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將△BPC繞點B逆時針方向旋轉至△BEA,連EP,所以EP=2根號2,又EA=3,AP=1,AD^2+EP^2=AE^2,故△AEP是直角三角形,故∠APE=90,所以∠APB=90+45=135,由余弦定理,AB^2=AP^2+BP^2-2*AP*BP*cos135=5+2√2,故AB=√(5+2√2)
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