证明:
作AH⊥AF,交CD延长线于H
正方形ABCD ==> AD⊥AB;AD=AB
∵AH⊥AF,AD⊥裂巧AB
∴∠HAD = ∠FAB //* 一个角的两边与另一个角的两边垂直,二角相历源郑等或互肢颂补
又∠ADH =ABF=90°,AD=AB
∴ΔADH ≌ΔABF ==> AH = AF,DH=BF
∵∠EAF=45°
AH⊥AF ==> ∠FAH=90°
∴∠EAH = ∠FAH - ∠EAF = 45° = ∠EAF
又 AH=AF,AE=AE
∴ΔEAH ≌ΔEAF ==> EF = EH = DE+DH
而 DH=BF
∴EF = DE +BF
证毕