BE⊥AC于E,CF⊥AB于F,BE、CF相交于点D,且CE=BF,求证:AD平分∠BAC
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"有题意可得：" BE⊥AC，BE⊥AC
"∵" ∠DFB=∠CED, ∠FDB=∠EDC, CE=BF
"∴" △CDE≌△BDF
"∴" CD=BD
"∵" ∠B=∠C, CD=BD,AD=AD
"∴" △ADB≌△ADC
"∴" ∠BAD=∠CAD
"∴" AD平分∠BAC
"∵" ∠DFB=∠CED, ∠FDB=∠EDC, CE=BF
"∴" △CDE≌△BDF
"∴" CD=BD
"∵" ∠B=∠C, CD=BD,AD=AD
"∴" △ADB≌△ADC
"∴" ∠BAD=∠CAD
"∴" AD平分∠BAC
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