诉求,在△ABC中,AB=AC,P是BC上任意一点,连AP,说明AC²=AP²+CP·BP

作辅助线AD 垂直BC于点D
可知：BD=CD AB=AC
勾股定理:AC^2 = AD^2 + CD^2 (△ACD)
AP^2 = AD^2 + PD^2 (△ADP)
用AD代入可得：AC^2 = AP^2 - PD^2 + CD^2
= AP^2 + (CD^2 - PD^2)
= AP^2 + (CD + PD)(CD - PD) ---A^2-B^2=(A=B)(A-B)
= AP^2 + CP·BP