如图,三角形ABC中,AB=AC,P为BC上一点,BP=2,CP=5,求AB²-AP²

如图,过点A作BC的垂线,垂足为D已知AB=AC则点D为BC中点所以,BD=CD由勾股定理有：AB^2=AD^2+BD^2；AP^2=AD^2+PD^2所以,AB^2-AP^2=(AD^2+BD^2)-(AD^2+PD^2)=BD^2-PD^2=(BD+PD)*(BD-PD)=(CD+PD)*PB=PC*PB即,AB^2-AP^2=PB*PC=10