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THANKSTHANKS THANKS分析：过D作DE⊥AB于E,根据等腰三角形性质推出AE=1/2AB,∠DEA=90°,求出AE=AC,根据SAS证△DEA≌△DCA,推出∠ACD=∠A

THANKS
THANKS
 

THANKS
分析：过D作DE⊥AB于E,根据等腰三角形性质推出AE=1/2AB,∠DEA=90°,求出AE=AC,根据SAS证△DEA≌△DCA,推出∠ACD=∠AED即可．
过D作DE⊥AB于E,
∵AD=BD  DE⊥AB
∴AE=1/2AB,∠DEA=90°,
∵AC=1/2AB
∴AE=AC
∵AD平分∠BAC
∴∠BAD=∠CAD,
在△DEA和△DCA中,

{AE＝AC    
{∠BAD＝∠CAD    
{AD＝AD,
∴△DEA≌△DCA,
∴∠ACD=∠AED,
∴∠ACD=90°,
∴AC⊥DC．

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