令PC>PB
过A点做AD垂直于BC于D,由△ABC是等腰三角形可知,BD=CD
PA^2=AD^2+PD^2=AB^2-BD^2+(PC-CD)^2=AB^2-BD^2+PC^2-2PC×CD+CD^2
=AB^2+PC^2-PC×BC=AB^2+PC(BC-PB)-PC×BC=AB^2-PC×PB
即PA^2+PC×PB=AB^2=36
令PC>PB
过A点做AD垂直于BC于D,由△ABC是等腰三角形可知,BD=CD
PA^2=AD^2+PD^2=AB^2-BD^2+(PC-CD)^2=AB^2-BD^2+PC^2-2PC×CD+CD^2
=AB^2+PC^2-PC×BC=AB^2+PC(BC-PB)-PC×BC=AB^2-PC×PB
即PA^2+PC×PB=AB^2=36