(8分)如圖,在△ABC中,AB=AC,點(diǎn)O為底邊上的中點(diǎn),以點(diǎn)O為圓心,
1為半徑的半圓與邊AB相切于點(diǎn)D.
(1)判斷直線AC與⊙O的位置關(guān)系,并說(shuō)明理由;
(2)當(dāng)∠A=60°時(shí),求圖中陰影部分的面積.
解:(1)直線AC與⊙O相切.···································································· 1分
理由是:
連接OD,過(guò)點(diǎn)O作OE⊥AC,垂足為點(diǎn)E.
∵⊙O與邊AB相切于點(diǎn)D,
∴OD⊥AB.·························································································· 2分
∵AB=AC,點(diǎn)O為底邊上的中點(diǎn),
∴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°,
【解析】略
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