延长CD至G,使DG=BE,连接AG
BE=DG AB=AD ∠B=∠ADG=RT∠
∴△ABE≌△ADG
∴∠BAE=∠DAG
∠GAG=∠GAD+∠DAF=∠BAE+∠DAF
∠AEF=45°=1/2∠BAD=∠BAE+∠DAF=∠GAG
AE=AG
∴△AFE≌△AFG
即得：EF=FG=FD+DG=FD+BE