Elements of geometry and trigonometry . paredwith ABC ; and then the demonstration given in the text wouldapply. GEOMETRY. PROPOSITION XXII. THEOREM. In any triangle, if a line he drawn parallel to the hase, then, alllines drawn from ike vertex will divide the base and the par-allel into proportional parts. Let DE be parallel to the base BC, andthe other lines drawn as in the figure ;then willDI : BF : : IK : FG : : KL : GIL For, since DI is parallel to BF, thetriangles ADI and ABF are equiangu-lar; and we have 1)1 : BF : : AI :AF ; and since IK is parallel to FG,w^e have in like manner AI : À
Elements of geometry and trigonometry . paredwith ABC ; and then the demonstration given in the text wouldapply. GEOMETRY. PROPOSITION XXII. THEOREM. In any triangle, if a line he drawn parallel to the hase, then, alllines drawn from ike vertex will divide the base and the par-allel into proportional parts. Let DE be parallel to the base BC, andthe other lines drawn as in the figure ;then willDI : BF : : IK : FG : : KL : GIL For, since DI is parallel to BF, thetriangles ADI and ABF are equiangu-lar; and we have 1)1 : BF : : AI :AF ; and since IK is parallel to FG,w^e have in like manner AI : ÀF : :IK : FG ; hence, the ratio AI : AF being common, we shallhave DI : BF : : IK : FG. In the same manner we shallfind IK : FG : : KL : GH ; and so with the other segments :hence the line D£ is divided at the points I, K, L, in the sameproportion, as the base BC, at the points F, G, H. Co7\ Therefore if BC were divided into equal parts at thepoints F, G, H, the parallel DE would also be divided into equalparts at the points I, K, PROPOSITION XXIII. THEOREM. If from the right angle of a right angled triangle, a peipendini-lar be let fall on the hypothenuse ; then, 1st. The two partial triangles thus formed, will be similar to eachoilier^ and to the whole triangle. 2d, Either side including the right angle will be a mean propor-tional between the hypothenuse and tJie adjacent segment. ^d. The perpendicular will be a mean proportional between thetwo segments of the hypothenuse. L(Jt BAC be a right angled triangle, and AD perpendicularto the hypothenuse BC; First. The triangles BAD and BAChave the common angle B. the rightangle BDA^BACj and therefore thethird angle BAD of the one, equal tothe third angle C, of the other (BookI. Prop. XXV. Cor 2.) : hence those
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