Two triangles ABC, DEF, which have an equal angle included between proportional sides, are similar. Let the angles A and D be equal; A D H B C E Take AG=DE, and draw GH parallel to BC. The angle AGH will be equal to the angle ABC; (24. 1;) and the triangles AGH, ABC will be equiangular; hence we shall have AB : AG:: AC: AH. But by hypothesis, we have AB : DE : : AC : DF; and by construction AG=DE; hence AH DF. The two triangles AGH, DEF have an equal angle included between equal sides; therefore they are equal. But the triangle AGH is similar to ABC; therefore DEF is also similar to ABC. Hence, If any two triangles have an angle of the one equal to an angle of the other, and the sides containing those angles proportional, those two triangles are similar. Two triangles, ABC, DEF, which have their homologous sides parallel, or perpendicular to each other, are similar. First. If the side AB is parallel to DE, and BC to EF, the angle ABC will be equal to DEF; (27. 1;) if AC is parallel to DF, the angle ACB will be equal to DFE, and also BAC to EDF; hence the triangles ABC, DEF are equiangular; therefore they are similar. (18. 4.) G H IE F B angles, (29. 1. Cor. 1,) the remaining two IAH, IDH will be together equal to two right-angles. But the two angles EDF, IDH are also equal to two right-angles: hence the angle EDF is equal to IAH or BAC. In like manner, if the third side EF is perpendicular to the third BC, it may be shown that the angle DFE is equal to C, and DEF to B: hence the triangles ABC, DEF, which have the sides of the one perpendicular to the corresponding sides of the other, are equiangular and similar. Hence, Any two triangles which have their homologous sides parallel, or perpendicular to each other, are similar. Scholium. When the sides are parallel, the homologous sides are the parallel ones: when they are perpendicular, the homologous sides are the perpendicular ones. Thus in the latter case DE is homologous with AB, DF with AC, and EF with BC. The lines AF, AG, &c. drawn at pleasure through the vertex of a triangle, will divide BC the base, and DE parallel to the base, in the same ratio; so that we shall have DI : BF: : IK FG :: KL GH, &c. For since DI is parallel to BF, the triangles ADI and ABF, are equiangular; and we have DI : BF : : AI : AF; and since IK is parallel to FG, in like manner AI: AF :: IK: FG; All lines drawn through the vertex of a triangle, will divide the base, and a line parallel to the base, in the same ratio. Cor. Therefore if BC were divided into equal parts at the points F, G, H, the parallel DE would also be divided into equal parts at the points, I, K, L. If from the right-angle A, the perpendicular AD be let fall on the hypothenuse of a right-angled triangle; then, 1. The two partial triangles ABD, ADC, will be similar to each other and to the whole triangle ABC. 2. Either side AB or AC will be a mean proportional between the hypothenuse BC and the adjacent segment BD or DC. 3. The perpendicular AD will be a. mean proportional between the two segments BD and DC. First. The triangles BAD and BAC have the common angle B, and the right-angle BDA=BAC, and therefore the third angle BAD of the one equal to the third C of the other: hence those two B D triangles are equiangular and similar. In the same manner it may be shown that the triangles DAC and BAC are similar; hence all the three triangles are similar and equiangular, Second. The triangles BAD, BAC being similar, their homologous sides are proportional. But BD in the small triangle and BA in the large one, are homologous, because they lie opposite the equal angles BAD, BCA; the hypothenuse BA of the small triangle is homologous with the hypothenuse BC of the large triangle; hence the proportion BD : BA :: BA : BC. By the same reasoning, we should find DC: AC :: AC: BC; hence each of the sides AB, AC is a mean proportional between the hypothenuse and the segment adjacent to that side. Third. Since the triangles ABD, ADC are similar, by comparing their homologous sides, we have BD: AD: : AD DC; hence, the perpendicular AD is a mean proportional between the segments DB, DC of the hypothenuse. Therefore, : In a right-angle triangle, if a perpendicular be let fall from the right-angle on the hypothenuse; then, 1. The two partial triangles will be similar to each other and to the whole triangle. 2. Either side will be a mean proportional between the hypothenuse and the segment adjacent to that side. 3. The perpendicular will be a mean proportional between the segments of the hypothenuse. Two triangles, ABC, ADE, having an equal angle in each, are to each other as the rectangles of the sides which contain that angle; that is, ABC ADE :: AB.AC: AD.AE. Multiply together the corresponding terms of these proportions, omitting the common term ABE; (18. 3. Cor. ;) we have ABC ADE: AB.AC: AD.AE. Hence, Any two triangles which have an equal angle in each, are to each other as the rectangles of the sides which contain that angle. Two similar triangles, ABC, DEF, are to each other as the squares of their homologous sides, AC, DF. Let the angle A be equal to D, and the angle B-E. Then, by reason of the equal angles A and D, according to the last proposition, we shall have ABC DEF:: AB.AC: DE.DF. A Also, because the triangles are similar, D B E F AB: DE:: AC: DF. And multiplying the terms of this proportion by the corres ponding terms of the identical proportion, AC: DF: AC: DF, |