三角形ABC三条角平分线相交于点O,证明:∠BOC=90°+二分之一∠BAC
问题描述:
三角形ABC三条角平分线相交于点O,证明:∠BOC=90°+二分之一∠BAC
答
因为∠BOC=180°-1/2∠ABC-1/2∠ACB=180°-1/2(∠ABC+∠ACB)=180°-1/2(180°-∠BAC)=90°+1/2∠BAC
三角形ABC三条角平分线相交于点O,证明:∠BOC=90°+二分之一∠BAC
因为∠BOC=180°-1/2∠ABC-1/2∠ACB=180°-1/2(∠ABC+∠ACB)=180°-1/2(180°-∠BAC)=90°+1/2∠BAC