Question

（8分）如图，在△ABC中，AB=AC，点O为底边上的中点，以点O为圆心，

1为半径的半圆与边AB相切于点D．

（1）判断直线AC与⊙O的位置关系，并说明理由；

（2）当∠A＝60°时，求图中阴影部分的面积．

Answer

【解析】
（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．

∵⊙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°，

【解析】略