证明：延长PC至D点,使得PA=PD,连接AD.
∵∠DPA=∠CBA=60°,∴⊿PAD是等边三角形,
∴DA=PA
∵AB=AC,PA=AD,∠BAP=∠CAB-∠PAC=∠DAP-∠PAC,
∴⊿APB≌⊿ACD∴BP=CD
∴PA=PC+CD=PC+PB,即PA=PB+PC