PROPOSITION IV. THEOREM. The sides about the equal angles of equiangular triangles are proportional; and the sides which are opposite to the equal angles are homologous sides, that is, they are the antecedents or the 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 the angle BAC equal to the angle CDE (2 Cor. 32. 1). The sides about the equal angles of the triangles ABC, DCE are proportional; and those are the homologous sides which are opposite to the equal angles. "Thus, the angle at A being equal to the angle at D, the sides about the former are proportional to the sides about the latter, namely, AB AC: ́: DC: DE; also the two antecedents AB, DC are opposite to the equal angles at C, E; and the two consequents AC, DE are opposite to the equal angles at B, C." be Let the triangle DCE be placed so that its side CE may contiguous to BC, and in the same straight line with it. "Because the angles ABC, ACB are together less than two right angles, the angles ABC, DEC are also less than two right angles; wherefore BA, ED produced will meet. Let them be produced, and meet in the point F. Because the angle ABC is equal to DCE, BF is parallel to CD (28. 1); and because the angle ACB is equal to DEC, AC is parallel to FE. Therefore FACD is a parallelogram; consequently AF is equal to CD (34. 1), and AC to FD. Because AC is parallel to FE, one of the sides of the triangle FBE, BA: AF BC CE (2. 6), or, because AF is equal to CD, BA: CD :: BC :CE; therefore, alternately, BA: BC B :: DC: CE (Propor. 36). Again, F D C E because CD is parallel to BF, BC: CE:: FD: DE, or, because FD is equal to AC, BC: CÉ :: AC: DE; therefore BC : CA :: CE: ED. But BA: BC: : DC : CE; : therefore BA: AC: CD: DE (Propor. 42, 43). Therefore, the sides &c. Q. E. D. COR. It follows from this prop. that equiangular triangles are of necessity similar figures, according to Def. 1; therefore similar triangles are often used to denote equiangular triangles. PROPOSITION C. THEOREM. LUDLAM. LUDLAM. If a straight line be drawn from the vertex of any triangle to the base, it will cut every line which is parallel to the base in the same ratio as the segments of the base. Let ABC be a triangle, BC the base, EF a line parallel to the base, ADM a line drawn from the vertex A, cutting EF in D, and the base in M; the segments of EF have the same ratio to each other as the segments of the base BC; DE: DF:: MB: MC. In the triangles AED, ABM the angles AED, ABM are equal (29. 1), and the angle FAD is common, therefore the remaining angles ADE, AMB are equal (2 Cor. 32. 1); therefore the triangles AED, ABM are equiangular. In the same man- E ner the triangles ADF, AMC are equiangular. Hence AD: AM :: DE: MB (4. 6), and AD: AM:: DF: MC, B M therefore DE: MB:: DF: MC (Propor. 34); therefore DE: DF:: MB: MC (Propor. 36). A F D COR. 1. If a straight line drawn from the vertex of a triangle to the base bisect the base, it will bisect every line parallel to the base, and terminating in the two sides of the triangle. COR. 2. Parallel lines BC, EF are cut proportionally by diverging lines AB, AM and AC, AM. For MB: MC:: DE: DF. COR. 3. Diverging lines are proportional to the corresponding segments into which they divide parallel lines; AB`: AE :: BM: ED (4. 6) : : MC: DF. PROPOSITION D. THEOREM. ED. If a straight line be drawn parallel to the base of a triangle, and another straight line be drawn from the vertex to meet the two parallels, the segments of that line intercepted between the vertex and the parallels will be to each other as the two parallels.-See last fig. Let a straight line EF be drawn parallel to the base BC of a triangle ABC, and from the vertex A let AM be drawn to meet the parallels BC, EF; AM: AD: : BC : EF. For the triangles AEF, ABC are equiangular, and also the triangles AED, ABM; therefore AE: EF:: AB: BC (4. 6), and AD AE::AM: AB; therefore AD: EF:: AM: BC (Propor. 42, 43), PROPOSITION V. THEOREM. If the sides of two triangles, about each of their angles, be proportional, the triangles will be equiangular, and will have their equal angles opposite to the homologous sides. Let the triangles ABC, DEF have their sides proportional, so that AB: BC :: DE ; EF, and BC: CA:: EF: FD, and BA: AC:: ED: DF; the triangle ABC is equiangular to the triangle DEF, and their equal angles are opposite to the homologous sides, namely, the angle ABC being equal to the angle DEF, and the angle BCA to EFD, and also the angle BAC to EDF. At the points E, F, in the line EF, make the angle FEG equal to the angle ABC, and the angle EFG equal to BCA; wherefore the angle BAC is equal to EGF (2 Cor. 32. 1); therefore the triangle ABC is equiangular to the triangle GEF; consequently they have their sides opposite to the equal angles proportional (4. 6). Wherefore DE: EF::GE: EF (Propor. 34). Therefore DE and GE are equal (Propor. 32). For the same reason DF is equal to FG, In the triangles DEF, GEF, DE is equal to EG, and EF common, and DF equal to GF; therefore the angle DEF is equal to GEF; wherefore the angle DFE is equal to GFE, and the angle EDF to EGF. Now the angle DEF is equal to GEF, and GEF to the angle ABC; therefore the angle ABC is equal to DEF. For the same reason the angle ACB is equal to DE, and the angle A to the angle D. Therefore the triangle ABC is equiangular to the triangle DEF. And they have their equal angles opposite to the homologous sides, namely, the equal angles ACB and DFE opposite to the homologous sides AB and DE, and the equal angles A and D opposite to the homologous sides BC and EF, and the equa angles ABC and DEF opposite to the homologous sides AC and DF.' Wherefore, if the sides &c. Q. E. D. PROPOSITION VI. THEOREM. If two triangles have one angle of one triangle equal to one angle of the other, and the sides about the equal angles proportional, the triangles will be equiangular, and will have those angles equal which are opposite to the homologous sides. Let the triangles ABC, DEF have the angle BAC in one triangle equal to the angle EDF in the other, and the sides about those angles proportional, that is, BA to AC as ED to DF; the triangles ABC, DEF are equiangular, and have those angles equal which are opposite to the homologous sides, namely, the angle ABC equal to the angle DEF, and the angle ACB equal to DFE. At the points D, F, in the line DF, make the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to ACB; wherefore the remaining angle B is equal to G (2 Cor. 32. 1); consequently the triangle ABC is equiangular to the triangle DGF; therefore BA: AC:: GD: DF (4.6). But, by hypothesis, BA: AC:: ED: DF; therefore ED: DF:: GD: DF (Propor. 34); wherefore ED is equal to DG (Propor. 32). The triangles EDF, GDF have the two sides ED, DF equal to the two sides GD, DF, and also the angle EDF equal to GDF; wherefore the base EF is equal to FG, and the angle DFG is equal to DFE, and the angle G to the angle E. But the angle DFG is equal to ACB; therefore the angle ACB is equal to DFE; also the angle BAC is equal to EDF; wherefore the remaining angle B is equal to E. Therefore the triangle ABC is equiangular to the triangle DEF. And they have the angles ABC, DEF equal, which are opposite to the homologous sides AC, DF, and the angles ACB, DFE equal, which are opposite to the homologous sides AB, DE.' Wherefore, if two triangles &c. Q. E. D.* PROPOSITION E. THEOREM. LEGENDRE. If the homologous sides of two triangles be parallel to each other, or perpendicular each to each; the triangles are similar. 1. Let ABC, DEF be two triangles which have the sides AB, BC, AC, parallel to the sides DE, EF, DF, each to each; the triangles ABC and DEF are similar. Because AB is parallel to DE, and BC to EF, the angle ABC is equal to DEF (D. 1); and because ÀC is parallel to DF, the angle ACB is equal to DFE; consequently the angle BAC is equal to EDF. Hence B the triangles ABC, DEF are equiangular. H A E G * See a more simple demonstration in Hutton's Mathematics, Geometry, Theor. 86; or West's Mathematics, Prop. V. B. V. ED. |