
如图所示,在△ABC中AB=AC,BD⊥AC于D,CE⊥AB于E,BD,CE交于F,说明AF平分∠BAC
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证明:
∵∠AEC = ∠ADB = 90 °
∠BAC为公共角
AB = AC
∴ △ADB≌△AEC 且为直角三角形
∵AB = AC
∴∠ABC = ∠ACB
又∵∠ABD = ∠ACE
=>∠DBC = ∠ECB
=>BF = CF
=>∠BAF = ∠CAF
即证:AF平分∠BAC
∵∠AEC = ∠ADB = 90 °
∠BAC为公共角
AB = AC
∴ △ADB≌△AEC 且为直角三角形
∵AB = AC
∴∠ABC = ∠ACB
又∵∠ABD = ∠ACE
=>∠DBC = ∠ECB
=>BF = CF
=>∠BAF = ∠CAF
即证:AF平分∠BAC
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