解答

設△ADE = x,DEGF = x/3,FGCB = x/2
△AFG = x+x/3 = 4x/3
△ABC = (4x/3)+(x/2) = 11x/6

DE² : FG² = △ADE : △AFG = x:(4x/3) = 3:4
所以DE : FG = √3:2
FG² : BC² = △AFG : △ABC = 8:11
FG : BC = √8:√11 = (2√2):11

所以DE : FG : BC = √6:(2√2):√11

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