BOOK VI. CDE have the fame ratio to the triangle ADE; and therefore the triangle BDE is equal to the triangle CDE: And they are f 9. 5. on the fame bafe DE; but equal triangles on the same base are g 39. 1. between the fame parallels; therefore DE is parallel to BC. Wherefore, if a ftraight line, &c. Q. E. D. a 31. 1. IF PROP. III. THEOR. F the angle of a triangle be divided into two equal angles, by a ftraight line which alfo cuts the bale; the fegments of the base fhall have the fame ratio which the other fides of the triangle have to one another: And if the fegments of the base have the fame ratio which the other fides of the triangle have to one another, the ftraight line drawn from the vertex to the point of fection, divides the vertical angle into two equal angles. Let the angle BAC of any triangle ABC be divided into two equal angles by the ftraight line AD: BD is to DC, as BA to AC. E A Through the point C draw CE parallel to DA, and let BA produced meet CE in E. Because the ftraight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate 29. 1. angle CAD: But CAD, by the hypothefis, is equal to the angle BAD; wherefore BAD is equal to the angle ACE. Again, because the ftraight line BAE meets the parallels AD, EC, the outward angle BAD is equal to the inward and oppofite angle AEC But the angle ACE has been proved equal to the angle BAD; therefore alfo ACE is equal to the angle AEC, and confequently the fide AE is equal to the fide AC: And becaufe AD is drawn parallel to one of the fides of the triangle BCE, viz. to EC, BD is to DC, as BA to AE : But AE is equal to AC; therefore, as BD to DC, fo is BA to AC *. c 6. 1. d 2. 6. e 7. 5. B D C Let now BD be to DC, as BA to AC, and join AD; the angle BAC is divided into two equal angles by the straight line AD. The fame conftruction being made; because, as BD to DC, fo is BA to AC; and as BD to DC, fo is BA to AE d, because f 11. 5. AD is parallel to EC; therefore BA is to AC, as BA to AE £; Confequently g 9. 5. h 5. 1. Confequently AC is equal to AE 8, and the angle AEC is there- Book. VI. fore equal to the angle ACE h: But the angle AEC is equal to the outward and oppofite angle BAD; and the angle ACE is equal to the alternate angle CAD: Wherefore alfo the angle BAD is equal to the angle CAD: Therefore the angle BAC is cut into two equal angles by the ftraight line AD. Therefore, if the angle, &c. Q. E. D. PROP. A. THEOR. F the outward angle of a triangle made by producing one of its fides, be divided into two equal angles, by a ftraight line which alfo cuts the bafe produced; the fegments between the dividing line and the extremities of the bafe, have the fame ratio which the other fides of the triangle have to one another And if the fegments of the base produced, have the fame ratio which the other fides of the triangle have, the ftraight line drawn from the vertex to the point of fection, divides the outward angle of the triangle into two equal angles. Let the outward angle CAE of any triangle ABC be divided into two equal angles by the ftraight line AD which meets the base produced in D: BD is to DC, as BA to AC. a E A b 29. I. Through C draw CF parallel to AD ; and becaufe the a 31. 1. ftraight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD: But CAD is equal to the angle DAE; therefore alfo DAE is equal to the angle ACF. clyp. Again, because the ftraight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward and oppofite angle CFA : But the angle ACF has been proved equal to the angle DAE; therefore alfo the angle ACF is equal to the angle CFA, and confequently the fide AF is equal to the fide AC: And becaufe AD is parallel to FC, a fide of the triangle BCF, BD is to DC, as BA to AF ; but AF is equal to AC; as therefore BD is to DC, fo is BA to AC. Let now BD be to DC, as BA to AC, and join AD; the angle CAD is equal to the angle DAE. B F e C D The fame conftruction being made, becaufe BD is to DC, as BA to AC; and that BD is alfo to DC, as BA to AF €; therefore $ 2 d 6. I. € 2.6. f. 5. BOOK VI. therefore BA is to AC, as BA to AF f; wherefore AC is equal to AF, and the angle AFC equal to the angle ACF: But the angle AFC is equal to the outward angle EAD, and the angle ACF to the alternate angle CAD; therefore alfo EAD is equal to the angle CAD. Wherefore, if the outward, &c. Q. E. D. 9. 5 h 5. 1. THE PROP. IV. THEOR. HE fides about the equal angles of equiangugular triangles are proportionals; and thofe which are oppofite to the equal angles, are homologous fides, that is, are the antecedents or confequents 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 a 32. 1. DEC, and confequently the angle BAC equal to the angle CDE. The fides about the equal angles of the triangles ABC, DCE are proportionals; and thofe are the homologous fides which are oppofite to the equal angles. Let the triangle DCE be placed, fo that its fide CE may be contiguous to BC, and in the same straight line with it: And becaufe the angle ABC is equal to the b F b 28. 1. angle DCE, BA is parallel to CD. e 2.6. f 7.5. B E to AF, as BC to CE; But AF is equal to CD; therefore, as BA to CD, fo is BC to CE; and alternately, as AB to BC, g 16. 5. fo is DC to CE: Again, because CD is parallel to BF, as BC to CE, fo is FD to DE ; 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 h 22.5. ED, by equality h, BA is to AC, as CD to DE. Therefore the fides, &c. Q. E. D. PROP. PROP. V. THEOR. F the fides of two triangles, about each of their angles, be proportionals, the triangles fhall be equiangular, and have their equal angles opposite to the homologous fides. Let the triangles ABC, DEF have their fides proportionals, fo that AB is to BC, as DE to EF; and BC to CA, as EF to FD; and confequently, by equality, BA to AC, as ED to DF; the triangle ABC is equiangular to the triangle DEF, and their equal angles are oppofite to the homologous fides, viz. the angle ABC equal to the angle DEF, and BCA to EFD, and alfo BAC to EDF. BOOK VI. At the points E, F, in the ftraight line EF, make a the angle a 23. 1. FEG equal to the angle ABC, and the angle EFG equal to A Db 32. 1. E F C Ꮐ C 4. T. BCA; wherefore the remaining f e 9. 5. PROP. Book VI. a 23. 1. 11. 5. e 9. 5. IF PROP. VI. THEOR. F two triangles have one angle of the one equal to one angle of the other, and the fides about the equal angles proportionals, the triangles fhall be equiangular, and fhall have thofe angles equal which are oppofite 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 fides about thofe 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. A D G CET At the points D, F, in the straight line DF, make a the angle FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB: Wherefore the remaining angle at B b 32. 1. is equal to the remaining one at G, and confequently the triangle ABC is equiangular to the triangle DGF; € 4. 6. and therefore, as BA to AC, fo is c GD to DF: But, by the hypothefis, B as BA to AC, fo is ED to DF; as, therefore, ED to DF, fo is d GD to DF; wherefore ED is equal to DG; and DF is common to the two triangles EDF, GDF: Therefore the two fides ED. DF are equal to the two fides GD, DF; and the angle EDF is equal to the angle f 4. 1. GDF; wherefore the bafe EF is equal to the bafe FG, and the angle DFG to the angle DFE: 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 8 Hyp angle EDF ; wherefore alfo the remaining 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. Q. E. D. e PROP. |
