等腰三角形ABC中,AB=AC=6,P为BC上一点,且PA=4,则PB×PC的值等于多少?

作AD垂直于BC交BC于D
PA^2=AD^2+PD^2 (勾股定理)
BD=CD (三线合一)
PB*PC=BD-PD)(CD+PD)
=（BD-PD)(BD+PD)
=BD^2-PD^2
=AB^2-AD^2-(AP^2-AD^2)
=AB^2-AP^2
=36-16=20