正方形ABCD中,F在BC上,E在CD上,AE=BC+CE,∠BAF=∠FAE 求证：BF=CF

正方形ABCD,AB=BC,
在AE上找1点C',使得C'E=CE,
则AC'=AE-C'E=BC+CE-C'E=BC=AB,
∠BAF=∠FAE
AF=AF,
△AFC'≌△AFB,[SAS]
FC'=FB,
∠AC'F=∠B=90°.
FC²=FE²-CE²=FE²-CE'²=FC'²
FC=FC'=FB
BF=CF.