在三角形ABC中,AB=AC=4,P为BC上任意一点,求AP平方加BP乘以CP

作AD垂直BC于D,BD=DC
AP^2=AD^2+PD^2=AC^2-DC^2+PD^2
=AC^2-(DC^2-PD^2)
=AC^2-(DC-PD)(DC+PD)
=AC^2-(BD-PD)(DC+PD)
=AC^2-BP*CP
AP^2+BP*CP=AC^2=16