(1)△ABP≌△DCP,
△BEP≌△CFP,
△BPF≌△CPE,
(2)选△ABP≌△DCP进行证明,
证:∵梯形ABCD,AD∥BC,AB=CD,
∴∠BAD=∠CDA,
又∵PA=PD,
∴∠1=∠2,
∴∠BAD-∠1=∠CDA-∠2,
∴∠3=∠4,
在△ABP与△DCP中:
AB=DC
∠3=∠4
PA=PD,
∴△ABP≌△DCP.
(1)△ABP≌△DCP,
△BEP≌△CFP,
△BPF≌△CPE,
(2)选△ABP≌△DCP进行证明,
证:∵梯形ABCD,AD∥BC,AB=CD,
∴∠BAD=∠CDA,
又∵PA=PD,
∴∠1=∠2,
∴∠BAD-∠1=∠CDA-∠2,
∴∠3=∠4,
在△ABP与△DCP中:
AB=DC
∠3=∠4
PA=PD,
∴△ABP≌△DCP.