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°+

1
2
∠A.

证明:∵∠ABC与∠ACB的平分线相交于点O,
∴∠OBC=

1
2
∠ABC,∠OCB=
1
2
∠ACB,
∴∠OBC+∠OCB=
1
2
(∠ABC+∠ACB),
在△OBC中,∠BOC=180°-(∠OBC+∠OCB)
=180°-
1
2
(∠ABC+∠ACB)
=180°-
1
2
(180°-∠A)
=90°+
1
2
∠A,
即:∠BOC=90°+
1
2
∠A.