Question

（1）如图所示，∠BAC=90°，AD⊥BC，垂足为D，BE平分∠ABC，交AC于E，交AD于F，试判断△AEF的形状，

Answer 1

证：

∠BAC=90°

∠ABF+∠AFB=90°

AD⊥BC

∠CBF+∠BED=90°

AF=AE

∠AEF=∠AFE

∠BED=∠AEF

∠AFE=∠BED

∠ABF=∠CBF

BE平分∠ABC