解:(1)直线AC与⊙O相切.························································· 1分 理由是: 连接OD,过点O作OE⊥AC,垂足为点E.
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105060040-96431.gif) ∵⊙O与边AB相切于点D, ∴OD⊥AB.·························································································· 2分 ∵AB=AC,点O为底边上的中点, ∴AO平分∠BAC····················································································· 3分 又∵OD⊥AB,OE⊥AC ∴OD= OE····························································································· 4分 ∴OE是 ⊙O的半径. 又∵OE⊥AC,∴直线AC与⊙O相切.·························································· 5分 (2)∵AO平分∠BAC,且∠BAC=60°,∴∠OAD=∠OAE=30°, ∴∠AOD=∠AOE=60°, |