(1)证明:∵在□ABCD中,AD ∥ BC,AE⊥BC于E ∴AE⊥AD于A,∠FPE=∠ADP ∵AD=AE,∠EAD=90° ∴将△AEF绕点A逆时针旋转90°得到△ADG ∴△AEF≌△ADG,∠FAG=90° ∴AG=AF,∠ADG=∠AEF ∵EF⊥PD,AE⊥BC ∴∠AEF+∠PEF=90°,∠FPE+∠PEF=90° ∴∠AEF=∠FPE ∵∠ADG=∠AEF,∠FPE=∠ADP ∴∠ADG=∠ADP ∴点G在PD上 ∵AF=AG,∠FAG=90° ∴ FG=
∵FG=DF-DG=DF-EF ∴ DF-EF=
(2)DF+EF=
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