is equal to the outward angle EAD, and the angle ACF to the Book VI. alternate angle CAD; therefore also EAD is equal to the angle CAD. Wherefore if the outward, &c. Q. E. D. T HE fides about the equal angles of equiangular triangles are proportionals; and those which arc opposite to the equal angles are homologous fides, that is, are the antecedents or consequents of the ratios. Let ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the angle DEC, and consequently a the angle BAC equal to the angle CDE. The a. 320 1. fides about the equal angles of the triangles ABC, DCE are proportionals; and those are the homologous fides which are oppofite to the equal angles. Let the triangle DCE be placed so that its fide CE may be contiguous to BC, and in the same straight line with it. and because the angles ABC, ACB are together less than two right angles b; b. 17, 1. ABC and DEC, which is equal to ACB, are also less than two right an, T gles. wherefore BA, ED produced shall meet ; let them be produced C.12. Ax.1 and meet in the point F. and because D the angle ABC is equal to the angle DCE, BF is parallel 4 to CD. Again, d. 28. le because the angle ACB is equal to B the angle DEC, AC is parallel to E FE d. therefore FACD is a parallelogram; and consequently AF is equal to CD, and AC to FDC, and because AC is parallel to FE e. 34. I. one of the sides of the triangle T'EX, BA is to AF, as BC to CE. f. 2. 6. but AF is equal to CD, therefore $ as BA to CD, fo is BC to CE; 8. 7. 3. and alternately, as AB to BC, so DG to CE. Again, because CD is parallel to BF, as BC to CE, fo is FD to DE f; but FD is equal to AC; therefore as BC to CE, fo is AC to DE. and alternately, as BC to CA, fo CE to ED. therefore because it has been proved · that AB is to BC, as DC to CE; and as BC to CA, fo CE to ED, ex aequali ", BA is to AC, as CD to DE, Therefore the sides, &c. h. 22. 50 Q. E. D. . 154 Book VI. IF (F the fides of two triangles, about each of their an gles, be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the homologous fides. a. 23. I. Let the triangles ABC, DEF have their fides proportionals, so that AB is to BC, as DE to EF; and BC to CA, as EF to FD; and consequently, ex aequali, BA to AC, as ED to DF. the triangle ABC is equiangular to the triangle DEF, and their equal angles are opposite to the homologous fides, viz. the angle ABC equal to the angle DEF, and BCA to EFD, and also BAC to EDF. At the points E, F in the straight line EF make a the angle FEG equal to the angle ABC, and the angle EFG equal to BCA; wherefore the remaining angle BAC is equal to the remaining angle A b. 32. I. EGF b, and the triangle ABC is therefore equiangular to the E wherefore as AB to BC, so is GE to EF; but as AB to BC, d. 11. 5. fo is DE to EF; therefore as DE to EF, fo : GE to EF. therefore DE and GE have the same ratio to EF, and consequently are equal e. for the same reason DF is equal to FG. and because, in the triangles DEF, GEF, DE is equal to EG, and EF common, the two sides DE, EF are equal to the two GE, EF, and the bafe DF f. 8. 1. is equal to the base GF; therefore the angle DEF is equal to the angle GEF, and the other angles to the other angles which are fubtended by the equal fides 6. Wherefore the angle DFE is equal to the angle GFE, and EDF to EGF. and because the angle DLF is equal to the angle GEF, and GEF to the angle ABC; therefore the angle ABC is equal to the angle DEF. for the same reason, the angle ACB is equal to the angle DFE, and the angle at A to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore if the sides, &c. Q. E. DA e. 9. 5. Book VI. yok PROP. VI. THEOR. IF F two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equi. angular, and fhall have those angles equal which are opposite to the homologous fides. Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the sides about those angles proportionals; that is, BA to AC, as ED to DF. The triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE. At the points D, F, in the straight line DF, make a the angle a. 23. T. FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB. wherefore the remaining angle at B is А. equal to the remaining one at Gb, and consequently the tri Gb. b. 32. I. angle ABC is equiangular to the triangle DGF; and therefore as BA to AC, so is c GD to Dr. but, by the Hypothesis, as BA to AC, fo is ED to DF;B CE F as therefore ED to DF, so is GD to DF; wherefore ED is equal to DG; and DF is common to the two triangles EDF, GDF. e. gu So therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF, wherefore the base EF is equal to the base FG f, and the triangle EDF to f. 4. I. the triangle GDF, and the remaining angles to the remaining angles, each to each, which are subtended by the equal fides. therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E. but the angle DFG is equal to the angle ACB ; therefore the angle ACB is equal to the angle DFE. and the angle BAC is equal to the angle EDF 8; wherefore also the remaining 8. Hype angle at B is equal to the remaining angle at E. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore if two triangles, &c. 0. E. D. C. 4. 6. d. il. 5. Book VI. PROP. VII. THEOR. See N. IF one angle of the other, and the sides about two other angles proportionals; then if each of the remaining angles be either less, or not less than a right angle; or if one of them be a right angle: the triangles shall be equiangular, and have those angles equal about which the sides are proportionals. Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz. the angle BAC to the angle EDF, and the sides about two other angles ABC, DEF proportionals, so that AB is to BC, as DE to EF; and, in the first case, let each of the remaining angles at C, F be less than a right angle. The triangle ABC is equiangular to the triangle DEF, viz. the angle ABC is equal to the angle DEF, and the remaining angle at C, to the remaining angle at F. For if the angles ABC, DEF be not equal, one of them is greater А D G ing angle DFE. therefore the C. 4. 6. triangle ABG is equiangular to the triangle DEF; wherefore e as A B is to BG, so is DE to EF; but as DE to EF, so, by Hypoched. 11. 5. lis, is AB to BC; therefore as AB to BC, so is AB to BG d; and because AB has the same ratio to each of the lines BC, BG; BC is equal e to BG, and therefore the angle BGC is equal to . f. 5.1. the angle BCG f. but the angle BCG is, by Hypothefis, less than a right angle; therefore also the angle BGC is less than a right angle, and the adjacent angle AGB must be greater than a right angle 8. But it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle. but, by the Hypothesis, it is less than a right angle; which is abm 2. 23. 1. e. 9. 5. $. 13. I. h. 17. 1. surd. Therefore the angles ABC, DEF are not unequal, that is, Book. VI. they are equal. and the angle at A is equal to the angle at D; wherefore the remaining angle at Cis equal to the remaining angle at F. therefore the triangle ABC is equiangular to the triangle DEF. Next, Let each of the angles at C, F be not less than a right angle. the triangle ABC is also in this case equiangular to the triangle DEF. The same construction being A made, it may be proved in like manner that BC is equal to BG, and the angle at C equal G to the angle BGC. but the angle at Cis not less than a right B E angle; therefore the angle F BGC is not less than a right angle. wherefore two angles of the triangle BGC are together not less than two right angles; which is impossible h; and therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case. 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 equiangu A lar, 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 B case, that BG is equal to BC. C D but the angle BCG is a right A angle, therefore the angle BGC is also a right angle; whence two of the angles of the triangle BGC E F are together not less than twoB right angles; which is impoflible h. therefore the triangle ABC is equiangular to the triangle DEF. Wherefore if two triangles, &c. Q. E. D. f. 5. I. |