已知∠B、∠C的平分线交于点O,求证:∠BOC=90°+二分之一∠A
问题描述:
已知∠B、∠C的平分线交于点O,求证:∠BOC=90°+二分之一∠A
答
证明:
∠BOC=180°-∠OBC-∠OCB
=180°-1/2∠B-1/2∠C
=180°-1/2(∠B+∠C)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A