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CDE similar to GFE CE : EG = DE : EF = 10 : 5 = 2 : 1 ABC similar to ADE AE : AC = DE : BC = 10 : 6 = 5 : 3 let EG = x, AC = y CE = 2x CG = 2x - x = x (y + 2x) : y = 5 : 3 3(y + 2x) = 5y 6x = 2y 3x = y AC : CG : EG = 3x : x : x = 3 : 1 : 1

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其他解答:

AC/BC = AE/DE (property of similar triangle) AC/6 = AE/10............(1) AE = AC + CG + GE.........(2) By mid- point theorem, CG = GE ......................(3) Sub. (2) and (3) into (1) AC/6 = (AC + 2CG)/10 10AC = 6AC + 12 CG 4AC = 12CG so AC : CG = 3 : 1 and AC : CG : GE = 3 : 1 : 1.
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