连接AD
∵在△ABC中,∠BAC=90°,AB=AC,D是BC的中线的中点
==>△ABC是等腰直角三角形,且AD是∠BAC的角平分线和AD⊥BC
==>∠FBD=∠FAD=∠EAD=45°,且∠ADB=∠ADC=90°
∴AD=BD.(1)
∠FBD=∠EAD.(2)
∵AE=BF.(3)
∴由(1)(2)(3)知,△ADE≌△BDF (边、角、边)
==>∠BDF=∠ADE
∵∠ADB=∠ADC=90°
∴∠ADF=∠CDE
故∠EDF=∠ADE+∠ADF
=∠ADE+∠CDE
=∠ADC
=90°.