*29. 1. f1 Ax. #6. 1. #2.6. equal to the inward and E quently the side AF is equal to the side AC: and because AD is parallel to FC, a side of the triangle BCF, therefore* BD : DC :: BA: AF: but AF is Def. 7. 5. equal to AC; therefore, as BD : DC† :: BA : AC. Next, let BD be to DC, as BA to AC, and join A, D: the angle CAD shall be equal to the angle DAE. The same construction being made, because BD: DC :: #2. 6. *2. 5. #9.5. BA: AC; and that also BD : DC :: BA : AF;* therefore* BA: AC :: BA: AF: wherefore AC is equal to AF, and the angle AFC equal to the Cor. 1. angle ACF:* but the angle AFC is equal to the outward angle EAD,† and the angle ACF to the alternate angle CAD; therefore also EAD is equalt to the angle CAD. Wherefore, if the outward, &c. #5. 1. 129. 1. +1 Ax. 2. E. D. PROP. IV. THEOR. The sides about the equal angles of equiangular triangles are proportionals; and those which are opposite to the equal angles are homologous sides; that is, are the antecedents or consequents of the proportions. 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 angle BAC #32. 1, equal to the angle CDE. The sides about the 3 Ax. equal angles of the triangles ABC, DCE shall be proportionals; and those shall be the homologous sides which are opposite to the equal angles. and #22. 1. +Hyp. 12 Ax. *17. 1. Let the triangle DCE be placed so that its side CE may be contiguous to BC, and in the same straight line with it: then the angle BCA will be equal to the angle CED: add to each the angle ABC; therefore the two angles ABC, BCA are equal to the two angles ABC, CED: but the angles ABC, BCA are together less than two right angles; therefore the angles ABC, CED are also less than two right angles: wherefore BA, ED if produced 12 Ax. 1. will meet:* let them be produced and meet in the point F: A then because the angle ABC is equal to the angle DCE, BF is parallel to CD: and because the angle ACB is equal to the *28. 1. #34. 1. F B *28. 1. 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 is parallel to FE, one of the sides of the triangle FBE, BC: CE: BA : AF;* *2.6. and, since CD is parallel to BF, BC CE:: FD: DE. But AF is equal to CD, and FD to AC, therefore BC CE: BA: CD BC CE: AC: DE *2. 5. Consequently* BA CD: AC: DE; #10. 5. and therefore, by alternating these proportions,* Scholium. It readily follows, from this proposition, that in similar triangles the bases are as the altitudes. For, let ABC, DEF be similar triangles, of which the altitudes are AG, DH, respectively. Then, since the angles at B and E are equal, as also the right angles at G and H, the triangles ABG, DEH are similar; therefore, but AB AG: DE: DH; BC AB EF: ED, therefore (115 or 16.5) BC: AG: EF: DH; hence, alternately, the bases are as the altitudes. PROP. V. THEOR. If the sides of two triangles, about two angles of each, be proportionals, the triangles shall be equiangular; and the equal angles shall be those which are opposite to the homologous sides. Let the triangles ABC, DEF have their sides, about two angles in each, as B, C in the one, and E, F in the other, proportionals, so that AB is to BC, as DE to EF; and BC to CA, as EF to FD; the triangle ABC shall be equiangular to the triangle DEF, and the angles which are opposite to the homologous sides shall be equal, viz. the angle ABC equal to the angle DEF, BCA to EFD, and consequently BAC to EDF. *23. 1. #32. 1. 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 equal to the remaining angle EGF, and the triangle ABC is therefore equiangular to the triangle GEF: consequently they *4.6. have their sides opposite to the equal angles pro aud 3 Ax. B A D portionals: wherefore, AB : BC :: GE: EF; but as AB: BC:: DE: EF; therefore DE: EF:: GE: EF; that is, DE and GE have the same consequents, and are therefore equal; for a like 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, each to each; and the base DF is equal to the base GF; there*8. 1. fore the angle DEF is equal to the angle GEF, and the other angles to the other angles which are subtended by the equal sides;* therefore the angle DFE is equal to the angle GFE, and EDF to EGF and because the angle DEF is equal to the +Const. angle GEF, and GEF equal to the angle ABC; therefore the angle ABC is equal to the angle DEF: for a similar reason, the angle ACB is equal to the angle DFE, and therefore the angle A to the angle D: therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q. E. D. #4.1. +1. Ax. 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 sides. 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. *23. 1. At the points D, F, in the straight line DF, make* the angle FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB: wherefore the remaining *32. 1. angle at B is equal to and 3 Ax. the remaining angle at G: angle ABC is equiangular to the B triangle DGF; therefore, as BA : *9.5. Cor. 1. D G CE AC:: GD: DF; but by the hypothesis, BA: AC :: ED: DF; therefore ED: DF:: GD: DF; wherefore ED is equal to DG; and DF is common to the two triangles EDF, GDF: therefore the two sides ED, DF are equal to the two sides GD, DF, each to Const. each; and the angle EDF is equal to the angle GDF; wherefore the base EF is equal to the base #4. 1. FG, and the triangle EDF to the triangle GDF, and 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 +Const. +1 Ax. *Hyp. 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: wherefore also +32. 1. 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. and 3 Ax. Q. E. D. |