摘要:△ABC中.点O为∠ABC和∠ACB角平分线交点.则∠BOC与∠A的关系是( ) A.∠BOC =2∠A B.∠BOC =180o-∠A C.∠BOC =90o+∠A D.∠BOC=900+∠A
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△ABC中,点O为∠ABC和∠ACB角平分线交点,则∠BOC与∠A的关系是( )
| A、∠BOC=2∠A | ||
B、∠BOC=180°-
| ||
C、∠BOC=90°+
| ||
| D、∠BOC=90°+∠A |
△ABC中,点O为∠ABC和∠ACB角平分线交点,则∠BOC与∠A的关系是( )
A.∠BOC=2∠A
B.∠BOC=180°
∠A
C..∠BOC=90°+∠A
D∠BOC=90°+
∠A
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△ABC中,点O为∠ABC和∠ACB角平分线交点,则∠BOC与∠A的关系是( )
查看习题详情和答案>>
| A.∠BOC=2∠A | B.∠BOC=180°-
| ||
C.∠BOC=90°+
| D.∠BOC=90°+∠A |