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来源：学生作业帮助网编辑：作业帮时间：2024/05/03 17:25:39

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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