Book VI. CDE have the same ratio to the triangle ADE; and therefore F. mthe triangle BDE is equal to the triangle CDE: And they are on the same base DE ; but equal triangles on the same base are g 39. 1. between the same parallels ; therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D. f g. 5 a 31. I. PROP. III. THEOR. equal angles, by a straight line which also cuts the bale; the segments of the base shall have the fame ratio which the other sides of the triangle have to one another: And if the fegments of the base have the same ratio which the other fides of the triangle have to one another, the straight line drawn from the vertex to the point of section, divides the vertical angle into two equal angles. Let the angle BAC of any triangle ABC be divided into two equal angles by the straight line AD: BD is to DC, as BA to AC. Through the point C draw CE parallel a to DA, and let BA produced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate D 29. 1. angle CAD b: But CAD, by the hypothesis, is equal to the angle BAD; wherefore BAD is equal to the angle ACE. A B D С parallel to one of the sides of the triangle BCE, viz. to EC, d 2. 6. BD is to DC, as BA to AE 4: But AE is equal to AC; there. foré, as BD to DC, fo is BA to AC e. 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 same construction being made ; because, as BD to DC, so is BA to AC; and as BD to DC, so is BA to AE , because f 11. 5. AD is parallel to EC; therefore BA is to AC, as BA to AEf: Consequently C 6. 1. e 7. 5. h Consequently 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 tom the outward and opposite angle BAD; and the angle ACE is equal to the alternate angle CADD: Wherefore also the angle BAD is equal to the angle CAD: Therefore the angle BAC is cut into two equal angles by the straight line AD. Therefore, if the angle, &c. Q. E. D. PROP. A. THEOR. F the outward angle of a triangle made by pro I , a 31.1. equal angles, by a straight line which also cuts the base produced ; the segments between the dividing line and the extremities of the bale, have the same ratio which the other sides of the triangle have to one another : And if the segments of the base produced, have the same ratio which the other sides of the triangle have, the straight line drawn from the vertex to the point of section, 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 straight line AD which meets the base produced in D: BD is to DC, as B A to AC. Through C draw CF parallel to AD a; and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CADO: But CAD is equal to the b 29.1. angle DAE "; therefore also DAE is equal to the angle ACF. c llyp. Again, because the straight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward and opposite angle CFA : I But the angle ACF has been proved equal to the angle DAE; therefore 1 also the angle ACF is equal to the angle CFA, and consequently the fide AF is equal to the fide AC : And be d 6. I. B C D cause AD is parallel to FC, a side of the triangle BCF, BD is to DC, as BA to AF ; but AF is e 2.6 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. The same construction being made, because BD is to DC, as BA to AC; and that BD is also to DC, as BA to AF; S 2 therefore A Book VI. therefore BA is to AC, as BA to AFf; wherefore AC is equal M to AF8, and the angle AFC equal h to the angle ACF: But f11. 5. the angle AFC is equal to the outward angle EAD, and the 9. 5. angle ACF to the alternate angle CAD; therefore also EAD is equal to the angle CAD. Wherefore, if the outward, &c. Q. E. D. h 5. I. PROP. IV. THEOR. THE fides about the equal angles of equiangu gular triangles are proportionals; and those which are opposite to the equal angles, are homologous fides, that is, are the antecedents or conse. quents 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 consequently the angle BAC equal to the angle CDE. The fides about the equal angles of the triangles ABC, DCE are proportionals; and those are the homologous fides which are opposite 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 becaufe the angle ABC is equal to the b 28. 1. angle DCE, BA is parallel to CD. Again, because the angle ACB is equal A to the angle DEC, AC is parallel to DE 6: Produce BA, ED, until they cCor.39.1. meeto in F: Therefore FACD is a pa rallelogram; and consequently AF is d 34. I. equal to CD, and AC to FD d: And because AC is parallel to FE, one of B as BA to CD, so is BC to CE ; and alternately, as AB to BC, g 16. 5. fo is DC to CE 8: Again, because CD is parallel to BF, as BC to CE, fo is FD to DE €; but FD is equal to AC; there. fore, as BC to CE, fo is AC to DE: And alternately 8, 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, so CE to h 23. 5. ED, by equality , BA is to AC, as CD to DE. Therefore the fides, &c. Q. E. D. e 2. 6. f 7. 5 PROP. PROP. V. THEOR. Book VI. IF the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the homologous fides. 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, by equality, 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 allo BAC to EDF. At the points E, F, in the straight line EF, make a the angle a 23. 5. A Db 32. 1. triangle ABC is therefore equiangular to the triangle GEF; and consequently they have their fides E F opposite to the equal angles pro- B portionals 6: Wherefore, as AB C4. 1. to BC, fo is GE to EF; but G AB to BC, so is DE to EF; therefore as DE to EF, so d GE to EF: Therefore DE and GE have the same ratio to EF, and consequently are equal o: e 9.5. 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 fides DE, EF are equal to the two GE, EF, and the base DF is equal to the base GF; therefore the angle DEF is equal † to the angle GEF: and GEF is equal to the angle ABC; f 8. I. 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. D. as dil. Se PROP. Book VI. PROP. VI. THEOR. IF two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous fides. a 23. 1. 4. 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 FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB : Wherefore the remaining angle at B D b 32. 1. is equal to the remaining one at G 5, G and consequently the triangle ABC is equiangular to the triangle DGF; 6. and therefore, as BA to AC, so is C GD to DF: But, by the hypothesis, B CE I as BA to AC, so is ED to DF; as, d 11. 5. therefore, ED to DF, so is d GD to DF; wherefore ED is e 9. 5. equal to DG; and DF is common to the two triangles EDF, GDF: Therefore the two fides ED. DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle f 4. 1. GDF; wherefore the base EF is equal to the base 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 & Hyp: angle EDF 5; wherefore also 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. |