连结BD∵AB=AD∴∠ABD=∠ADB∵∠ABC=∠ADC∴∠CBD=∠CDB∴BC=CD∵AB=AD,AC=AC∴⊿ABC≌⊿ADC﹙SSS﹚∴∠BAC=∠DAC即AC平分∠BAD
证明：连接BD∵AB=AD ，∠ABC=∠ADC∴∠ABD=∠ADB∴∠DBC=∠BDC∴CB=CD又∵AC=CA∴△ADC≌△ABC∴∠DAC=∠BAC∴AC平分∠BAD
