三角形abc的角b、角c的外角平分线交于点o,则角boc等于多少?(答案与角a有关)

问题描述:

三角形abc的角b、角c的外角平分线交于点o,则角boc等于多少?(答案与角a有关)

∠BOC=90°-1/2∠A
证明:∠BOC=180°-(∠OBC+∠OCB)
=180°-(1/2(∠180°-∠ABC)+1/2(∠180°-∠ACB))
=180°-180°+1/2(∠ABC+∠ACB)
=1/2(180°-∠A)
=90°-1/2∠A