如图,AD是Rt三角形ABC的斜边BC上的高,角B的平分线BE交AD于F,交AC于E,求证:AE=AF

问题描述:

如图,AD是Rt三角形ABC的斜边BC上的高,角B的平分线BE交AD于F,交AC于E,求证:AE=AF

:∵∠BAC=90°∴∠ABE+∠AEF=90°
∵∠ADB=90°,∴∠DBE+∠DFB=90°
∵∠ABE=DBE∴∠AEF=∠DFB
∵∠DFB=∠AFE∴∠AEF=∠AFE
∴AE=AF

证明:∵∠BAC=90°∴∠ABE+∠AEF=90°
∵∠ADB=90°,∴∠DBE+∠DFB=90°
∵∠ABE=DBE∴∠AEF=∠DFB
∵∠DFB=∠AFE∴∠AEF=∠AFE
∴AE=AF