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數學考古題求計算過程

Jan 10th 2014, 14:33

首先三角形BGC與ABC互為 (SSA)相似 A為"角C"
接著
FE線段平行BG線段
FG線段也平行BC
BGC與FEG這兩個三角形 (SAS)

所以FG線段其實也等於FE線段

建立起所求與FE線段之間的關係之後

在看到DEF與FGB這兩個相似三角形
BG=BC=12 FG=FE=X DE=6

DE : FE = FG : BC
6 : X = X :12

X^2=72 X=根號72

參考資料有問題再補充

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