Elements of geometry and trigonometry . fwe had AB : AD : : AE : AC ; w hich would happen if DCwere parallel to BE. PROPOSITION XXV. THEOREM. Tiio similar triangles are to each other as the squares describedon their homologous sides. Let ABC, DEF, be two similar trian-gles, having the angle A equal to D, andthe angle B = E. Then, first, by reason of the equal an-gles A and D, according to the last pro-position, we shall have ABC : DEF : : : , because the triangles are similar, AB : DE : : AC : DF,And multiplying the terms of this proportion by the corresponding terms of the ide
Elements of geometry and trigonometry . fwe had AB : AD : : AE : AC ; w hich would happen if DCwere parallel to BE. PROPOSITION XXV. THEOREM. Tiio similar triangles are to each other as the squares describedon their homologous sides. Let ABC, DEF, be two similar trian-gles, having the angle A equal to D, andthe angle B = E. Then, first, by reason of the equal an-gles A and D, according to the last pro-position, we shall have ABC : DEF : : : , because the triangles are similar, AB : DE : : AC : DF,And multiplying the terms of this proportion by the corresponding terms of the identical proportion, AC : DF : : AC : DF,there v/ill result : : : AC- : , ABC : DEF : : AC^ : , two similar triangles ABC, DEF, are to eachother as the squares described on their homologous sides AC,DF, or as t!ie squares of any other two homologous sides. PROPOSITION XXVI. THEOREM. TVo similar polygons are composed of the same number of tri-angles, similar each to each, and similarly situated. mmmm. 92 GEOMETRY. Let ABCDE, FGHIK, be two similar polygons. From any angle A, in ç the polygon ÀBCDE,draw diagonals AC, ADto the other angles. Fromthe homologous angle F,in the other polygonFGHIK, draw diagonalsFH, FI to the other an-gles. These polygons being similar, the angles ABC, FGH, whichare homologous, must be equal, and the sides AB, BC, mustalso be proportional to FG, Gil, that is, AB : FG : : BC ;GH (Def. 1.). Wherefore the triangles ABC, FGH, have eachan equal angle, contained between proportional sides ; hencethey are similar (Prop. XX.) ; therefore the angle BC A is equalto GHF, Take away these equal angles from the equal anglesBCD, GUI, and there remains ACD=FHI. But since thetriangles ABC, FGH, are similar, we have AC : FH : : BC :GH ; and, since the polygons are similar, BC : GH : : CD :III ; hence AC : FH : : CD : HI. But the angle ACD, wealready know, is equal to FHI ; hence the triangles ACD, FHL,have an equal angle in each, in
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