∠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°.
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