已知:如图,在△ABC中,∠ABC与∠ACB的平分线相交于点O. 求证:∠BOC=90°+1/2∠A.
已知:如图,在△ABC中,∠ABC与∠ACB的平分线相交于点O.
求证:∠BOC=90°+
∠A.
求证:∠BOC=90°+
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数学人气:575 ℃时间:2019-11-04 22:14:02
优质解答
证明:∵∠ABC与∠ACB的平分线相交于点O,∴∠OBC=12∠ABC,∠OCB=12∠ACB,∴∠OBC+∠OCB=12(∠ABC+∠ACB),在△OBC中,∠BOC=180°-(∠OBC+∠OCB)=180°-12(∠ABC+∠ACB)=180°-12(180°-∠A)=90°+12∠A...
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