Therefore the triangle ABC is equiangular with the triangle DEF. Therefore etc. Q. E. D. Todhunter points out, after Walker, that some more words are necessary to make the enunciation precise: "If two triangles have one angle equal to one angle, the sides about other angles proportional <so that the sides subtending the equal angles are homologous>....' This proposition is the extension to similar triangles of the ambiguous case already mentioned as omitted by Euclid in relation to equality of triangles in all respects (cf. note following 1. 26, Vol. 1. p. 306). The enunciation of vi. 7 has suggested the ordinary method of enunciating the ambiguous case where equality and not similarity is in question. Cf. Todhunter's note on 1. 26. Another possible way of presenting this proposition is given by Todhunter. The essential theorem to prove is: If two triangles have two sides of the one proportional to two sides of the other, and the angles opposite to one pair of corresponding sides equal, the angles which are opposite to the other pair of corresponding sides shall either be equal or be together equal to two right angles. For the angles included by the proportional sides must be either equal or unequal. If they are equal, then, since the triangles have two angles of the one equal to two angles of the other, respectively, they are equiangular to one another. We have therefore only to consider the case in which the angles included by the proportional sides are unequal. The proof is, except at the end, like that of VI. 7. Let the triangles ABC, DEF have the angle at A equal to the angle at D; let AB be to BC as DE to EF, but let the angle ABC be not equal to the angle DEF. G The angles ACB, DFE shall be together equal to two right angles. Let ABC be the greater; and make the angle ABG equal to the angle DEF. Then we prove, as in vi. 7, that the triangles ABG, DEF are equiangular, whence But AB is to BG as DE is to EF. AB is to BC as DE is to EF, by hypothesis. and the angle BGC is equal to the angle BCA. Now, since the triangles ABG, DEF are equiangular, the angle BGA is equal to the angle EFD. Add to them respectively the equal angles BGC, BCA; therefore the angles BCA, EFD are together equal to the angles BGA, BGC, i.e. to two right angles. It follows therefore that the angles BCA, EFD must be either equal or supplementary. But (1), if each of them is less than a right angle, they cannot be supplementary, and they must therefore be equal; (2) if each of them is greater than a right angle, they cannot be supplementary and must therefore be equal; (3) if one of them is a right angle, they are supplementary and also equal. Simson distinguishes the last case (3) in his enunciation: "then, if each of the remaining angles be either less or not less than a right angle, or if one of them be a right angle...." The change is right, on the principle of restricting the conditions to the minimum necessary to enable the conclusion to be inferred. Simson adds a separate proof of the case in which one of the remaining angles is a right angle. "Lastly, let one of the angles at C, F, viz. the angle at C, be a right angle ; in this case likewise the triangle ABC is equiangular to the triangle DEF. For, if they be not equiangular, make, at the point B of the straight line AB, the angle ABG equal to the angle DEF; then it may be proved, as in the first case, that BG is equal to BC. But the angle BCG is a right angle; therefore the angle BGC is also a right angle; whence two of the angles of the triangle BGC are together not less than two right angles: which is impossible. A Therefore the triangle ABC is equiangular to the triangle DEF.” PROPOSITION 8. If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another. Let ABC be a right-angled right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC; I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another. H. E. II. 14 For, since the angle BAC is equal to the angle ADB, for each is right, and the angle at B is common to the Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD, so is AB itself which subtends the angle. at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD, and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4] Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional. Therefore the triangle ABC is similar to the triangle ABD. Similarly we can prove that [vi. Def. 1] the triangle ABC is also similar to the triangle ADC; therefore each of the triangles ABD, ADC is similar to the whole ABC. I say next that the triangles ABD, ADC are also similar to one another. For, since the right angle BDA is equal to the right angle ADC, and moreover the angle BAD was also proved equal to the angle at C, [1. 32] therefore the remaining angle at B is also equal to the remaining angle DAC; therefore the triangle ABD is equiangular with the triangle ADC. Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal [VI. 4] to the angle at B, and so also is BA to AC, these sides subtending the right angles; therefore the triangle ABD is similar to the triangle ADC. Therefore etc. [VI. Def. 1] PORISM. From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base. Q. E. D. Simson remarks on this proposition: "It seems plain that some editor has changed the demonstration that Euclid gave of this proposition: For, after he has demonstrated that the triangles are equiangular to one another, he particularly shows that their sides about the equal angles are proportionals, as if this had not been done in the demonstration of prop. 4 of this book: this superfluous part is not found in the translation from the Arabic, and is now left out." This seems a little hypercritical, for the "particular showing" that the sides about the equal angles are proportionals is really nothing more than a somewhat full citation of VI. 4. Moreover to shorten his proof still more, Simson says, after proving that each of the triangles ABD, ADC is similar to the whole triangle ABC, "And the triangles ABD, ADC being both equiangular and similar to ABC are equiangular and similar to one another," thus assuming a particular case of vI. 21, which might well be proved here, as Euclid proves it, with somewhat more detail. We observe that, here as generally, Euclid seems to disdain to give the reader such small help as might be afforded by arranging the letters used to denote the triangles so as to show the corresponding angular points in the same order for each pair of triangles; A is the first letter throughout, and the other two for each triangle are in the order of the figure from left to right. It may be in compensation for this that he states at such length which side corresponds to which when he comes to the proportions. In the Greek texts there is an addition to the Porism inserted after "(Being) what it was required to prove," viz. "and further that between the base and any one of the segments the side adjacent to the segment is a mean proportional." Heiberg concludes that these words are an interpolation (1) because they come after the words oπep edeɩ deîgai which as a rule follow the Porism, (2) they are absent from the best Theonine MSS., though P and Campanus have them without the onep dei deiga. Heiberg's view seems to ὅπερ ἔδει δεῖξαι. be confirmed by the fact noted by Austin, that, whereas the first part of the Porism is quoted later in vi. 13, in the lemma before x. 33 and in the lemma after XIII. 13, the second part is proved in the former lemma, and elsewhere, as also in Pappus (111. p. 72, 9-23). PROPOSITION 9. From a given straight line to cut off a prescribed part. thus it is required to cut off from AB a prescribed part. 5 ΤΟ Let the third part be that prescribed. Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC, and let DE, EC be made equal to AD. [1. 3] E Let BC be joined, and through D let DF be drawn parallel to it. [1. 31] B Then, since FD has been drawn. parallel to BC, one of the sides of the triangle ABC, therefore, proportionally, as CD is to DA, so is BF to FA. 15 But CD is double of DA; therefore BF is also double of FA; therefore BA is triple of AF. [VI. 2] Therefore from the given straight line AB the prescribed third part AF has been cut off. Q. E. F. 6. any angle. The expression here and in the two following propositions is TuxоÛσа ywvía, corresponding exactly to Tuxòv onueîor which I have translated as "a point (taken) at random"; but "an angle (taken) at random" would not be so appropriate where it is a question, not of taking any angle at all, but of drawing a straight line casually so as to make any angle with another straight line. Simson observes that "this is demonstrated in a particular case, viz. that in which the third part of a straight line is required to be cut off; which is not at all like Euclid's manner. Besides, the author of that demonstration, from four magnitudes being proportionals, concludes that the third of them is the same multiple of the fourth which the first is of the second; now this is nowhere demonstrated in the 5th book, as we now have it; but the editor assumes it from the confused notion which the vulgar have of proportionals." The truth of the assumption referred to is proved by Simson in his proposition D given above (p. 128); hence he is able to supply a general and legitimate proof of the present proposition. "Let AB be the given straight line; it is required to cut off any part from it. From the point A draw a straight line AC making any angle with AB; in AC take any point D, and take AC the same multiple of AD that AB is of the part which is to be cut off from it; join BC, and draw DE parallel to it: then AE is the part required to be cut off. E |