Question

如图,在Rt△ABC中,∠BAC=90°,AD⊥BC,点D为垂足,∠ABC的平分线分别交AD,AC与点E,F,求证：AE

Answer 1

在Rt△ABC中,∠BAC=90°,AD⊥BC

∠BAE=∠BDF=90

BE是∠ABC的平分线

∴∠ABE=∠DBF

∠ABE+∠AEB=90

∠DBF+∠DFB=90

∠AEB=∠DFB

∠DFB=∠AFE

∠AEB=∠AFE

AF=AE