Re: [中學] 幾何一題
設FD交GE於H
CG : GA = CE : EB
=> GE // AB 且 GE = (4/5)AB
=> FD垂直AB
設AD = 2x => DB = 3x,GE = 4x,HD = 2x = GH
GH = AD且HD垂直GH、AD
=> GHDA為正方形
=> GA = HD = 2x,GC = 8x
(10x)^2 + (5x)^2 = 15^2
正方形面積 = (2x√2)^2 = 8x^2 = 72/5
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