在△ABC中,AB=AC=4,p是BC上异于B,C的点,求AP²＋BP×PC的值.

Question

在△ABC中,AB=AC=4,p是BC上异于B,C的点,求AP²＋BP×PC的值.

Answer 1

证明：作AD⊥BC于D,则BD=CD,由勾股定理可得
AP^2=PD^2+AD^2
AD^2=AB^2-BD^2=16-BD^2
所以AP^2+PB×PC=PD^2+AD^2+PB×PC
=PD^2+16-BD^2+(BD-PD)(CD+PD)
=PD^2+16-BD^2+BD^2-PD^2
=16