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See complete Problem 97Similar Triangles, Areas. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
We've AC = AM + MC = DF+ MC = DE + EF + MCor 1 = DE/AC + EF/AC + MC/AC ---(1). Noting that triangles GMC, GEF, BDE & BAC are all similar we can say that their areas are proportional to the squares of their corresponding sides. Hence DE/AC = V(S1/S) where V(x)=square root of x. Similarly, EF/AC=V(S2/S) and MC/AC=V(S3/S). BY (1) we can say that 1 = V(S1/S)+V(S2/S)+V(S3/S) or VS= V(S1)+ V(S2)+ V(S3) QEDAjit
Solution to problem 97.Let h, h1, h2 and h3 be the altitudes of triangles ABC, DBE, FGE and MGC, corresponding to sides BC, BE, EG and GC, respectively.As these four triangles are similar,BC/h = BE/h1 = EG/h2 = GC/h3 = k.For the areas we haveS = BC.h/2 = (k/2).h*2,S1 = BE.h1/2 = (k/2).h1*2,S2 = EG.h2/2 = (k/2).h2*2and S3 = GC.h3/2 = (k/2).h3*2.Then sqr(S) = sqr(k/2).h,sqr(S1) = sqr(k/2).h1,sqr(S2) = sqr(k/2).h2and sqr(S3) = sqr(k/2).h3.So sqr(S1) + sqr(S2) + sqr(S3) == sqr(k/2).(h1 + h2 + h3) == sqr(k/2).h = sqr(S).We have used here the result of problem 91,h = h1 + h2 + h3.