证明:将AF与BE的交点设为O
∵AD⊥BC
∴∠C+∠CAD=90
∵∠BAC=90
∴∠C+∠ABC=90, ∠BAF+∠CAF=90
∴∠CAD=∠ABC
∵BE平分∠ABC
∴∠ABE=∠CBE=∠ABC/2=∠CAD/2
∵AF平分∠CAD
∴∠CAF=∠CAD/2
∴∠ABE=∠CAF
∴∠AOE=∠BAF+∠ABE=∠BAF+∠CAF=90
∴∠AOB=∠FOB=90
∵BO=BO
∴△ABO≌△FBO (ASA)
∴AB=FB
∵BE=BE
∴△BAE≌△BFE (SAS)
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