解:(1)直線AC與⊙O相切.························································· 1分
理由是:
連接OD,過點O作OE⊥AC,垂足為點E.
∵⊙O與邊AB相切于點D,
∴OD⊥AB.·························································································· 2分
∵AB=AC,點O為底邊上的中點,
∴AO平分∠BAC····················································································· 3分
又∵OD⊥AB,OE⊥AC
∴OD= OE····························································································· 4分
∴OE是
⊙O的半徑.
又∵OE⊥AC,∴直線AC與⊙O相切.·························································· 5分
(2)∵AO平分∠BAC,且∠BAC=60°,∴∠OAD=∠OAE=30°,
∴∠AOD=∠AOE=60°,