如图,三角形ABC的三条角平分线AD、BE和CF交于一点O,且OH垂直BC于H.
试说明角COH等于角ABO加角BAO.
人气:342 ℃ 时间:2019-08-20 12:45:59
解答
∠COH=90°-1/2∠ACB=90°-1/2[180°-(∠ABC+∠CAB)]
=1/2(∠ABC+∠CAB)=1/2∠ABC+1/2∠CAB=∠ABO+∠BAO.
所以∠COH=∠ABO+∠BAO.
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