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圖你就自己畫.我設正方形邊長為1,FC=x.則FD=1-x. 由直角三角形ACF,有:AF^2=AE^2+EF^2 又AF^2=1^2+(1-x)^2 AE^2=1+1/4 EF^2=x^2+1/4 都帶入畢氏定理式 解出x=1/4 由數據關係可得:AF=AD+FC
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