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 1

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．

∵⊙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