The same construction being made because as BD to DA, so is CE to EA; and as BD to DA, so is the triangle BDE to the triangle ADE*: and as CE to . 1. 6. EA, so is the triangle CDE to the triangle ADE; therefore t the triangle BDE is to the triangle ADE, † 11. 5. as the triangle CDE to the triangle ADE; that is, the triangles BDE, CDE have the same ratio to the triangle ADE: therefore * the triangle BDE is equal to * 9.5. the triangle CDE: and they are on the same base DE: but equal triangles on the same base are between the same parallels* therefore DE is parallel to BC. Where- * 39. 1. fore, if a straight line, &c. Q. E. D. PROP. III. THEOR. If the angle of a triangle be divided into two equal angles, See N. by a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another : and if the segments of the base have the same ratio which the other sides 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 ABC be a triangle, and let the angle BAC be divided into two equal angles by the straight line AD: BD shall be to DC, as BA to AC. Through the point C draw CE parallel * to DA, * 31. 1. 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 angle CAD: but CAD, * 29. 1. by the hypothesis, is equal to the angle BAD; wherefore BAD is equal t to the angle ACE. Again, be- + 1 Ax. cause the straight line BAE meets the parallels AD, EC, the outward angle BAD is equalt to the inward + 29, 1, and opposite angle AEC: but the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal t to the angle AEC, + 1 Ax. and consequently the side AE is equal * to the side AC: * 6. 1. and because AD is drawn parallel to one of the sides of the triangle BCE, viz. to EC, therefore * BD is to * 2. 6. DC, as BA to AE: but AE is equal to AC; therefore*, *7.5. as BD to DC, so is BA to AC. E B * 2. 6. * 11. 5. * 9. 5. * 5. 1. Next, let BD be to DC, as BA to AC, and join AD : the angle BAC shall be 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 AD is parallel to EC; therefore BA is to AC, as BA to AE: consequently AC is equal * to AE, and therefore the angle A ĖC is equal * to the angle ACE; but the angle AEC is equal to the outward and opposite angle BAD; and the angle ACE is equal to the alternate angle CAD*: wherefore also the angle BAD is equal t to the angle CAD; that is, the angle BAC is cut into two equal angles by the straight line AD. Therefore, if the angle, &c. * 29. 1. 71 Ax. Q. E. D. PROP. A. THEOR. If the outward angle of a triangle made by producing one of its sides, be divided into two equal angles, by a straight line which also cuts the base produced ; the segments between the dividing line and the extremities of the base 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 ABC be a triangle, and let one of its sides BA be produced to E; and let the outward angle CAE be divided into two equal angles by the straight line AD which meets the base produced in D; BD shall be to DC, as BA to AC. Through C draw* CF parallel to AD: and because the straight line AC meets the parallels AD, FC, the angle ACF is equal * to the alternate angle CAD: but CĂD is equal* io the angle DAE; therefore also DAE is equal to the angle ACF. Again, because the straight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward and opposite angle CFA: but the angle ACF has been proved equal to the angle DAE; therefore also the angle ACF is equal t to the angle CFA; and consequently the E 1. А +1 Ax. 2. 6. side AF is equal * to the side AC: and because AD * 6.1. is parallel to FC, a side of the triangle BCF, therefore* BD is to DC, as BA to AF: but AF is equal * 2. 6. to AC; therefore as BD is to DCt, so is BA to AC. +7.5. Next, let BD be to DC, as BA to AC, and join AD: the angle CAD shall be 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; therefore * BA is to AC, as BA to AF; * 11. 5. wherefore AC is equal* to AF, and the angle AFC 9. 5. equal * to the angle ACF: but the angle AFC is equal * 5. 1. ' to the outward angle EAD t, and the angle ACF to the + 29. 1. alternate angle CAD; therefore also EAD is equal + +1 Ax. to the angle CAD. Wherefore, if the outward, &c. R. E. D. PROP. IV. THEOR. gles are proportionals; and those which are opposite 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 * the 92. 1. & angle BAC equal to the angle CDE." The sides about the equal angles of the triangles ABC, DCE shall be proportionals; and those shall be the homologous sides, which are opposite to the equal angles. Let the triangle DCE be placed *, so that its side * 22. 1. CE may be contiguous to BC, and in the same straight line with it: then, because the angle BCA is equal + + Hyp. to the angle CED: add to each the angle ABC; therefore the two angles ABC, BCA are equal t to the +2 Ax. two angles ABC, CED: but the angles ABC, BCA are together less * than two right angles; therefore * 17. 1. the angles ABC, CED are also less than two right angles: wherefore BA, ED if produced will meet * : let * 12 Ax. 1. them be produced and meet in the point F; then because the angle ABC F is equal to the angle DCE, BF is parallel * to CD; and because the angle A D · ACB is equal to the angle DEC, AC is parallel to FE*: therefore FACD is a parallelogram; and consequently AF is equal to CD, and AC to FD: and because AC 3 Ax. 28. 1. * 28. 1. B * 34. 1. * 9. 6. * 7.5. * + 16.5. 2.6. + 7.5. † 16. 5. is parallel to FE, one of the sides of the triangle FBE, BA is to AF*, as BC to CE: but AF is equal to CD; therefore as BA to CD, so is BC to CE; and alternately t, as AB to BC, so is DC to CE: again, because CD is parallel to BF, as BC to CE *, so is FD to DE: but FD is equal to AC; therefore t, as BC to CE, so is AC to DE; and alternately t, as BC to CA, so 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 ED, ex æquali *, BA is to AC, as CD to DE. Therefore the sides, &c. Q. E. D. * 22. 5. PROP. V. THEOR. 23. 1. If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular ; Let the triangles ABC, DEF have their sides pro- At the points E, F, in the straight line EF, make * the angle FEG equal to the angle ABC, and the angle EFG equal to BCA; wherefore the remaining angle BAC is * 32. 1. & equal * to the remaining angle EGF, and the triangle ABC is therefore equiangular to the triangle GEF: con- wherefore, as AB to BC, so is GE to + Hyp. EF: but as AB to BCt, so is DE to 3 Ax. D * 11.5. EF F B * 9. 5. * 8. 1. the angle DFE is equal to the 'angle GFE, and EDF PROP. VI. THEOR. angle of the other, and the sides about the equal angles 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, shall be equiangular, and shall have the angle ABC equal to the angle DEF, and ACB to DFE. At the points D, F, in the straight line DF, make* • 43. 1. the angle FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle D AČB : wherefore the remaining angle at B is equal* to the remaining • 32. 1.& angle at G: and consequently the CE triangle ABC is equiangular to the triangle DGF: therefore as BA to AC, so is * GD to • 4. 6. DF: but by the hypothesis, as BA to AC, so is ED to DF; therefore as ED to DF, so is* GD to DF; • 11. 5. wherefore ED is equal * to DG: and DF is common • 9. 5. to the two triangles EDF, GDF: therefore the two sides ED, DF, are equal to the two sides GD, DF, each to each; and the angle EDF is equal + to the an- + Constr. gle GDF; wherefore the base EF is equal to the base FG*, and the triangle EDF to the triangle GDF, and • 4. 1. the remaining angles to the remaining angles, each to each, which are subtended by the equal sides : 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 t to the angle ACB; therefore the angle ACB + Constr. is equal to the angle DFE: and the angle BAC is +1 Ax. equal * to the angle EDF: wherefore also ihe remain- * Hyp. А B F 3 Ax. |