI. 29.) 
 
 In this case the construction is always possible for all magnitudes of the given 
 
 area.
 
 152] EUCLID, BOOKS V. AND VI. 109 
 
 If P be one position of the required point, and a point P be taken on FB 
 produced through B so that P'F = PF, then it may be shown that P' is another 
 position of the required point. 
 
 It results from Cases I and II that if the given area be less than half the 
 given triangle OAB there are four solutions of the problem, viz. P may be 
 between A and F or between F and B, or on BA produced through A, or on AB 
 produced through B. 
 
 If the given area be equal to half the triangle OAB there are three solutions, 
 viz. P may be at F, or on BA produced through A, or on AB produced through B. 
 
 If the given area be greater than half the triangle OAB there are two 
 solutions, viz. P may be on BA produced through A, or on AB produced 
 through B.
 
 SECTION X. 
 
 THE REMAINING IMPORTANT PROPOSITIONS IN THE THEORY OF 
 RATIO, WITH GEOMETRICAL APPLICATIONS. 
 
 PROPOSITIONS 5764. 
 
 Art. 153. PROPOSITION LVII. (Euc. V. 23.) 
 
 ENUNCIATION. If A, B, C are three magnitudes of the same kind, 
 if T, U, V are three magnitudes of the same kind, 
 if A:B=U:V, 
 
 and B:C=T:U, 
 
 prove that A : C = T : V. 
 
 o 
 
 Compare A : G with any rational fraction - . 
 
 It is necessary to consider only the two cases [Art. 48. 
 
 (1) A:C>*. 
 
 r 
 
 (2) A:C<*. 
 
 r 
 
 In case (1) A : C > - , 
 
 r 
 
 /. rA > sC. 
 /. rA sC is a magnitude of the same kind as B.
 
 153] EUCLID, BOOKS V. AND VI. Ill 
 
 Hence by Archimedes' Axiom an integer n exists, such that 
 
 n (rA -sT)>B, 
 :. nrA > nsC + B. 
 Hence an integer t exists, such that 
 
 nrA >tB> nsC. [Prop. 8. 
 
 .'. nrA > tB, 
 
 .'. A:B>-; 
 nr 
 
 but A:B=U:V, 
 
 /. U:V> t 
 nr 
 
 .: nrU>tV. (I)* 
 
 Since tB > nsC, 
 
 p.p. 
 .. B.C>j, 
 
 but B:C=T:U, 
 
 . 
 
 /. tT>nsU\ (II) 
 
 .*. rtT>rnsU, 
 
 and from (I) snrU > stV, 
 
 .'. rtT>stV, 
 .'. rT>sV, 
 
 Hence if A:C>-, then T: V > - . 
 
 r r 
 
 In case (2) A : C < - , 
 
 r 
 
 .-. rA<sC. 
 .'. sC rA is a magnitude of the same kind as B. 
 
 * In the above proposition the inequalities (I) and (II) are transformed so that the multiples of U, 
 which occurs in both, are the same. The inequalities (III) and (IV) below are treated in the same way.
 
 112 EUCLID, BOOKS V. AND VI. [153 
 
 Hence by Archimedes' Axiom an integer n exists, such that 
 
 n (sG - rA) > B, 
 : . nsC > nrA + B. 
 Hence a multiple of B exists, say tB, such that 
 
 nrA <tB< nsG. [Prop. 8. 
 
 Since nrA < tB, 
 
 ,'. A:B<-- 
 nr 
 
 but A:B=U:V, 
 
 .-. U:V<-, 
 nr 
 
 .'. nrU<tV. (Ill) 
 
 Since tB < nsC, 
 
 K.n^ 118 
 . . x : L> < . 
 
 E 
 
 But B:C=T:U, 
 
 .:T:U<*, 
 
 .'. tT<nsU. (IV) 
 
 snrU < stV. 
 
 rtT< rnsU, 
 :. rtT<stV t 
 .'. rT<sV, 
 
 Hence if A:C<-, then 
 
 r 
 
 But if A:C>-, then T:F>-. 
 
 r r 
 
 Consequently A : C = T : V. [Art. 48.
 
 154] EUCLID, BOOKS V. AND VI. 113 
 
 Art. 154. PROPOSITION LVIII. (Euc. V. 18.) 
 
 ENUNCIATION. If two ratios are equal, the ratio of the sum of the antecedent 
 and consequent of the first ratio to the consequent of the first ratio is equal to the 
 ratio of the sum of the antecedent and consequent of the second ratio to the consequent 
 of the second ratio ; 
 
 i.e. if A:B = X:Y, 
 
 to prove that A + B:B = X+Y:Y. 
 
 g 
 Compare the ratio A + B : B with the rational fraction - . 
 
 There are three alternatives 
 
 (1) A+B:B> 8 -, 
 
 (2) 
 
 (3) A+B:B< S -. 
 
 In case (1) A+B:B>-, 
 
 .-. r(A+B)>sB. 
 
 In this case it is necessary to consider separately the cases 
 
 r < s, r = s, r >s. 
 
 If r < s, the last inequality can be written 
 
 rA > (s r) B. 
 
 r 
 but A:B = X:Y, 
 
 .-. rX>(s-r)Y, 
 .-. r(X+Y)>sY, 
 
 15
 
 114 EUCLID, BOOKS V. AND VI. [154 
 
 If r = , then rY=s7 
 
 and .'. r(X+Y)>sY, 
 
 .*. X+Y:Y>-. 
 r 
 
 If r>s, then rY > sY, 
 
 /. X+Y:Y>-. 
 r 
 
 In case (2) A+B:B = -, 
 
 .'. r(A+B)=sB. 
 This is impossible unless r < s. 
 
 Hence it can be written 
 
 rA = (s - r) B, 
 
 r 
 
 but A:B = X:Y, 
 
 s r 
 
 /. X:Y = 
 
 r 
 = (s-r)Y, 
 
 ,'.X+Y:Y= S -. 
 
 In case (3) A+B:B<-, 
 
 r 
 
 .'. r(A+B)<sB. 
 
 This is impossible unless r < s. 
 
 .'. rA<(s-r)B, 
 
 D S ~ r 
 
 .'. A :B < . 
 
 r 
 
 But A:B = X:Y, 
 
 Q __ ** 
 
 .-. X:Y< S -, 
 r 
 
 /. rX<(s-r)Y,
 
 156] EUCLID, BOOKS V. AND VI. 115 
 
 .-. r(X+Y)<sY, 
 
 /. X+Y:Y<-. 
 r 
 
 Hence, if A+B:B>-, then X+Y:Y>-, 
 
 r r 
 
 if A+B:B = -, then X+F:F=-, 
 
 r r 
 
 if A + B:B<-, then X+Y:Y<' -. 
 
 r r 
 
 Art. 155. EXAMPLES. 
 
 77. If ABC be a triangle right-angled at C, and if AD bisect the angle BAG 
 cutting BC at D, prove that 
 
 AC: CD = AC + AB-.SC. 
 
 (By means of this proposition Archimedes showed that the length of the circum- 
 ference of a circle was less than 3^- times its diameter.) 
 
 78. If ABC be a triangle right-angled at C, if AD bisect the angle BAG and cut 
 BC at D, and if BE be drawn perpendicular to AD cutting it at E, prove that 
 
 AE: EB = AC + AB-.BC. 
 
 (By means of this proposition Archimedes showed that the length of the circum- 
 ference of a circle was greater than 3^ times its diameter.) 
 
 Art. 156. PROPOSITION LIX. (Euc. V. 17.) 
 
 ENUNCIATION. If two ratios are equal, then the ratio of the difference of the 
 antecedent and consequent of the first ratio to the consequent of the first ratio is equal 
 to the ratio of the difference of the antecedent and consequent of the second ratio to the 
 consequent of the second ratio ; 
 
 i.e.if A:B = X:Y, 
 
 then A ~B:B = X~ Y:Y. 
 
 It is necessary to consider separately the cases 
 
 A>B, 
 A = B, 
 A<B. 
 
 152
 
 116 EUCLID, BOOKS V. AND VI. [156 
 
 If A > B, and therefore X > T, then it is required to prove that 
 
 A-B:B=X-Y: Y. 
 
 o 
 
 Compare A B :B with the rational fraction - . 
 
 r 
 
 There are three cases 
 
 (1) A-B:B>-. 
 
 (2) A-B:B = -. 
 
 r 
 
 (3) A-B:B<-. 
 
 r 
 
 Incase(l) A-B:B>-, 
 
 T 
 
 .'. r(A-B)>sB, 
 
 /. rA > (r+s)B, 
 
 r 
 but A:B = X:Y, 
 
 r 
 
 .'. rX > (r + s) F, 
 r(X-Y)>sY, 
 
 X-Y:Y>-. 
 
 In case (2) A-B:B=-, 
 
 r 
 
 .'. r(A-B) = sB, 
 
 .'. rA = 
 
 r 
 but A:B = X:Y, 
 
 Y V T+S 
 ~
 
 156] EUCLID, BOOKS V. AND VI. 117 
 
 /. r(X-Y) = sY, 
 
 /. X-Y:Y=-. 
 
 r 
 
 In case (3) A - B : B < - , 
 
 r 
 
 .-. r(A-B)<sB, 
 
 :. rA 
 
 r 
 But A : B = X : F, 
 
 .'. rX<(r + s)Y, 
 .'. r(X-Y)<sY, 
 
 .-. X-Y:Y<-. 
 r 
 
 Hence if A-B:B>-, then X-Y: Y>-, 
 
 r r 
 
 if A-B:B = -, then X-Y:Y = -, 
 
 r r 
 
 if A-B:B<-, then X- F:F<-. 
 
 r r 
 
 /. A-B:B = X-Y:Y. 
 HA=B, then X = Y. 
 
 Hence the difference of A and B, and the difference of X and F, are both zero. 
 
 Hence the first term of each of the ratios A B : B and X Y: Y is zero, and 
 the ratios may be considered to be the same. 
 
 If A <B, and therefore X < Y, then it is required to show that 
 
 B-A:B= Y-X:Y. 
 
 This may be proved independently as in the first case, or may be deduced 
 from it. 
 
 For if A:B=X:Y, 
 
 then B:A = Y:X. [Prop. 21. 
 
 .-. B-A:A = Y-X:X, by Case 1. 
 But A : B = X : Y. 
 
 .-. B-A:B= F-Z:F. [Prop. 37.
 
 118 EUCLID, BOOKS \ AND VI. [157 
 
 Art. 157. EXAMPLES. 
 
 79. Prove the last case of Proposition 59 directly in a manner similar to that 
 adopted for the first case. 
 
 80. If A : B= X : T, and A > B, prove that 
 
 81. If A : B = B : C, and A > B, prove that 
 
 A -C:A -B = B + C : B. 
 
 Art. 158. PROPOSITION LX. 
 
 ENUNCIATION. // A : B = X : Y, 
 
 prove that A ~ B:A + B = X ~ Y:X + Y. 
 
 If A:B = X:Y, 
 
 then A+B:B=X + Y:Y. [Prop. 58. 
 
 .'. B:A+B= Y:X+ Y. [Prop. 21. 
 
 Also A ~ B : B = X ~ Y: Y. [Prop. 59. 
 
 .-. A~B:A+B = X ~ Y:X + Y. [Prop. 37. 
 
 Art. 159. PROPOSITION LXI. 
 
 ENUNCIATION. If A, B, C, D be four harmonic points, A and C being conjugate, 
 and if be the middle point of AC, then OC is a mean proportional between 
 OB and OD. 
 
 A Q B C D 
 
 Fig. 66. 
 
 Since A divides BD in the same ratio as C does, 
 
 AB : AD = BC : CD. 
 .'. AB:BC=AD: CD. [Prop. 24. 
 
 .-. 205: 20(7= 20(7 :WD. 
 
 /. OB: OC =00: OD. [Prop. 15. 
 
 .' . OC is a mean proportional between OB and OD.
 
 162] EUCLID, BOOKS V. AND VI. 119 
 
 Art. 160. EXAMPLES. 
 
 82. Prove that if a circle be drawn through two points which are inverse with 
 regard to a second circle, then the two circles cut each other at right angles. 
 
 83. If A, B, C, D be four harmonic points, and if be the middle point of AC, 
 show that a circle can be drawn with centre so as to cut at right angles any circle 
 that can be drawn through B and D in the plane of the circle whose centre is 0. 
 
 84. If the diagonals AC, BD of the quadrilateral A BCD intersect at E and a 
 straight line EG be drawn parallel to one of the sides AB meeting the opposite side 
 CD in G and the third diagonal (i.e. the straight line joining H the intersection of AB 
 and CD to / the intersection of AD and BC) in J, then EJ is bisected at G. 
 
 [If EG cut AD in K and BC in L, prove that 
 
 AB:BH = KL: LJ= EL:LG= KE : EG.] 
 Hence by the aid of Ex. 50 show that HA, HE, HC, HI are four harmonic lines. 
 
 Art. 161. PROPOSITION LXII. (Euc. V. 24.) 
 
 ENUNCIATION. If A : C = X : Z, 
 
 and B:C=Y:Z, 
 
 prove that A+B:C = X+Y:Z. 
 
 Since B:C=Y:Z, 
 
 .-. C:B = Z:Y, [Prop. 21. 
 
 but A:C = X:Z, 
 
 .: A:B = X:Y, [Prop. 37. 
 
 .-. A + B:B = X+Y:Y, [Prop. 58. 
 
 but B:C=Y:Z, 
 
 .'. A+B:C = X+Y:Z. [Prop. 37. 
 
 Art. 162. PROPOSITION LXIII. (Euc. V. 4.) 
 ENUNCIATION. If A:B = X:Y, 
 
 to prove that rA:sB = rX:sY. 
 
 P 
 
 Compare the ratio r A : sB with any rational fraction - .
 
 120 EUCLID, BOOKS V. AND VI. [162 
 
 There are three cases 
 
 (1) rA:sB>, 
 
 (2) rA:sB = ?, 
 
 (3) rA:sB<. 
 
 In case (1) rA:sB>-, 
 
 .'. q(rA)>p( 8 B), 
 .'. qrA >psB, 
 
 qr 
 But A:B = X:Y, 
 
 qr 
 
 .'. qrX >psY, 
 .'. q(rX)>p(sY), 
 
 Hence if rA : sB >^ , then rX : 
 
 V V 
 
 In like manner it can be shown that 
 
 if rA:sB = , then rX:sY=, 
 
 V 9 
 
 and if rA : sB < %-, then rX : sY< 2 . 
 
 q q 
 
 /. rA :sB = rX:sY. 
 
 Art. 163. EXAMPLE 85. 
 
 Prove the converse of Proposition 63, viz. 
 If rA:aB = rX:sY, 
 
 then A:B = X: Y.
 
 165] EUCLID, BOOKS V. AND VI. 121 
 
 Art. 164. PROPOSITION LXIV. 
 
 ENUNCIATION. // K, L, M, P be four straight lines in proportion, if the 
 lengths of L and M be fixed, if the length of K can be made smaller than that 
 of any line however small, to show that the length of P can be made greater than 
 that of any line Q, however great Q may be. 
 
 By Archimedes' Axiom (Art. 22), it is always possible to find an integer r 
 
 such that 
 
 rM>Q. 
 
 Now divide L into r equal parts, and take K smaller than one of these 
 equal parts. 
 
 Then rK < L. 
 
 Now K:L = M:P. 
 
 .'. rK:L = rM:P. [Prop. 63. 
 
 Now rK < L. 
 
 :. rM<P. 
 
 But Q < rM. 
 
 :. Q<P. 
 
 Art. 165. EXAMPLE 86. 
 
 If K, L, M, P be four straight lines in proportion, if the lengths of L, M be fixed, 
 and if the length of K can be made greater than that of any line however great, show 
 that the length of P can be made smaller than that of any line Q however small. 

 
 SECTION XI. 
 
 OTHER PROPOSITIONS IN THE THEORY OF RATIO. 
 PROPOSITIONS 65, 66. 
 
 Art. 166. PROPOSITION LXV. (Euc. V. 19.) 
 
 ENUNCIATION. If A , B, C, D are magnitudes of the same kind, and 
 
 A:B=C:D, 
 
 prove that A ~ C :B ~ D = A :B. 
 
 A:B=C:D, 
 
 .-. A:C = B:D, [Prop. 24. 
 
 .-. A ~ C: C = B~D:D, [Prop. 59. 
 
 .-. A~C:B~D = C:D, [Prop. 24. 
 
 .-. A~C:B~D = A:B. 
 
 Art. 167. EXAMPLE 87. 
 
 If X, A, S, A' are four harmonic points, A and A' being conjugate, and if C be the 
 middle point of AA\ prove that 
 
 SA : AX=CS: CA = CA: CX. 
 
 Art. 168. PROPOSITION LXVI. (Euc. V. 25.) 
 
 ' ENUNCIATION. // four magnitudes of the same kind are proportional, then the 
 greatest and least of them together are greater than the sum of the other two. 
 
 Let A:B=G:D. 
 
 Suppose A the greatest of the four magnitudes.
 
 169] EUCLID, BOOKS V. AND VI. 123 
 
 Then A > B. 
 
 .-. A:B>\. 
 
 .-. C>D. 
 
 Again V A:B = C:D. 
 
 Now A > G. 
 
 .-. B>D. [Art. 70. 
 
 Hence D is the least of the four magnitudes. 
 
 Hence it is required to prove that 
 
 A+D>B + G. 
 
 Since A:B^C:D, 
 
 ... A-B:B = G-D:D. [Prop. 59. 
 
 .-. A-B:C-D = B:D. [Prop. 24. 
 
 Now B > D. 
 
 ... A-B>G-D. 
 
 Art. 169. EXAMPLE 88. 
 
 If three quantities be in proportion, show that the sum of the extremes will exceed 
 double the mean. 
 
 162
 
 NOTES. 
 
 Art. 170. NOTE 1. ON PROPS. 16, 15 AND ART. 31. 
 
 Props. 1 5 relate to certain simple cases of the application of the Com- 
 mutative, Associative and Distributive Laws, with which the reader who has 
 commenced elementary Algebra is already familiar. 
 
 Prop. I. r (A + B) = rA + rB. 
 
 Treating A + B as the multiplicand, 
 and r as the multiplier, 
 
 it is seen that the multiplicand is divided (or distributed) into its parts A, B. 
 
 Prop. 1 1. (a + b) R = aR + bR. 
 
 Treating a + b as the multiplier, 
 and R as the multiplicand, 
 
 it is seen that the multiplier is distributed into its parts a, b. 
 
 Prop. III. If A > B, 
 
 r(A-B) = rA- rB. 
 
 Here the multiplicand A B is distributed into its parts A, B. 
 
 Prop. IV. If a > b, 
 
 (a-b)R = aR- bR. 
 
 Here the multiplier a b is distributed into its parts a, b. 
 
 Prop. V. r (sA) = (rs) A = (sr) A =s (rA). 
 
 This illustrates both the Commutative and Associative Laws. 
 
 The fact (rs) A = (sr) A 
 
 illustrates the Commutative Law. 
 
 The fact that r(sA) = (rs) A 
 
 and the fact that (sr) A = s (rA) 
 
 both illustrate the Associative Law.
 
 172] EUCLID, BOOKS V. AND VI. 125 
 
 If Propositions 1 4, 6, 15 and Art. 31 be arranged in parallel columns, thus: 
 I. r(A+B) = rA + rB. II. (a-f b)R = aR + bR. 
 
 III. r(A-B) = rA-rB. IV. (a-b)R = aR-bR. 
 
 VI (i). If A^B, Vl(iii). If a 1 6, 
 
 then rA | rB. then aR | bR. 
 
 VI (ii). If rA^rB, VI (iv). If aR^bR, 
 
 then ^|. then a 1 6. 
 
 XV. A:B = nA:nB. Art. 31 a:b = aN:bN. 
 
 Then the two propositions in any one line are related to each other in such a 
 manner that magnitudes in either are replaced by whole numbers in the other. 
 
 NOTE 2. ON PROPS. I. AND II. 
 
 Art. 171. A more formal proof of Proposition I. will now be given. 
 
 The Commutative Law is 
 
 X+Y=Y+X. (I) 
 
 The Associative Law is 
 
 (X+ Y) + Z = X + (Y+Z). (II) 
 
 The argument will be followed more easily if the associative law be also written 
 X + (Y + Z) = (X+Y) + Z. (Ill) 
 
 When used in the form (II) the first term on the left X + Y is broken into 
 two parts, and the term Y is added to the second term Z on the left. 
 
 When used in the form (III) the second term on the left Y+Z is broken into 
 two parts, and the term F is added to the first term X on the left. 
 
 Art. 172. LEMMA. 
 
 To prove (rA + rB) + (A + B) = (r + 1) A + (r + 1 ) B. 
 
 Putting X = rA+rB, Y=A, Z=B iu (III), 
 
 (rA + rB) + (A + B) = [(rA + rB) + A] + B. 
 Putting X = rA, Y=rB, Z = A in (II), 
 
 [(rA + rB) + A] = rA + (rB + A ) 
 
 rB) by (I).
 
 126 EUCLID, BOOKS V. AND VI. [172 
 
 Putting X = rA, Y=A, Z = rB in (III), 
 
 rA + (A + rB) = (rA + A) + rB. 
 
 Putting X = (rA + A), Y= rB, Z = B in (II), 
 
 [(rA+A) + rB]+B = (rA+A) 
 
 .-. (rA+rB) + (A+B) = (rA+A) 
 
 But by Art. 3 rA+A=(r + l)A, 
 
 and 
 
 Art. 1 73. To prove r (A + B) = rA + rB. 
 
 Now since 2Z = X + X, 
 
 But by putting r = 1 in Art. 172 
 
 (A + B) + (A+B) = 2A + 25, 
 .-. 2 (A+ B) = 2A + 25. 
 Again since 3Z = 2X + X, 
 
 .-. 3(4 + B) = 2(^+J5)+ (A+B), 
 
 = 3^+35 by putting r = 2 in Art. 172. 
 Proceeding thus suppose it has been proved that for some positive integer t 
 
 t(A+B) = tA+ tB. 
 Since (t+I)X = tX + X, 
 
 = (tA + tB) + (A + B) by hypothesis 
 = (t + I)A + (t+l)B, 
 
 as is seen by putting r = t in Art. 172, 
 
 i.e. (t + 1) (A + B) = (t + 1) A + (t + 1) B. 
 
 Hence if r (A + B) = rA + rB, for any integral value of r, it is true for the 
 next integral value. 
 
 But it is true for r = 2 and r = 3, therefore it is true for r = 4, therefore for 
 r 5 and so on it is true for every positive integral value.
 
 175] EUCLID, BOOKS V. AND VI. 127 
 
 Art. 174. A more formal proof of Prop. II. will now be given. 
 To prove aR + bR = (a + b} R, 
 
 aR + R = (a + l)R by Art. 3, 
 /. (aR + R) + R = (a + l)R + R = (a + 2) R by Art. 3. 
 
 Now (aR + R) + R = aR + (R + R) by (II) of Art. 171 
 
 = aR + 2R, 
 
 by Art. 3. 
 
 Now (aR + 2R) + R = aR + (2R + R) by (II) of Art. 171 
 
 = aR + 3R, 
 
 Suppose it has been proved by successive steps that 
 
 aR + cR = (a + c) R. 
 
 Then (aR + cR) + R = (a + c)R + R = (a + c + 1) R. [Art. 3. 
 
 But (aR + cR) + R = aR + (cR + R) by (II) of Art. 171 
 
 Hence if aR + cR = (a + c) R, 
 
 then aR+(c+l)R = (a + c 
 
 Now it has been proved that 
 
 aR + 3R = (a + 3) R, 
 
 and so on it follows that 
 
 a# + bR = (a + 6) R, 
 
 where a, b are any positive integers. 
 
 Art. 175. NOTE 3. ON ART. 46. 
 
 Euclid's Eleventh Proposition of his Fifth Book is as follows : 
 
 If the ratio of A : B is the same as that of C : D, 
 
 and if the ratio of A : B is the same as that of E : F, 
 
 then the ratio of C : D is the same as that of E : F. 
 
 As remarked in the preface, if a ratio is treated as a magnitude, then this 
 merely expresses that if X = Y, and if X = Z, then Y = Z.
 
 128 EUCLID, BOOKS V. AND VI. [175 
 
 Euclid in his proof takes a different point of view ; which may be expressed 
 thus : 
 
 If the magnitudes A, B, C, D satisfy the conditions of the Fifth Definition of 
 the Fifth Book ; and if the magnitudes A, B, E, F also satisfy them, then will the 
 magnitudes C, D, E, F satisfy them. 
 
 The proof, stated in the manner of this book, will be as follows : 
 
 o 
 
 Take any rational fraction - . 
 Compare with it the ratio C : D. 
 
 o 
 
 Then (1) C : D may be greater than - , 
 
 o 
 
 or (2) C : D may be equal to - , 
 
 o 
 
 or (3) C : D may be less than - . 
 
 Take the case (1) C:D>-. 
 
 r 
 
 Then since A, B, C, D satisfy the conditions of Euclid's Fifth Definition 
 
 A:B>-, 
 r 
 
 and then since A, B, E, F satisfy the conditions of Euclid's Fifth Definition, 
 
 -. 
 r 
 
 - 
 r r 
 
 Hence if C: D>- , then E: F> 
 
 r 
 
 Similarly if G:D = -, then E:F = -, 
 
 r r 
 
 and if C:D<-, then E:F<-. 
 
 r r 
 
 :. C:D=E:F. [Art. 46. 
 
 Art. 176. NOTE 4. EUCLID'S DEFINITION OF RATIO. 
 
 Euclid's Definition of Ratio (the third Definition of the Fifth Book) is as 
 
 follows : 
 
 A6fyo<? eVrl 8vo peyedwv 6/J,o<yev(av ij Kara TrvjXiKOTrjTa Trpbs a\\tj\a TTOIO,
 
 177] 
 
 EUCLID, BOOKS V. AND VI. 
 
 129 
 
 De Morgan translates it thus : 
 
 " Ratio is a certain mutual habitude of two magnitudes of the same kind 
 depending on their quantuplicity." 
 
 The word "quantuplicity" which represents the Greek " irrf^iicor^ " is 
 especially difficult. It contains the idea of relative magnitude. 
 
 De Morgan defines Ratio as Relative Magnitude on page 63 of his Treatise on 
 the Connexion of Number and Magnitude. 
 
 Art. 177. NOTE 5. INCOMMENSURABLE MAGNITUDES. (See Art. 43.) 
 
 To prove that incommensurable magnitudes exist, it is sufficient to show this 
 in a particular case. 
 
 The case which will be selected is that of a side and a diagonal of a square. 
 Let X be the side, and F the diagonal of the square OABG. 
 
 Along the diagonal BO take BD = BA. Draw DE perpendicular to BO to 
 meet OA at E. 
 
 Then the right-angled triangles BE A, BED have the 
 hypotenuse BE in common, and one side BA =one side BD. 
 
 Hence they are congruent. 
 
 .-. EA=ED. 
 Again DOE is an isosceles right-angled triangle, 
 
 .-. ED = DO. 
 OB = BD + DO 
 
 Now 
 
 Now let 
 
 Call 
 
 Then 
 
 E 
 Fig. 67. 
 
 = OE+OD. 
 
 OB=Y. 
 
 Hence if X and F have a common measure, 
 
 Xi is measured by it, 
 and .*. FI is measured by it. 
 
 Hence if X and F have a common measure, this common measure also measures 
 X, and F,. 
 
 17
 
 130 EUCLID, BOOKS V. AND VI. [177 
 
 Now X l and Y l are the side and diagonal of a smaller square. 
 But this is a process which can be continued without limit. 
 
 It will now be shown that any common measure of X and Y is also a common 
 measure of the side and diagonal of a square, whose side can be made as small as 
 we please. 
 
 Suppose that by repeating the construction on the square whose side is X it 
 another square is obtained whose side is X 2 and diagonal F 2 and so on. 
 
 Now X = X,+ Y,, 
 
 but F 1 >Z 1 . 
 
 .-. X>2X 1 . 
 
 Similarly Z x > 2X 2 , 
 
 and so on. 
 
 Hence after repeating the process n times, it follows that 
 
 X > 2 n X n . 
 
 But 2 n = (1 + l) w > 1 + n > n, 
 
 .'. X > nX n . 
 
 Hence the common measure of X and F, if it exist, is not greater than X n and 
 
 
 .'. less than however large n may be. 
 
 Now if X y Y have a common measure G, it is always possible by Archimedes' 
 Axiom to find an integer n, such that 
 
 nG>X, 
 
 n X 
 
 .-. >-, 
 
 n 
 
 which is contrary to what is proved above. 
 Hence X and F have no common measure. 
 
 Hence the side and diagonal of a square have no common measure. 
 Hence incommensurable magnitudes exist. 
 
 Art. 178. NOTE 6. ON PROP. 24. 
 
 In order to complete the proof of Prop. 24 without using Prop. 14 it is necessary 
 
 to show directly that 
 
 if A:B=C:D, 
 
 and if rA = sC, then rB = sD.
 
 EUCLID, BOOKS V. AND VI. 
 
 If rA=sC, 
 
 then rA:B=sG:B. 
 
 Now A:B=G:D, 
 
 .'. rA:B = rC:D. [Prop. 63. 
 
 .-. sG:B = rG:D. 
 
 The required result follows from this by so altering the terms of the ratios that 
 the 1st term is the same in each. 
 
 Now sC:B = rsC: rB, [Prop. 15. 
 
 and rC:D = srC: sD. [Prop. 1 5. 
 
 . '. rsC : rB = srC : sD. 
 
 ,'. rB = sD. [Prop. 23. 
 
 ANOTHER PROOF. 
 Since A:B=C:D, 
 
 V rA:rB = A:B, [Prop. 15. 
 
 and sC:sD=C:D, [Prop. 15. 
 
 /. rA: rB = sG:sD. 
 But rA = sC, 
 
 .-. rB = sD. [Prop. 23. 
 
 Art. 179. NOTE 7. ON PROP. 26. 
 
 The 26th Proposition is a very suggestive one. It not only leads naturally 
 to the consideration of the point at infinity on a straight line, which is briefly 
 mentioned below, but also to the consideration of negative ratios (which are not 
 treated in this book). 
 
 It has been shown that if K : L is not a ratio of equality, then there is one way 
 of dividing AB internally and one way of dividing it externally in the ratio K:L. 
 
 Let the internal point of division be C, and the external point of division C", 
 then it appears from the proof of Prop. 26 that G and C' always lie on the same 
 side of 0, the middle point of AB. 
 
 Further it follows from Prop. 61 that 
 
 the rect. OC . OC' = the square on OA. 
 
 17-2
 
 132 EUCLID, BOOKS V. AND VI. [179 
 
 Now suppose that the length of K is fixed, and let the effect of diminishing 
 the length of L down to equality with that of K be investigated. 
 
 Then AC and CB tend to become equal, and therefore approaches 0, and by 
 making the difference of K and L sufficiently small the length OC may be made 
 smaller than any length however small ; and therefore by Prop. 64 the length of 
 00' can be made greater than any length however great. 
 
 Similar conclusions can be drawn from the case in which K is greater than L, 
 except that C and G' are on the same side of as B. 
 
 When however K = L, the internal point of division is the middle point, but 
 the external point of division does not exist from Euclid's point of view ; because 
 Euclid regards parallel straight lines as never meeting and so the construction 
 fails in this case. 
 
 Hence from Euclid's point of view it is impossible to state generally that to 
 every point C on AB between A and B there corresponds a point G' such that C' 
 divides AB externally in the same ratio as C divides AB, because there is no 
 point corresponding to the middle point of AB. 
 
 From the point of view of Modern Geometry in which a straight line is 
 supposed to have one point at infinity, when G is at the middle point of AB, 
 C' is at infinity ; and the theorem can be stated quite generally that there is 
 one way of dividing AB internally and one way of dividing it externally in any 
 given ratio K : L. 
 
 Art. 180. NOTE 8. ON ART. 91. 
 
 The contents of Art. 91 may perhaps be more easily appreciated by con- 
 sidering the following numerical case. 
 
 Consider the triangle whose sides are a, b, c ; and the triangle whose sides 
 k 2 k* 1? 
 
 Q T*i __ 
 
 J a' 1' c' 
 
 Then 6:c = ^ : | 2 ; 
 
 c 6 
 
 k 2 k* 
 
 c: a= : ; 
 a c 
 
 k* k 2 
 a:b= b : a> 
 
 so that any two sides of one triangle are proportional to some two sides of 
 the other.
 
 181] EUCLID, BOOKS V. AND VI. 133 
 
 But the two triangles do not satisfy the condition in the enunciation of 
 Prop. 29 implied in the words " taken in order." 
 
 k 2 
 
 E.g. whilst b corresponds to in the 1st proportion, 
 
 c 
 
 & 2 
 it corresponds to - in the 3rd proportion. 
 
 As a numerical example take a = 5, 6 = 3, c = 4 and A? = 60. 
 
 to i* M 
 
 Then -=12, 7= 20, - = 15; 
 a b c 
 
 and we have 3 : 4 = 15 : 20 
 
 4:5 = 12:15 
 
 3 : 5 = 12 : 20. 
 
 But since 5 2 = 3 2 + 4 8 , 
 
 but (20) 2 ^(12) 2 + (15) 2 ; 
 
 the first triangle is right-angled, the second is not ; and therefore the triangles are 
 not similar. 
 
 Art. 181. NOTE 9. ON PROP. 37. 
 
 In order to complete the proof of Prop. 37 without using Prop. 14 it is necessary 
 to show directly that 
 if A:B=T:U, 
 
 if B:C=U:V, 
 
 and if rA = sC, then rT = sV. 
 
 If rA = sC, 
 
 then rA:B = sC:B. 
 
 Since A:B=T:U, 
 
 .-. rA:B = rT:U. [Prop. 63. 
 
 Since B:C=U:V, 
 
 .-. B:sC=U:sV. [Prop. 63. 
 
 .-. sC:B = sV:U. [Prop. 21. 
 
 .-. \frA= sG, then rT = s V.
 
 134 
 
 EUCLID, BOOKS V. AND VI. 
 
 [182 
 
 Art. 182. NOTE 10. ON PROP. 43. 
 
 The areas of similar triangles are proportional to the areas of the squares 
 described on corresponding sides* 
 
 Let ABC, DEF be similar triangles. 
 
 Let AB, DE be corresponding sides. 
 
 On AB, DE describe the squares ABLK, DENM. 
 
 It is required to prove that 
 
 &ABC : &DEF = square ABLK : square DENM. 
 
 Q 
 
 R F 
 
 L 
 Fig. 68. 
 
 Draw CG perpendicular to AB, and FH perpendicular to DE. 
 
 Describe on AB the rectangle ABQP having the same altitude as the triangle 
 ABC, and on DE the rectangle DESR having the same altitude as the triangle 
 DEF. 
 
 The triangles AGO, DFH are similar, for 
 
 CGA = FED, 
 .-. ACG = DFH. 
 
 On account of the importance of this proposition an independent proof is given.
 
 183] EUCLID, BOOKS V. AND VI. 135 
 
 Hence the triangles are similar by Prop. 28. 
 
 .-. CG : CA = FH : FD, 
 
 .'. CG:FH=CA:FD, [Prop. 24. 
 
 but GA:FD = AB: DE, 
 
 since the triangles are similar ; 
 
 .-. CG:FH=AB:DE, 
 i.e. AP:DR = AK:DM, 
 
 .-. AP : AK = DR : DM, [Prop. 24. 
 
 .-. rect. ABQP : square ABLK = rect. DESK : square DENM, 
 .'. rect. ABQP : rect. DESR = square ABLK : square DENM. 
 
 [Prop. 24. 
 N ow rect. A BQP = 2^ ABC 
 
 and rect. D^Sfl = 2&DEF, 
 
 /. rect. ^15QP : rect. DESR = &ABC : &DEF, [Prop. 15. 
 
 .-. A ABC : &DEF= square ABLK : square DENM. 
 
 Art. 183. NOTE 11. ON PROP. 57. 
 
 (a) In order to complete the proof of Prop. 57 without using Prop. 14 it is 
 necessary to prove directly that 
 if A : B = U : F, 
 
 if B:C=T:U, 
 
 and if rA = sC, then rT = sV. 
 
 If rA = 5(7, 
 
 then rA:B = sC:B (I). 
 
 Now A:B=U:V, 
 
 .'. rA:B = rU:V (II). [Prop. 63. 
 
 Further B:C=T:U, 
 
 .-. C:B=U:T, [Prop. 21. 
 
 .-. sC:B = sU:T (III). [Prop. 63. 
 
 From (I), (II), (III) it follows that 
 
 rU:V = sU:T (IV). 
 
 In the last result it is possible to transform the terms of the ratios so that the 
 
 first term is the same in each. 
 
 rU:V = srU:sV (V), [Prop. 15. 
 
 sU:T=rsU:rT (VI). [Prop. 15.
 
 136 EUCLID, BOOKS V. AND VI. [183 
 
 From (IV), (V), (VI) it follows that 
 
 S rU:sV=rsU:rT. 
 But srU=rsU, 
 
 .'. sV=rT. [Prop. 23. 
 
 .-. if rA = sC, then rT = sV. 
 
 (ft) Propositions 37 and 57 are so nearly alike in form that the difference 
 between them should be carefully noted. 
 
 Arranging them in parallel columns : 
 
 Prop. 37. Prop. 57. 
 
 If A:B = T:U, If A:B=U:V, 
 
 and if B:C=U:V, and if B:C=T:U, 
 
 then A:C=T:V. then A:C=T:V, 
 
 it appears that the positions of the ratios T: U, U :V in Prop. 37 are interchanged 
 in Prop. 57. 
 
 Art. 184. NOTE 12. ON PROP. 59. 
 
 The statement on lines 18 and 19 of Page 117 that two ratios in which the first 
 term is zero may be considered to be the same may present some difficulty, inasmuch 
 as a ratio presupposes the existence of two magnitudes, and the ratio of zero to a 
 magnitude has not been defined. 
 
 Without going fully into the subject, which here touches upon the difficulties 
 of the Infinitesimal Calculus, it may be sufficient to remark that since 
 
 A 
 
 : A = A : nA, [Prop. 15. 
 
 71 
 
 and v A :nA =- , 
 
 n 
 
 A -A- 1 
 
 -L 
 
 n n 
 
 A 
 Now imagine the integer n to increase without limit, then tends to the 
 
 // 
 
 ^ 
 
 limit zero, and therefore the ratio : A is one in which the first term tends to 
 
 n 
 
 the limit zero, whilst at the same time the ratio, viz. -, tends to the limit zero. 
 
 n
 
 184] EUCLID, BOOKS V. AND VI. 137 
 
 In this connection the following proposition is of interest. 
 
 If the ratio of A to R is the same as that of C to D ; if A can be made as small 
 as we please, and if B and D be fixed magnitudes, then C can be made smaller than 
 any magnitude E, however small E may be. 
 
 It is possible by Archimedes' Axiom to choose n so that 
 
 nE>D- 
 
 D 
 
 and by hypothesis A can be taken smaller than , 
 
 71 
 
 i.e. nA < B, 
 
 1 
 
 n' 
 
 But A:B = C:D, 
 
 1 
 
 n' 
 
 .'. nC<D, 
 
 .'. nC<nE, 
 
 .'. C<E. 
 
 So that when the first term of the ratio of A to B tends to zero, so also does 
 the first term of the ratio of C to D. 
 
 Another proposition of a similar kind is this : 
 
 If the ratio of A to B is the same as that of C to D ; if A can be made as small 
 as we please, if C and D be fixed magnitudes, then B can be made smaller than any 
 magnitude E, however small E may be. 
 
 It is possible by Archimedes' Axiom to choose n so that 
 
 nC>D. 
 1 
 
 n ' 
 
 Then since A:B=C:D, 
 
 .'. A : B > - , 
 n 
 
 .'. nA > B. 
 Now by hypothesis A can be made as small as we please. 
 
 H. E. 
 
 18
 
 138 EUCLID, BOOKS V. AND VI. [184 
 
 Choose therefore 
 
 E 
 
 A<-, 
 n 
 
 i.e. nA < E. 
 
 .'. B<E. 
 
 Hence if the ratio of A to B be given, and one term tend to zero, so does 
 the other. 
 
 Art. 185. NOTE 13. THE COMPOUNDING OF RATIOS. 
 
 The 37th Proposition is a very important one. 
 
 It is as follows : 
 
 if A:B=T:U, 
 
 and B:C=U:V, 
 
 then A:C=T:V. 
 
 Euclid describes the connection of the ratio A : C with the ratios A : B and 
 B : C by saying that the first ratio is compounded of the other two. 
 
 He uses the result in connection with Prop. 40, in which the ratio of the areas 
 of two equiangular parallelograms is determined. 
 
 This terminology has not been employed in the text, because it is found 
 difficult by beginners. OD the other hand the object of the terminology, so far as 
 the determination of the ratio of the areas of two equiangular parallelograms 
 is concerned, is completely attained by the mode in which Prop. 40 is proved. 
 
 But the value of the terminology extends beyond its use in this proposition. 
 It will therefore be considered more at length. 
 
 Art. 186. The development of the process of compounding ratios is made in 
 four stages. 
 
 STAGE 1. When it is necessary to determine the relative magnitude of two 
 magnitudes, A and C, of the same kind, it is often convenient not to make 
 the comparison directly, but indirectly by taking another magnitude B of the 
 same kind as A and C ; and then comparing A with B, and afterwards B with C. 
 
 From this point of view the relative magnitude of A and C is considered to be 
 determined by the relative magnitude of A and B and the relative magnitude of 
 B and C.
 
 186] EUCLID, BOOKS V. AND VI. 139 
 
 STAGE 2. Euclid expresses the general idea stated in the first stage by saying 
 that the ratio of A to C is compounded of the ratio of A to B and the ratio of 
 BtoC. 
 
 STAGE 3. Def. 19. THE PROCESS OF COMPOUNDING RATIOS. 
 
 Let the ratios to be compounded be P : Q and T : U. 
 Take any arbitrary magnitude A, and then find B so that 
 
 P:Q = A:B (I), 
 
 and then find C so that T: U=B:C (II). 
 
 Then the ratio compounded of P : Q and T : U is the ratio compounded 
 of A : B and B : C, and is therefore A : C by the statement in the second 
 stage. 
 
 This process* contains an arbitrary element, viz. A. 
 
 STAGE 4. In order to. justify the process described in the preceding stage, it 
 is necessary to show that the presence of the arbitrary element in the third stage 
 has no influence on the value of the resulting ratio. 
 
 Suppose that instead of A, the magnitude A' had been selected, and that 
 B' and C' had then been found so that 
 
 P:Q = A':B' (Ill), 
 
 T:U=B':C' (IV). 
 
 Then the resulting ratio would be that compounded of A ' : B' and B' : C', 
 and would therefore be A' : C'. 
 
 In order that this may agree with the previous result, it is necessary to 
 
 show that 
 
 A:C=A':C' (V). 
 
 From (I) and (III) 
 
 A:B=A':R (VI). 
 
 From (II) and (IV) 
 
 B-.C^R-.C' (VII). 
 
 From (VI) and (VII) by Prop. 37, the proportion (V) follows. 
 Hence the process in the third stage always leads to the same value of the 
 resulting ratio, whatever be the value of the arbitrary element. 
 
 This is the justification of the process described in the third stage. 
 
 * It should be noted that this process assumes the existence of B and C, when A has been chosen 
 arbitrarily, the proof of which depends on the Fundamental Proposition in the Theory of Ratios 
 
 (Art. 215). 
 
 182
 
 140 EUCLID, BOOKS V. AND VI. [187 
 
 Art. 187. ARITHMETICAL APPLICATION OF THE PROCESS FOR 
 
 COMPOUNDING RATIOS. 
 
 To compound the ratio r : s with the ratio u : v where r, s, u, v are positive 
 integers. 
 
 r : s = ru : su 
 
 : s = ru : su } , 
 
 [ by Prop. 15. 
 
 : v = su : sv ) 
 
 u 
 
 Hence r : s compounded with u : v 
 
 
 
 = ru : su compounded with su : sv 
 = ru:sv 
 by the definition of the ratio compounded of other ratios in Art. 186, Stage 2. 
 
 T 
 
 Now r : s = - ; 
 
 u:v=-; \ [Art. 31. 
 
 ru 
 
 and ru:sv = . 
 
 sv J 
 
 TTU 7* 
 
 Observe further that - ' is defined to be the Arithmetical Product of 
 
 sv s 
 
 at 
 
 and - , the result being written 
 v 
 
 r u ru 
 - x - = . 
 
 S V SV 
 
 Hence this arithmetical theorem corresponds to the theorem that 
 r : s compounded with u:v = ru: sv. 
 
 Art. 188. Def. 20. DUPLICATE RATIO. 
 
 If a ratio be compounded with itself the resulting ratio is called the 
 duplicate ratio of the original ratio. 
 
 Thus if A : B be compounded with A : B, the resulting ratio is called the 
 duplicate ratio of A : B.
 
 191] EUCLID, BOOKS V. AND VI. 141 
 
 Art. 189. PROPOSITION LXVII. 
 
 ENUNCIATION. // three magnitudes be in proportion the first has to the third 
 the duplicate ratio of the first to the second. 
 
 Let A : B = B : C. 
 
 Then if A : B be compounded with A : B, the result is the same as if A : B be 
 compounded with B : C, and is therefore A : C. [Art. 186, Stage 2. 
 
 Hence A : C is the duplicate ratio of A : B, if A : B = B:C. 
 
 Art. 190. PROPOSITION LXVIII. 
 
 ENUNCIATION. If two ratios be equal, their duplicate ratios are also equal. 
 
 If A:B=C:D .................................... (1), 
 
 it is required to prove that the duplicate ratio of A : B is equal to that of C: D. 
 
 Take E so that A:B = B:E .................................... (2), 
 
 and Fso that G:D = D:F .................................... (3). 
 
 Then by (1), (2), (3) B:E=D:F .................................... (4). 
 
 Hence from (1) and (4) A:E=C: F, [Prop. 37. 
 
 But by (2) A : E is the duplicate ratio of A : B [Prop. 67. 
 
 and by (3) C : F is the duplicate ratio of C : D, [Prop. 67. 
 
 .'. the duplicate ratio of A : B is equal to the duplicate ratio of C: D. 
 
 Art. 191. EXAMPLES. 
 
 89. Using the symbol # as an abbreviation for the words "compounded with," 
 prove that 
 
 (i) (A:B)*(C'.D)=(C'.D)*(A-.B). 
 
 (ii) [(A:B)*(C:D)]*(E:F) 
 
 = (A:B)*[(C:D)*(E:F)]. 
 
 90. Prove that (A :B) *[(B: A) * (C :D)] = C : D. 
 
 Hence show how to find the ratio which must be compounded with A : fi to give 
 the ratio C : D. 
 
 91. Prove that [(A : B) * (C : D)] *(D:C) = (A: B).
 
 142 EUCLID, BOOKS V. AND VI. [191 
 
 92. If E:C = A:P, 
 
 E:D = :Q, 
 prove that (A : B} * (C : D] = (P : Q}. 
 
 93. What is the result of compounding any ratio with a ratio of equality ? 
 What is the result of compounding a ratio of equality with any ratio ? 
 
 94. If a ratio be compounded with its reciprocal show that the result is a ratio 
 of equality. 
 
 95. If A, , C, D be magnitudes all of the same kind, prove that 
 
 A : B compounded with C : D 
 gives the same result as 
 
 A : D compounded with C : B. 
 
 96. Prove that A : B compounded with C : D gives L : B where D : C = A : L. 
 
 97. If A, B, C, D are all magnitudes of the same kind, and 
 
 D : A compounded with D : B 
 is equal to D : C, find the relation between A, B, C, D. 
 
 If A, , C, D are all magnitudes of the same kind, and if A : D compounded with 
 B : D give C : D, find the relation between A, B, G, D. 
 
 98. (i) What ratio must be compounded with A : C to give B : C ? 
 (ii) What ratio must be compounded with C : A to give C : B 1 
 
 99. If the duplicate ratio of A : B be equal to the duplicate ratio of C : D, then prove 
 that A:B=C:D. 
 
 100. (i) Prove the theorem of Menelaus ; viz. if a straight line A'B'C' cut the sides 
 of the triangle ABC, viz. BC in A', GA in B', AB in C' ; then the ratio compounded of 
 
 the ratios 
 
 BA':A'G, CB':BA, AC' : C'B 
 is a ratio of equality. 
 
 (ii) Prove the converse of the theorem of Menelaus : if the sides of the triangle 
 A'B'C' be divided, BG in A', CA in B', AB in C", so that two of the points of division 
 are internal and one external, or else all three are external ; and if further the ratio 
 compounded of 
 
 BA' : A'C, CB 1 : BA, AC' ' : C'B 
 
 is a ratio of equality, then the points A', ', C' lie on one straight line. 
 
 101. Prove that the six centres of similitude* of three circles lie three by three 
 on four straight lines (called the axes of similitude of the circles). 
 
 [Apply the last example, taking A, B, C at the centres of the circles.] 
 
 * See Ex. 37, p. 73.
 
 193] EUCLID, BOOKS V. AND VI. 143 
 
 102. (i) Prove the theorem of Ceva, viz. : 
 
 If O be a point in the plane of the triangle ABC, if AO cut BC at A', if BO cut CA 
 at B', if CO cut AB at C", then the ratio compounded of BA' : A'C, Gff : HA, AC' : C'B 
 is a ratio of equality. 
 
 (ii) Prove the converse of the theorem of Ceva, viz. : 
 
 If the sides of the triangle ABC be divided, BC at A', CA at B', AB at C' t so that 
 two of the points of division are external and one internal, or else all three are internal, 
 and if further the ratio compounded of BA':A'C, CB 1 : B 1 A, AC': C'B is a ratio of 
 equality, then the three straight lines A A', BB', CC' are concurrent. 
 
 Art. 192. Def. 21. CROSS OR ANHARMONIC RATIO. 
 
 If A, B, C, D be four points on a straight line, they determine six 
 segments on that line. 
 
 Take any one of these six segments, say JSD. 
 
 Then A divides it in the ratio AB:AD; and C divides it in the ratio 
 CB:CD. 
 
 Then the ratio which must be compounded with either of these ratios 
 to produce the other is a value of the cross or anharmonic ratio of the 
 four points*. 
 
 Art. 193. EXAMPLES. 
 
 103. If A, B, (7, D be four points on a straight line, and any point not on 
 that straight line, and if through B a straight line be drawn parallel to OD to cut OA at 
 E and OC at F, then prove that 
 
 (CB : CD) # (BE : BF) = (AB : AD). 
 
 104. By means of the preceding example prove that if four fixed straight lines 
 passing through a point be cut by any fifth straight line, then the cross-ratio of the four 
 points of intersection is independent of the position of the fifth straight line. 
 
 105. (i) If A, B, C are three points arranged in this order on a straight line L ; 
 if A', B', C' are any three points arranged in this order on another straight line L' ; if 
 the points A, B, C be joined to any point ; then prove that it is possible to draw a 
 transversal A"B"C" across the lines OA, OB, OC such that 
 
 * This definition is sufficient for solving the problems set in this book. It is not however 
 a complete one as the signs of the segments have not been specified. When the signs are specified, 
 it can be shown that there are six values of the cross-ratio, all of which are determined when any 
 
 one is given. One of these six values is (AB : AD) -)(- (CD : CB) and is usually written - -
 
 144 EUCLID, BOOKS V. AND VI. [193 
 
 (ii) If through a point any four straight lines be drawn, and these be cut by any 
 transversal at A, B, C, D ; if the straight lines joining any point O l to A, B, C, D be 
 cut by any transversal at A lt B,, C lt D,; if the straight lines joining any point 2 to 
 ^u -#i> Ci, A be cut by any transversal at A 2 , B 2 , C 2 , D 2 ; and if this process be repeated 
 any number of times, so that four points A n , B n , C n , D n are finally obtained ; then 
 prove that it is possible to draw a transversal A'B'G'D' across OA, OB, OC, OD such that 
 
 A'ff = A n B n , B'C' = B n C n , C'D' = C n D n . 
 
 (This is the geometric property which is unaltered by projection and corresponds to 
 the constancy of the anharmonic ratio of four points on a straight line.) 
 
 106. If A!, l} d be any three points on a straight line ; and A 2 , B 2 , C 2 any three 
 points on another straight line, show how to determine points D l , D 2 on the two straight 
 lines so that the cross-ratio of A 1} B lt C^ D^ shall be equal to that of A 2 , B 2 , C 2 , D 2 , 
 [On the straight line A t A 2 take any two points O iy 2 . Let 0^, 2 B 2 meet at B; 
 let 0$!, 2 C 2 meet at C ; then show that the points D 1 , Z> 2 are such that O^D 2 D 2 
 intersect on BC.~\ 
 
 107. If ABC be a triangle, if D be the middle point of BC, if any straight line 
 through D cut AB at E, AC at F, and a parallel through A to BC at G, then find 
 the value of the cross-ratio of E, D, F, G. 
 
 108. If A, B, G, D be four fixed points on a circle, and P, Q any two other points 
 on the same circle, prove that the cross-ratio of the four straight lines PA, PB, PC, PD 
 is equal to that of the four straight lines QA, QB, QC, QD ; and also to that of the four 
 points in which the tangent at P to the circle is cut by the tangents at A, B, C, D. 
 
 109. Let the points P, P' on the straight line OX be said to correspond when the 
 rectangle OP . OP' is equal to a given rectangle. Then prove that the cross-ratio of any 
 four points is equal to the cross-ratio of their corresponding points. 
 
 110. If from any point two tangents be drawn to a circle, the points of contact 
 and the points of intersection of any secant from the same point are such that the 
 straight lines joining them to any fifth point on the circle form a harmonic pencil. 
 
 111. (i) Through any point a tangent OU and a secant ORS are drawn to a 
 circle ; OPQ is another secant passing through the centre of the circle (P, Q being 
 the extremities of a diameter). Show that if QR, QU, QS cut the tangent at P at 
 R', U', S', respectively, then 
 
 PR' : PU' = PU' : PS'. 
 
 (ii) If the point be inside the circle, and U be taken as the extremity of the 
 shortest chord through 0, and the rest of the construction be as above, show that 
 
 PR' :PU'
 
 194] EUCLID, BOOKS V. AND VI. 145 
 
 Art. 194. NOTE 14. THE AGGREGATING OF RATIOS. 
 
 On proposition 62 depends the process, called in this book the Aggregating 
 of Ratios, which corresponds to the addition of the measures of the ratios. 
 
 The development of this process is made in four stages. 
 
 STAGE 1. The general idea at the root of the process of aggregating ratios 
 is this : 
 
 When it is desired to find the ratio of one magnitude to a second, it is 
 permissible to break up the first magnitude into two parts, then to find the 
 ratio of each part to the second magnitude, and then to add the two ratios 
 thus found. 
 
 (It should be carefully noticed that it is the first magnitude, not the second, 
 which may be broken up.) 
 
 STAGE 2. To make the general idea stated in the first stage quite precise the 
 following definition is necessary. 
 
 Let the ratio X + Y : Z be said to be aggregated from the ratios X : Z and 
 Y : Z. It is known when the magnitudes X, Y, Z are known. 
 
 (This may be compared with Euclid's 22nd Datum.) 
 
 Let the symbol placed between two ratios denote that they are to be 
 aggregated. 
 
 Then X+ Y:Z = (X :Z) ~ (Y : Z). 
 
 STAGE 3. In the second stage the two ratios which are aggregated both have 
 the same second term, and therefore do not at first sight appear^ to be entirely 
 independent. 
 
 It is necessary therefore to explain what is meant by aggregating any two 
 ratios, i.e. two ratios whose terms are all independent. 
 
 Def. 22. THE PROCESS OF AGGREGATING RATIOS. 
 
 Let the ratios to be aggregated be A : B and C : D. 
 
 Take any arbitrary magnitude Z. 
 
 Then find* two others X and Fsuch that 
 
 C:D=Y:Z. 
 
 * This assumes the Fundamental Proposition in the Theory of Ratios (Art. 215). 
 H. E.
 
 146 EUCLID, BOOKS V. AND VI. [194 
 
 Theii (A:B)~(C:D) 
 
 = (X:Z)~(Y:Z) 
 = X+Y:Z. 
 
 STAGE 4. The form of the resulting ratio found in the third stage depends 
 on the value of the arbitrary magnitude Z. If the process is to be of any use it is 
 necessary to show that the value of the resulting ratio does not depend on the 
 value of Z. 
 
 This will be accomplished when it is shown that if any other magnitude 
 be taken, say Z' t instead of Z> and the process repeated, then the value of the 
 resulting ratio is unaltered. 
 
 Let therefore X', Y' be found so that 
 
 A:B=X':Z f , 
 
 C:D= Y':Z'. 
 Then (A:B)~(C:D) 
 
 Since A:B = X .Z, 
 
 and A:B = X':Z', 
 
 .; X:Z = X':Z f . 
 
 Since C:D=Y:Z, 
 
 and C:D=Y':Z f , 
 
 .'. Y:Z = Y':Z'. 
 
 Since X:Z=X':Z', 
 
 and Y:Z = Y':Z', 
 
 .-. X+Y:Z=X'+Y':Z'. [Prop. 62. 
 
 i.e. the value of the resulting ratio is unaltered. 
 
 This is the justification of the process, and shows that it always leads to 
 consistent results. 
 
 Art. 195. EXAMPLES. 
 
 112. Prove that 
 
 (A:E)~(C:D) = (C:D)~(A:B}. 
 
 113. Prove that
 
 198] EUCLID, BOOKS V. AND VI. 147 
 
 Art. 196. ARITHMETICAL APPLICATION OF THE PROCESS OF 
 AGGREGATING RATIOS. 
 
 If r, s, u, v are integers, prove that 
 
 (r : s) * (u : v) = (vr + us : vs). 
 
 This corresponds to the Arithmetical Theorem - + - = vr + us 
 
 [_ S V V8 
 
 Now r:s = vr:vs [Prop. 15. 
 
 and u : v = us : vs. [Prop. 15. 
 
 .* . (r : s) (u : v) 
 = (vr : vs) ~ (us : vs) 
 = vr + us:vs. 
 
 Art. 197. EXAMPLE 114. 
 Prove that [(A : B) ~ (C : D)] * (E : F ) 
 
 Art. 198. NOTE 15. INTRODUCTION OF THE IRRATIONAL 
 NUMBER INTO ANALYSIS* 
 
 There are two direct operations which can always be performed on integers, 
 viz. addition and multiplication; and the results are always positive integers. 
 
 If however an attempt is made to reverse these processes, the result is not 
 always a positive integer, e.g. if any two positive integers a and b being given, 
 it is desired to find the number, which when added to a will give b, the result will 
 be a positive integer only when b is greater than a. 
 
 Again if it is desired to find the number which when multiplied by a will 
 give b, the result will be a positive integer only when b is an integral multiple 
 of a. 
 
 In order that it may be always possible to reverse the process of addition it is 
 necessary to invent the negative integer, and in order that it may be always 
 possible to reverse the process of multiplication, it is necessary to invent the 
 positive rational fraction. 
 
 * The ideas on which this note is based are due to Dedekind, as stated in the Preface. 
 
 192
 
 148 EUCLID, BOOKS V. AND VI. [198 
 
 If two positive rational fractions are added together, the result is always a 
 positive rational fraction ; but if it is desired to reverse the process of addition of 
 positive rational fractions, it is necessary to invent the negative rational fraction. 
 
 The positive integers, the negative integers, the positive rational fractions and 
 the negative rational fractions, are together said to form the system of rational 
 numbers, and the system may be denoted by the letter R. 
 
 ,1J A* 
 
 Art. 199. If-, - be any two rational numbers, then it is known that 
 q s 
 
 p _ps 
 q~qs' 
 
 r _qr 
 s qs' 
 
 There are now three possibilities. 
 
 (1) ps>qr, 
 
 (2) ps = qr, 
 
 (3) ps < qr. 
 
 ps qr 
 
 In case (1) > , 
 
 qs qs 
 
 . >?: 
 
 ' q s' 
 
 Incase (2) ! = , 
 
 qs qs 
 
 p _r 
 ~ ' 
 
 Incase (3) <, 
 
 qs qs 
 
 /V) IV* 
 
 In case (2), - and - are not regarded as essentially distinct. 
 
 Art. 200. The first property of rational fractions to be proved is that 
 
 if >r, 
 
 q s' 
 
 r t 
 and if - > - , 
 
 s u
 
 202] EUCLID, BOOKS V. AND VI. 149 
 
 then - > - 
 
 q u 
 
 Since - > - 
 
 q s' 
 
 .'. ps > qr. (I) 
 
 o- r t 
 
 Since - > - 
 
 s u 
 
 /. ur > ts. (II) 
 
 .-. by (II) pur>pts, 
 
 .'. pur> t(ps). (HI) 
 
 Now from (I) t (ps) > t (qr). (IV) 
 
 Hence from (III) and (IV) pur > tqr, 
 
 .'. pu>tq. 
 
 Art. 201. A geometrical analogy to this proposition is as follows: 
 
 If A, B, C be three points on a straight line, and if A be to the right of B and 
 if B be to the right of C, then A is to the right of C. 
 
 Art. 202. The second property of the system of rational numbers to be proved 
 is that between any two distinct rational numbers, another rational number 
 always exists. 
 
 nf\ ft* 
 
 Let - and - be two distinct rational numbers, then the rational number 
 q s 
 
 *-!. " lies between them. 
 
 2qS 
 
 p r 
 Suppose - > - , 
 
 .'. ps>qr. 
 It is to be proved that 
 
 p ps + rq r 
 
 q 2qs s ' 
 
 To prove - > , . it is necessary to show 
 q zqs 
 
 which is the case. 
 
 *2ps >ps + rq, 
 ps > rq,
 
 150 EUCLID, BOOKS V. AND VI. [202 
 
 r ... 
 
 To prove ^ ^ * > - . it is necessary to show 
 2qs s 
 
 ps + rq> 2rq, 
 
 .'. ps >rq, 
 which is the case. 
 
 Art. 203. A geometrical analogy to the above result is as follows : 
 
 If A and B be any two distinct points on a straight line, then another point 
 G lies between them. 
 
 T 
 
 Art. 204. Now let - be any positive rational number. Then all the numbers 
 
 belonging to the system of rational numbers, viz. R may be separated into two 
 groups or classes as follows. 
 
 Into the first class, which may be called the lower class and which may be 
 
 V 
 
 denoted by R 1} let there be put all the positive rational numbers less than - , and 
 
 S 
 
 all the negative rational numbers. 
 
 Into the second class, which may be called the upper class, and which may be 
 
 7* 
 
 denoted by R 2 , let there be put all the rational numbers greater than - . 
 
 s 
 
 T 
 
 Let - be put into either class, it does not matter which. 
 s r 
 
 7* 
 
 If - be put into R lt it is the greatest number in R lf 
 s 
 
 T 
 
 If - be put into R 2 , it is the least number in R 2 . 
 s 
 
 In either case the separation of the system of all the rational numbers R into 
 the two classes R lt RZ is such that every number of the lower class R l is less than 
 every number of the upper class R 2 . 
 
 Art. 205. The following is a geometrical analogy to the preceding. 
 
 If P be a point in any straight line then all the points in the straight line may 
 be separated into two classes P,, P 2 , as follows. 
 
 The first class, P 1} contains all the points that lie on one side, say the left, of P. 
 The second class, P 2 , contains all the points that lie on the right of P. 
 The point P itself may be put into either class, it does not matter which. 
 
 If P be put into the class P x , then the class Pj has a point, viz. P, which is the 
 farthest to the right, but the class P 2 has no point which is farthest to the left.
 
 207] EUCLID, BOOKS V. AND VI. 151 
 
 If P be put into the class P 2 , then the class 1\ has a point, viz.P, which is the 
 farthest to the left, but the class Pj has no point which is farthest to the right. 
 
 In either case the separation of all the points on the straight line into the 
 two classes P u P a is of such a nature that every point of the first class P, is on the 
 left of every point of the second class P 2 . 
 
 Art. 206. It will now be shown that it is possible to separate the system of 
 all the rational numbers E into two classes such that 
 
 (a) every number in one class (the lower class) is less than every number 
 in the upper class ; 
 
 (6) the lower class has no greatest number ; 
 (c) the upper class has no least number. 
 Note carefully the distinction between this enunciation and that of the 
 
 nt* 
 
 separation of the system of rational numbers effected by means of - , in which 
 
 s 
 
 either the lower class had a greatest number, or the upper class had a least 
 number. 
 
 Art. 207. Algebraic illustration of the separation of the system of rational 
 numbers into two classes possessing the characteristics (a), (6) and (c) of Art. 206. 
 
 Let D be a rational number, which is not the square of any rational number. 
 
 Let all positive rational numbers whose squares are greater than D be placed 
 in the upper class. 
 
 Let all positive rational numbers whose squares are less than D, and all 
 negative rational numbers, be placed in the lower class. 
 
 It will be proved that the upper class has no least number, and that the lower 
 class has no greatest number. 
 Let x be any rational number. 
 
 a* + 
 
 Then, since x and D are rational, it follows that y is also rational. 
 From (I) it follows that
 
 152 EUCLID, BOOKS V. AND VI. [207 
 
 Now choose any number x in the upper class, so that 
 
 a?>D, 
 
 then y < x, by (II), 
 
 and y*>D, by (III). 
 
 Hence y is in the upper class, but is less than x. 
 
 Hence the upper class has no least number. 
 
 Nvext let x be in the lower class 
 
 :. y>x, by (II), 
 
 and y*<D, by (III). 
 
 Hence y is in the lower class, but is greater than x. 
 Hence the lower class has no greatest number. 
 
 The numbers in the lower class are all less than the numbers in the upper 
 class. 
 
 Since no rational fraction exists whose square is equal to D, it follows that all 
 the rational fractions have been separated into two classes, which possess the 
 characteristics (a), (6), (c) of Art. 206. 
 
 Art. 208. Geometric illustration of the separation of the whole system of 
 rational numbers into two classes, possessing the characteristics (a), (6), (c) of 
 Art. 206. 
 
 Let A and B be two segments of straight lines, which have no common 
 measure. 
 
 
 
 Let - be any rational fraction. 
 r 
 
 Let Q be a magnitude such that B = rQ. 
 
 Now let the magnitude nQ be compared with A. 
 
 Then A cannot be equal to nQ, for then Q would be a common measure of 
 A and B, which is assumed not to be the case. 
 
 Hence either A < nQ, or A > nQ. 
 If A < nQ, 
 
 then A:B<nQ:B, [Art. 43. 
 
 /. A:B<nQ:rQ.
 
 209] EUCLID, BOOKS V. AND VI. 153 
 
 71 
 
 This being so, let - be assigned to the upper class. 
 
 But if A > nQ, 
 
 then A:B>nQ:B, 
 
 /. A:B>nQ:rQ, 
 
 r. W 
 
 .'. A :B> -. 
 r 
 
 1 
 
 In this case let - be assigned to the lower class. 
 
 Hence every rational fraction falls into one of the two classes. 
 Those in the upper class are greater than A : B. 
 Those in the lower class are less than A : B. 
 
 Hence every rational fraction in the lower class is less than every rational 
 fraction in the upper class. 
 
 Art. 209. It will next be proved 
 
 (i) that the lower class contains no greatest number. 
 (ii) that the upper class contains no least number. 
 
 (i) Let - be any rational number in the lower class. 
 
 Choose Q so that B = rQ. 
 
 Then by the preceding argument A > nQ. 
 
 Since A is not a multiple of Q, it must lie between two consecutive multiples 
 of Q. 
 
 Let sQ < A < (s + 1) Q. 
 
 Now n cannot be greater than s, for sQ is the greatest multiple of Q which is 
 less than A. 
 
 If n be less than s, then since 
 
 B = rQ, A>sQ. 
 
 Therefore - is a rational number in the lower class, which is greater than - . 
 r 
 
 Li: !).-!: 
 
 If however n = s, then 
 
 A-sQ<Q. 
 
 Choose t so that 
 
 t(A-sQ)>Q. 
 
 20
 
 154 EUCLID, BOOKS V. AND VI. [209 
 
 Let V be a magnitude such that 
 
 Q=tV, 
 
 .: t(A-stV)>tV, 
 .'. A-stV> V, 
 
 but B 
 
 .'. A:B>(st 
 
 st + 
 
 .'. A:B> 
 
 rt 
 
 st + l . . ., , 
 .*. - - is in the lower class ; 
 rt 
 
 st + l st 
 
 but r- > - , 
 
 rt rt 
 
 st + l s_ 
 ~rtT > r' 
 
 st + 1 n 
 rt r' 
 
 /n 
 
 Hence in each case a rational number greater than - exists which is in the 
 lower class. 
 
 Hence the lower class has no greatest number. 
 
 ft 
 
 (ii) Let - be any rational number in the upper class. 
 
 Then if B = rQ, 
 
 A<nQ. 
 As before let 
 
 sQ<A<(s + l)Q. 
 Then n cannot be less than (s + 1). 
 If n>s + l. 
 
 mi s + I n 
 
 Then - < - , 
 
 r r 
 
 j s + I . 
 and is in the upper class. 
 
 But if n = (s + l), 
 
 then since (s + 1 ) Q A is less than Q,
 
 210] EUCLID, BOOKS V. AND Vt. 155 
 
 find t so that t [(s + 1) Q - A] > Q. 
 
 As before let Q = tV, 
 
 .'. t[(s+l)tV-A]>tV, 
 (s+l)tV-A> V, 
 
 Now 
 
 .'. A :B<[(s+l)t- l]V:rtV, 
 
 (s + l)t-I 
 . , A. : n < 
 
 ri 
 
 ( + !)*- 1 . 
 
 Hence - is m the upper class. 
 
 rt 
 
 Now 
 
 rt rt 
 
 rt 
 
 - w 
 r r* 
 
 Hence a rational number smaller than - has been found, which is in the 
 
 r 
 
 upper class. 
 
 Hence the upper class has no least number; and the lower class has no greatest 
 number. Also every number in the lower class is less than every number in the 
 upper class. 
 
 Hence the existence of the two incommensurable magnitudes A, B renders it 
 possible to separate the whole system of rational numbers into two classes 
 possessing the characteristics, marked (a), (6), (c) in Art. 206. 
 
 Art. 210. Correspondence between the rational numbers and the points on a 
 straight line. 
 
 If we select any origin on the line, and a unit of length, then to the rational 
 
 number - there will correspond a point P whose distance from is equal to 
 
 S 
 
 - units of length, and which is on a definite side of 0, previously chosen. In the 
 
 S 
 
 same way, the negative rational numbers will correspond to points on the line on 
 the other side of 0. 
 
 202
 
 156 EUCLID, BOOKS V. AND VI. [210 
 
 Thus all the positive and negative rational numbers correspond to definite 
 points on the straight line. But it is not true conversely that all the points on the 
 straight line correspond to rational numbers. For example, if the side of a square 
 be taken as the unit of measurement, and a length OP be measured from the 
 origin on the line equal to the diagonal of the square, then to the point P on the 
 line no rational number will correspond, because as has been shown (Note 5, 
 Art. 177), the side and diagonal of a square are incommensurable. 
 
 Now it may be proved that there are an infinitely great number of lengths, 
 which are incommensurable with the unit of length. 
 
 Therefore the straight line has an infinite number of points which do not 
 correspond to the rational numbers. 
 
 If therefore it is desired to construct a number which shall correspond to each 
 point on a straight line, then it is clear that the system of rational numbers will 
 not suffice, and it is therefore necessary to invent new numbers, so that the system 
 of numbers invented shall possess the same degree of completeness as the straight 
 line. This property of completeness is called the continuity of the straight line. 
 
 The preceding comparison of the system of rational numbers R with the 
 points on a straight line has led to the recognition of the fact that whilst all 
 rational numbers correspond to points on a straight line, all points on a straight 
 line do not correspond to rational numbers. 
 
 Consequently there exist gaps in the system of rational numbers ; there is an 
 incompleteness or discontinuity in this system ; whilst the straight line is con- 
 sidered to be free from gaps and to possess completeness or continuity. 
 
 It is necessary to explain in what this continuity consists. 
 
 It was shown in Art. 205 that every point P of a straight line separates the 
 straight line into two parts such that every point of one part lies to the left of 
 every point of the other part. 
 
 The essence of continuity consists in the converse of the above, viz. in the 
 following principle. 
 
 " If all points of the straight line fall into two classes such that every point of 
 the first class lies to the left of every point of the second class, then there exists 
 one and only one point which produces this separation of all points of the straight 
 line into two classes." 
 
 This principle is known as the Cantor-Dedekind Axiom. It cannot be proved, 
 It is the Axiom by means of which we attribute continuity to the straight line. 
 
 Compare this with the following statement regarding the system of rational 
 numbers.
 
 210] EUCLID, BOOKS V. AND VI. 157 
 
 " If all rational numbers are separated into two classes such that every number 
 of the first class is less than every number of the second class, then only when the 
 first class has a greatest number or the second class a least number does there exist 
 a rational number which produces this separation of all the rational numbers into 
 two classes." 
 
 We can now proceed to complete the discontinuous system of rational numbers 
 so as to obtain therefrom a continuous system. 
 
 Just as negative and fractional numbers are invented and introduced into 
 analysis, and as the laws of operating with these numbers may be reduced to the 
 laws of operating with positive integers, so it is necessary to invent irrational 
 numbers and to define them by means of the rational numbers alone. 
 
 It appears from the preceding discussion that if any rule be given for separating 
 all rational numbers into two classes such that every number in the lower class 
 is less than every number in the upper class, the following are the possible 
 alternatives : 
 
 (i) the lower class has a greatest number and the upper class has no 
 least number ; 
 
 (ii) the upper class has a least number and the lower class has no 
 greatest number ; 
 
 (iii) the lower class has no greatest number and the upper class has no 
 least number. 
 
 r 
 Suppose that in case (i) the greatest number in the lower class is - . 
 
 . S . y ** 
 
 7* 
 
 If this number - is transferred to the upper class, there will be formed another 
 
 S 
 
 separation of all the rational numbers into two classes such that every number in 
 the lower class is less than every number in the upper class, but this separation of 
 the system of rational numbers is regarded as essentially the same as the one from 
 
 7* 
 
 which it was produced by the transfer of - from the lower to the upper class. 
 
 S 
 
 7* 7* 
 
 Both separations are regarded as produced by - ; and - is the number which 
 
 S S 
 
 corresponds to the separation. 
 
 In case (iii) the separation is not produced by any rational number. It is 
 regarded as produced by an irrational number, which is said to correspond to the 
 separation. 
 
 Dedekind called a separation a ' cut.' It will be called here a ' section.' 
 
 It is in this property that all sections of the system of rational numbers cannot
 
 158 EUCLID, BOOKS V. AND VI. [210 
 
 be produced by rational numbers that the incompleteness or discontinuity of the 
 system of rational numbers consists. 
 
 An irrational number is regarded as known, whenever a rule is given for 
 determining whether any given rational number belongs to the lower or upper 
 class of the section of the system of rational numbers to which the irrational 
 number corresponds because all the properties of the irrational number can be 
 deduced from a knowledge of the numbers contained in these two classes. 
 
 It is usual to denote the irrational number itself by a single symbol. 
 
 Hence to every section of the system of rational numbers there corresponds a 
 definite rational or irrational number. 
 
 For example, the irrational number corresponding to the section described in 
 Arts. 208 and 209 is equal to the ratio of A to B. 
 
 Two numbers are regarded as different or unequal only when they correspond 
 to essentially different sections. 
 
 The aggregate of all the rational and irrational positive and negative numbers 
 is called the aggregate of real numbers. 
 
 Art. 211. Operations with real numbers. 
 
 To reduce any operation with two real numbers a, ft to operations with rational 
 numbers, it is necessary to show how from the section (A lt A 2 ), which defines 
 o, and the section (5 U B z ) which defines ft, to define another section (C^, (7 2 ) 
 which is to correspond to the result of the operation. 
 
 As examples consider the cases of addition and multiplication of two irrational 
 numbers a, ft. 
 
 *Let pi be any number in A lt 
 
 ^2 ..................... -Ba- 
 
 it will be proved that by a suitable choice of these numbers it is possible 
 to make 
 
 and 
 
 as small as we please. 
 
 Write down in order of magnitude the rational fractions whose denominator is n. 
 
 There must be some stage at which we pass from the lower to the upper class 
 of a. 
 
 * In this and some of the following articles small letters with suffixes denote rational fractions.
 
 212] EUCLID, BOOKS V. AND VI. 159 
 
 m m 
 
 Let - < a < 
 
 n n 
 
 Divide the interval ( j into t equal parts, and consider the fractions 
 
 whose denominators are tn, viz. 
 
 tm tm + 1 tm + t 
 
 tn' tn ' ...... tn ' 
 
 The first belongs to the lower class of a, the last belongs to the upper class 
 of a. 
 
 Hence an integer k exists, such that 
 
 tm + k tm + k+1 , 
 
 - < a < - - . (* + 1 5 ft 
 
 tn tn 
 
 In like manner we can find 
 
 r n r + 1 
 
 . 
 
 tn tn 
 
 Take now 
 
 tm + k tm + k + 1 
 
 P 1= -toT> p * = -* 
 
 rt + l rt + 1 + 1 
 
 tm + tr + k + l 
 
 Now k < t, 
 
 1+1*2 t, 
 
 m + r + 2 1 
 .'. p*q,-piqi< tf-- 'H' 
 
 Now suppose m, r, n to be fixed ; then by increasing t without limit, it follows 
 that (p 3 + q z )-(pi + qi) and pq t - ptfi can each be made as small as we please. 
 
 Art. 212. Now consider the case of addition. 
 
 Form all possible sums of the form p 3 + q 2 , and put every rational number which 
 is greater or equal to any sum of this form into a class, which may be called the 
 upper class.
 
 160 EUCLID, BOOKS V. AND VI. [212 
 
 Form all possible sums of the form p l + q lt and put every rational number which 
 is less or equal to any sum of this form into another class, which may be called the 
 lower class. 
 
 Let Si be a number of the lower class. 
 
 Let s 2 be a number of the upper class. 
 
 Then it is always possible to find p it p^, q 1} q 2 such that 
 
 But 
 
 .'. Sj< S 2 . 
 
 Hence every number in the lower class is less than every number in the upper 
 class. 
 
 It will next be shown that the two classes include all the rational numbers with 
 the possible exception of one rational number. 
 
 If possible let there be two distinct rational numbers a, b which belong to 
 neither class. 
 
 If a do not belong to the upper class, a is less than every sum of the form 
 Pi + q*- 
 
 If a do not belong to the lower class, a is greater than every sum of the form 
 Pi + qi- 
 
 Hence if a belong to neither class p^ + q l < a < p z + q z . 
 
 Similarly p l + q l < b < p 2 + <&> 
 
 '. (p-2 + ? 2 ) - (Pi + q 1 )>a~b. 
 
 But it has been shown that (p. 2 + q 2 ) (p l + q^) can be made as small as we 
 please. Hence it can be made smaller than a ~ 6, which is contrary to what has 
 been proved. 
 
 Hence two distinct rational numbers, each of them belonging to neither class. 
 cannot exist. 
 
 It is therefore proved that there can exist at most one rational number which 
 belongs to neither class, but it is not proved that there is one such rational number. 
 
 Consequently there may be no rational number which belongs to neither class. 
 
 If one rational number exist which does not belong to either class, it is greater 
 than all the numbers in the lower class and less than all the numbers in the upper 
 class. Put it into either class, and a section will be defined by the two classes so 
 formed. The number corresponding to this section is the single rational number 
 just mentioned, and it is called the sum of a and ft.
 
 214] EUCLID, BOOKS V. AND VI. 161 
 
 If no rational number exist, which does not belong to either class, then all the 
 rational numbers belong to either the lower or upper class. Hence the two classes 
 define a section. The number corresponding to this section is an irrational number. 
 This irrational number is defined to be the sum of a and $. 
 
 Art. 213. Next consider the case of multiplication. 
 
 Now the upper class consists of all the rational numbers which are greater or 
 equal to any product of the form p^q z . 
 
 The lower class consists of all the rational numbers which are less or equal to 
 any product of the form p^. 
 
 The demonstration may be obtained from the demonstration in the case of 
 addition by replacing throughout the word ' sum ' by the word ' product ', p^ + q t by 
 Piqi, and p z + q z by p^. 
 
 Art. 214. On the Compounding of Ratios. 
 Let the ratio of A to B be equal to a-. 
 Let the ratio of B to C be equal to /9. 
 
 Let pi*, p 2 be rational numbers in the lower and upper class of a. 
 Let q ly q 2 be rational numbers in the lower and upper class of $. 
 Then p l B<A<p. 2 B, 
 
 q l C<B<q,C, 
 .'. p l q l C<p l B 
 <A 
 
 < A 
 
 * If p.= ,and A : B>p lt 
 
 r 
 
 then ^ :B>- t 
 
 . .: rA>sB, 
 
 i * B 
 
 :. A> . 
 
 r 
 
 This is also written ^ > ^-* 
 
 If therefore A : B >j>, , then 
 
 21
 
 162 EUCLID, BOOKS V. AND VI. [214 
 
 Hence if the ratio of A to C be equal to 7, then every p 2 q y and all greater 
 numbers are in the upper class of 7, whilst every p^ and all smaller numbers 
 are in the lower class of 7. 
 
 Consequently 7 = a/8, by Art. 213. 
 
 Hence if A : B = a, and if B : C= /3, 
 
 then A :C=a/3. 
 
 Art. 215. PROPOSITION LXIX. 
 
 (The Fundamental Proposition in the Theory of Ratio.) 
 
 ENUNCIATION. If A, B be two magnitudes of the same kind, and if C be any 
 third magnitude, to prove that there exists a fourth magnitude Z, such that 
 
 A:B=C:Z. 
 
 It is sufficient to prove that there exists a magnitude Z such that Z : C = B : A, 
 (1) If B and A be commensurable, le't 
 
 Then it is necessary to find Z, so that 
 
 r 
 Z:C=-. 
 
 s 
 
 rC 
 Form the magnitude . 
 
 Then ^ :C = S ( ) : sG [ pr P- 15 ' 
 
 = rC:sC 
 
 r 
 
 = . 
 s 
 
 rC 
 
 Hence Z = . 
 
 s 
 
 (2) If B and A have no common measure, let B : A be equal to the irrational 
 number p. 
 
 Let p lt pi, p", ... represent rational numbers in the lower class of p in 
 ascending order of magnitude. 
 
 Let p 2 , p 2 ', p 2 ", ... represent rational numbers in the upper class of p in 
 descending order of magnitude.
 
 215] EUCLID, BOOKS V. AND VI. 163 
 
 Let the magnitudes 
 
 piC, Pl 'C, P1 "C, .... 
 
 be constructed. 
 
 Then Pl C < Pl '0 < Pl "C < . . . < pJ'C < p*C < p,C. 
 
 Consider now the set of all the magnitudes of the same kind as C. They fall 
 into two classes. 
 
 (i) Those which are greater than every magnitude of the form Pl G. Call any 
 magnitude of this form a magnitude F. 
 
 (ii) Those which are not greater than every magnitude of the form Pl C. 
 Call any magnitude of this form a magnitude X. 
 
 In the first place it will be proved that every magnitude Y is greater than 
 every magnitude X. 
 
 From the definition of the magnitudes X, it appears that any X, say X lt does 
 not exceed every magnitude of the form Pl C. 
 
 Suppose that X^ does not exceed the magnitude Pl 'C. 
 
 Then by the definition of the magnitudes Y, every Y exceeds every Pl G and 
 therefore every F exceeds P i'C. 
 
 But X l does not exceed P i'G. 
 
 Therefore F exceeds X^ 
 
 Therefore every F exceeds every X. 
 
 In the second place it will be shown that the set of magnitudes X include no 
 greatest magnitude. 
 
 The characteristic of the magnitudes X is that they are not greater than 
 every magnitude Pl C. 
 
 Suppose X' one of the magnitudes X, and let X' ^ p\G. 
 
 Now there is no greatest Pi . 
 
 Suppose Pl ' < Pl ", 
 
 then P1 'C< PI "C. 
 
 Take then X" = P i"C, which is possible. 
 
 .-. X">X' 
 and so on other magnitudes X can be found in increasing order of magnitude. 
 
 Thus the magnitudes X include no greatest magnitude. 
 
 Now the magnitudes X and F together include all the magnitudes of the same 
 kind as C. In regard to these magnitudes we assume an axiom corresponding 
 to the Cantor-Dedekind Axiom for the straight line, as follows: 
 
 212
 
 164 EUCLID, BOOKS V. AND VI. [215 
 
 " If all the magnitudes of the same kind as G be separated into two classes 
 such that every magnitude of the one class is less than every magnitude of the 
 other class, then there is one and only one magnitude of the same kind as G which 
 produces this separation, and it is either the greatest magnitude of one class, or 
 the least magnitude of the other class." 
 
 Now we have proved that the magnitudes X include no greatest magnitude. 
 
 Therefore the magnitudes F include a least magnitude. 
 
 Call this least magnitude Z. 
 
 Since the magnitude Z is a magnitude Y, therefore Z is greater than every 
 magnitude of the form p$. 
 
 Write this thus : Z > every p^, 
 
 .'. Z : G > every p^. 
 
 We have next to prove that 
 
 Z : G < every p 2 , 
 
 i.e. Z < every p^G. 
 Suppose if possible 
 
 Z = some p 2 G. 
 
 Then since the rational numbers p 2 include no least rational number, let 
 
 P*>p*- 
 Form the magnitude p 2 'G. 
 
 Then p, f G<p,G, 
 
 /. fr'C<Z. 
 
 Now every p 2 > every p lt 
 
 .'. p 2 '> every p,, 
 
 /. p. 2 'C > every pfi, 
 therefore pG is a magnitude F. 
 
 But plC < Z. 
 
 Hence there is a magnitude F, which is less than Z; which is contrary to the 
 definition of Z, viz. that it was the least of the magnitudes F. 
 Hence Z < every p 2 C, 
 
 /. Z : G < every p 2 . 
 
 And it was proved that 
 
 Z : G > every p t . 
 
 Hence Z : C determines the same section as the irrational number p, i.e. the 
 same section as B : A. 
 
 .'. B : A = Z : C, 
 
 .-. A:B=C:Z.
 
 INDEX 
 
 The references are to the Articles. 
 
 Aggregating Ratios 194 
 
 Angles at centre of circle proportional to arcs 
 
 on which they stand 41, 62 
 Anharmonic Ratio 192 
 Antecedent term of a Ratio 31 
 Archimedes' Axiom 22 
 Area of Rectilinear Figure, Approximate Measure 
 
 of 36 
 
 Areas 111-143 
 of similar triangles proportional to squares 
 
 on corresponding sides 127, 182 
 Arithmetical Applications of Process for Aggre- 
 gating Ratios 196 
 of Process for Com- 
 pounding Ratios 187 
 Associative Law for Addition 4, 171 
 
 Centre of Inversion 117 
 Circle, Radical Axis of 119 
 Commensurable Magnitudes 29 
 
 Ratio of 30 
 Common Measure 28 
 Commutative Law for Addition 4, 171 
 
 - for Multiplication 12, 170 
 Compounding Ratios 185-187, 214 
 
 Definition of Process for 
 
 186 
 
 Congruent figures 79 
 Congruent triangles 84 
 Consequent term of a Katio 31 
 Corresponding points on two straight lines 39 
 segments on two straight lines 
 
 39 
 sides of similar figures 77 
 
 Corresponding terms in a proportion 33 
 Cross Ratio 192 
 
 Definitions 
 
 Aggregating Ratios 194 
 Anharmonic Ratio 192 
 Antecedent term of a Ratio 31 
 Centre of Inversion 117 
 Commensurable Magnitudes 29 
 Common Measure 28 
 Compounding Ratios 185 
 Consequent term of a Ratio 31 
 Corresponding points on two straight lines 
 
 39 
 Corresponding segments on two straight 
 
 lines 39 
 
 Corresponding terms in a proportion 33 
 Cross Ratio 192 
 Duplicate Ratio 188 
 Equal Ratios, Euclid's Test for 46 
 Equality, Ratio of 32 
 Extremes of a Proportion 33 
 Figures with sides reciprocally proportional 
 
 139 
 
 Fourth Proportional 33 
 Harmonic Lines 108 
 Points 72 
 
 Homologous terms in a proportion 33 
 Inverse Locus 117 
 Inversion 117 
 Mean Proportional 33 
 Measure 19, 27 
 Multiple 3 
 Polar 115
 
 166 
 
 INDEX 
 
 Definitions (continued) 
 
 Pole 115 
 
 Proportion 33 
 
 Radical Axis of two circles 119 
 
 Ratio, Euclid's definition of 176 
 of Similitude 78 
 
 Ratios, Euclid's Test for Equal 46 
 
 Reciprocal Ratios 63 
 
 Similar Figures 77 
 
 Similarly divided straight lines 60, 61 
 
 Terms of a Ratio 31 
 
 Third Proportional 33, 58 
 
 Unequal Ratios, Euclid's Test for distin- 
 guishing 49 
 Distributive Law for Multiplication 
 
 Distribution of Multiplicand 5, 9, 170 
 
 of Multiplier 7, 10, 170 
 Division of Segment of Straight Line into equal 
 
 parts 38 
 Duplicate Ratio, Definition of 188 
 
 Equal Ratios, Euclid's Test for 46 
 
 Equality, Ratio of 32 
 
 Equiangular parallelograms, ratio of areas of 
 
 123 
 
 Euclid's Definition of Ratio 176 
 Euclid's Test for distinguishing the greater of 
 
 two unequal ratios 49 
 Euclid's Test for Equal Ratios 46 
 Extreme and Mean Ratio 147 
 Extremes of a Proportion 33 
 
 Figures, Congruent 79 
 Similar 77-102 
 
 Definition of 77 
 with sides reciprocally proportional 
 
 139-142 
 Fourth Proportional 33, 56 
 
 to three straight lines 56 
 Fraction, Rational 30 
 
 Harmonic Lines, Definition of 108 
 Points, Definition of 72 
 Homologous terms in a proportion 33 
 
 Incommensurable Magnitudes 177 
 
 Infinity, point at, on a straight line 179 
 
 Inverse Locus 117 
 
 Inversion 117 
 
 Irrational Number 198-215 
 
 Irrational Number, Addition of 212 
 
 Multiplication of 213 
 
 Line divided in extreme and mean ratio 147 
 Lines, Harmonic 108 
 
 Magnitude, Notation for a 2 
 Magnitudes, Commensurable 29 
 
 of the same kind 21 
 
 Mean Proportional between two straight lines 103 
 Means of a Proportion 33 
 Measure 19, 27 
 
 Common 28 
 Multiple, Definition of a 3 
 Multiplicand, Distribution of 5, 9, 170 
 Multiplication, Commutative Law for 12, 170 
 Multiplier, Distribution of 7, 10, 170 
 
 Notation for Aggregating Ratios 194 
 
 Compounding Ratios 191 
 
 Magnitude 2 
 
 - Positive Whole Numbers 1 
 
 - Ratio 31 
 Number, Irrational 198-215 
 
 Positive whole, Notation for 1 
 
 Parallelograms, Equiangular 123 
 
 of equal altitude proportional 
 
 to their bases 34, 52 
 Part 19 
 
 Point at infinity on a straight line 179 
 Points, Harmonic 108 
 Polar, Definition of 115 
 Pole, Definition of 115 
 Positive whole number, Notation for 1 
 Process of Aggregating Ratios 194 
 
 of Compounding Ratios 186 
 Proportion, Corresponding Terms in 33 
 
 Definition of 33 
 Extremes of a 33 
 Homologous terms in a 33 
 Means of a 33 
 Proportional, Fourth 33, 56 
 
 Third 33 
 Proportionality of angles at centre of a circle 
 
 to their arcs 41, 62 
 of parallelograms of equal alti- 
 tude to their bases 34, 52 
 of triangles of equal altitude to 
 their bases 34, 52
 
 INDEX 
 
 167 
 
 Radical axis of two circles 119 
 
 Ratio 42-50 
 
 Ratio, Anharmonic 192 
 
 Antecedent Term of a 31 
 
 Approximate Value of 45 
 
 Comparison with rational fraction 44 
 
 Consequent Term of a 31 
 
 Cross 192 
 Definition of 30, 210 
 Duplicate 188 
 Extreme and Mean 147 
 Measure of 30, 210 
 Notation for 31 
 
 of areas of Equiangular Parallelograms 
 123 
 
 of Similar Rectilineal Figures 
 
 133 
 
 Triangles 127, 182 
 
 of two Squares 126 
 
 of Commensurable Magnitudes 30 
 of Equality 32 
 
 of Similitude 78 
 Rational Fraction 30 
 Ratios, Aggregating of 194 
 
 Compounding 185-187 
 Equal 46-48 
 
 Reciprocal 63 
 Unequal 49 
 
 Reciprocal Ratios 63 
 
 Right-angled triangles divided into similar tri- 
 angles 101 
 
 Segments, Corresponding 39 
 Similar figures 77-102 
 
 Corresponding sides of 77 
 
 Definition of 77 
 
 Ratio of Similitude of 78 
 
 similarly described on sides of 
 
 a right-angled triangle 136 
 
 when similarly described 80 
 Similar rectilineal figures, ratio of area of 133 
 
 Triangles 83, 87-95 
 
 Areas of, proportional to squares 
 described on corresponding 
 sides 127, 182 
 Similarly described figures 80 
 
 divided straight lines 60, 61 
 Similitude, Ratio of 78 
 
 Symbols employed 191, 194 
 
 Terms of a Ratio 31 
 
 Test for Equal Ratios, Euclid's 46 
 
 greater of two unequal ratios, Euclid's 
 
 49 
 
 Third proportional to two straight lines 58 
 Triangle, right-angled, divided into similar tri- 
 angles 101 
 Triangles, congruent 84 
 
 of equal altitude proportional to their 
 
 bases 34, 52 
 similar 83, 87-95 
 
 Unequal Ratios, Euclid's Test for distinguishing 
 the greater of two 49 
 
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