请看连结OQ,过O做ON垂直于QR交QR于N设圆半径为r欲证明2(QP2+PR2)=AB2即证明2QP2+2PR2+4PQ*PR=AB2+4PO*PR由相交弦定理知:PQ*PR=PA*PB即证明:2(PQ+PR)2=AB2+4PA*PB由于PQ+PR=2QN代入即证明8NQ2=AB2+4PA*PB由于NQ2=r2-ON2代入即证明:8r2-8ON2=AB2+4PA*PB由于在三角形ONP中角QPA=45度,有2ON2=OP2;又知道AB2=4r2代入得:8r2-4OP2=4r2+4PA*PB由于OP=(PA-PB)/2代入即证明:4r2-(PA-PB)2=4PA*PB即证明:4r2=(PA+PB)2即证明:AB2=AB2显然成立原式得证