在正方形ABCD中,点P是CD上一动点.连接PA,分别过点D、B作BE⊥PA、DF⊥PA,垂足分别为E、F

(1)证明:∵∠BAE+∠DAF=90°;
∠ADF+∠DAF=90°.
∴∠BAE=∠ADF(同角的余角相等);
又AB=AD;∠BEA=∠AFD=90°.
∴⊿BAE≌⊿ADF(AAS),AE=DF;BE=AF.
故:BE-DF=AF-AE=EF.
(2)当P点在DC延长线上时,DF-BE=EF.
证明:∵∠BAE=∠ADF(均为角DAF的余角);
AB=AD;∠BEA=∠AFD=90°.
∴⊿BAE≌⊿ADF,AE=DF;BE=AF.
∴DF-BE=AE-AF=EF.