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.
问题描述:
已知:如图,在△ABC中,∠ABC与∠ACB的平分线相交于点O.
求证:∠BOC=90°+
∠A.1 2
答
证明:∵∠ABC与∠ACB的平分线相交于点O,
∴∠OBC=
∠ABC,∠OCB=1 2
∠ACB,1 2
∴∠OBC+∠OCB=
(∠ABC+∠ACB),1 2
在△OBC中,∠BOC=180°-(∠OBC+∠OCB)
=180°-
(∠ABC+∠ACB)1 2
=180°-
(180°-∠A)1 2
=90°+
∠A,1 2
即:∠BOC=90°+
∠A.1 2