答：△ABC为等边三角形
证明：∵∠CPB与∠CAB同弧且∠CPB=60°
∴∠CAB=∠CPB=60°
又∵∠APC与∠ABC同弧且∠APC=60°
∴∠ABC=∠APC=60°
则在△ABC中,∠ABC=∠CAB=60°
∴△ABC为等边三角形.