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如图,O 是△ABC的3条角平分线的交点,OG垂直于BC,垂足为G.
1)猜想
人气:437 ℃ 时间:2019-10-25 11:55:33
解答
(1)∠BOC=∠BOD+∠COD=(1/2∠BAC+1/2∠ABC)+(1/2∠BAC+1/2∠ACB)=1/2∠BAC+1/2(∠BAC+∠ABC+∠ACB)=90+1/2∠BAC、(2)∠DOB=1/2∠BAC+1/2∠ABC=1/2(∠BAC+∠ABC+∠ACB)-1/2∠ACB=90-1/2∠ACB,∠GOC=90-1/2...
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