PROPOSITION XV. THEOREM. If, in any triangle, a line be drawn parallel to the base, it will divide the two other sides proportionally. Let ABC be a triangle, and DE a straight line drawn parallel to the base BC; then will AD : DB ᎠᏴ :: AE EC. Draw the lines BE and CD. Then, the triangles ADE, BDE, having a common vertex, E, have the same altitude, and are to each other as their bases (P. 6, c.); hence we have A D E ADE : BDE :: AD : DB. The triangles ADE, DEC, having a B common vertex D, also have the same altitude, and are to each other as their bases; hence, ADE : DEC :: AE EC. But the triangles BDE, DEC, are equivalent, having the same base DE, and their vertices B and C in a line parallel to the base: and therefore, we have (B. II., P. 4, C.) AD : DB :: AE : EC. Cor. 1. Hence, by composition, we have (B. II., P. 6), AD+DB: AD :: AE+EC: AE, or AB: AD:: AC: AE; and also, AB: BD :: AC: CE. Cor. 2. If any number of parallels AC, EF, GH, BD, be drawn between two straight lines AB, CD, those straight lines will be cut proportionally, and we shall have EG: FH : GB AE CF CF :: For, let O be the point where AB and CD meet. In the triangle OEF, the line AC being drawn parallel to the base EF, we shall have OE : AE : OF: CF. In the triangle OGH, we shall like- OE : EG Conversely: If two sides of a triangle are cut proportionally by a straight line, this straight line will be parallel to the third. side. In the triangle BAC, let the line DE be drawn, cutting the sides BA and CA proportionally in the points D and E; that is, so that BD : DA :: CE : EA: BDE: DAE :: DEC DAE, and since BDE and DEC have the same ratio to DAE, they have the same area, and hence are equivalent (D. 4). They also have a common base BC; hence, they have the same altitude (P. 6, c.); and consequently, their vertices D and E lie in a parallel to the base BC (B. I., P. 23). PROPOSITION XVII. THEOREM. The line which bisects the vertical angle of a triangle, divides the base into two segments, which are proportional to the adjacent sides. In the triangle ACB, let AD be drawn, bisecting the angle CAB; then will BD : CD :: AB : AC. Through the point C draw CE parallel to AD, and prolong it till it meets BA produced in E. In the triangle BCE, the line AD is parallel to the base CE; hence, we have the proportion (P. 15), E A But the triangle ACE is isosceles: for, since AD, CE, are parallel, we have the angle ACE=DAC, and the angle AEC=BAD (B. I., P. 20, C. 2, 3); but, by hypothesis, DAC =DAB; hence, the angle ACE=AEC, and consequently, AE=AC (B. I., P. 12). In place of AE in the above proportion, substitute AC, and we shall have, Cor. If the line AD bisects the exterior angle CAE of the triangle BAC, we shall have, BD : CD : AB : AC. For, through C draw CF par PROPOSITION XVIII. THEOREM. Equiangular triangles have their homologous sides proportional, and are similar. Let BCA and CED be two equiangular triangles, having the angle BAC-CDE, ABC=DCE, and ACB =DEC; then, the homologous sides will be proportional, viz.: BA: :: B A F CD :: AC : DE. D E BC : CE Place the homologous sides BC, CE in the same straight line; and prolong the sides BA, ED, till they meet in F. Since BCE is a straight line, and the angle BCA equal to CED, it follows that AC is parallel to DE (B. I., P. 19, c. 2). In like manner, since the angle ABC is equal to DCE, the line AB is parallel to DC. Hence, the figure ACDF is a parallelogram, and has its opposite sides equal (B. I., P. 28). In the triangle BEF, the line AC is parallel to the base FE; hence, we have (P. 15,) BC : CE :: BA AF; : or putting CD in the place of its equal AF, BC: CE :: BA : CD. In the same triangle BEF, CD is parallel to BF'; and hence, BC CE :: FD: DE; : or putting AC in the place of its equal FD, And finally, since both these proportions contain the same ratio BC to CE, we have (B. II., P. 4, c.), BA CD: AC: DE Thus, the equiangular triangles CAB, CED, have their homologous sides proportional. But two figures are similar when they have their angles equal, each to each, and their homologous sides proportional (D. 1, 2); consequently, the equiangular triangles BAC, CED, are two similar figures. Cor. Two triangles which have two angles of the one equal to two angles of the other, are similar; for, the third angles are then equal, and the two triangles are equiangular (B. I., P. 25, c. 2.) Scholium. Observe, that in similar triangles, the homologous sides are opposite to the equal angles; thus, the angle BCA being equal to CED, the side AB is homologous to DC; in like manner AC and DE are homologous, because opposite to the equal angles ABC, DCE. PROPOSITION XIX. THEOREM. Conversely: Triangles, which have their sides proportional, are equiangular and similar. If, in the two triangles BAC, EDF, we have, BC : EF :: BA : ED AC: DF; :: then will the triangles BAC, EDF, have their angles equal, namely, A=D, B=E, C=F. At the point E, make the angle FEG=B, and at F, the angle EFG =C; the third angle G will then be equal to the third angle A (B. I., P. 25, c. 2). Therefore, by the last theorem, we shall have B BC : EF :: AB: EG : but, by hypothesis, we have BC EF:: AB: DE; hence, EG-DE. By the same theorem, we shall also have BC EF :: AC : FG: and by hypothesis, we have BC : EF AC: DF; :: hence, FG=DF. Hence, the triangles EGF, FED, having |