延长CD到点G,使得 DG = BE .
因为,在△ADG和△ABE中,AD = AB ,∠ADG = 90° = ∠ABE ,DG = BE ,
所以,△ADG ≌ △ABE ,
可得:AG = AE ,∠DAG = ∠BAE ;
则有:∠GAF = ∠DAG+∠DAF = ∠BAE+∠DAF = ∠BAD-∠EAF = 45° = ∠EAF ;
因为,在△AEF和△AGF中,AE = AG ,∠EAF = ∠GAF ,AF为公共边,
所以,△AEF ≌ △AGF ,
可得:EF = GF = DF+DG = DF+BE .