∠BPF=120°,
证明:∵在等腰梯形ABCD中,AD=CD=AB,∠BAE=∠D,DE=CF,
∴AE=DF
∴△ABE≌△DAF(SAS)
∴∠ABE=∠DAF,∠AEB=∠DFA,
∵∠ABC=∠C=60°,
∴∠BAD=∠CDA=120°,
∵∠ABE+∠AEB+∠BAD=180°,
∴∠ABE+∠AEB=60°,
∵∠DAF+∠AEB+∠APE=180°,
∠BPF=∠APE,
∴∠BPF=180°-(∠DAF+∠AEB)
=180°-(∠ABE+∠AEB)
=180°-60°
=120°.