如图,△ABC的三条角平分线交于点O,过点O作OE⊥BC于点E,求证:∠BOD=∠COE.
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如图,△ABC的三条角平分线交于点O,过点O作OE⊥BC于点E,求证:∠BOD=∠COE.
![](http://img.wesiedu.com/upload/b/f3/bf35492bcc62374a20e41f900b623723.jpg)
![](http://img.wesiedu.com/upload/b/f3/bf35492bcc62374a20e41f900b623723.jpg)
![如图,△ABC的三条角平分线交于点O,过点O作OE⊥BC于点E,求证:∠BOD=∠COE.](/uploads/image/z/3535664-32-4.jpg?t=%E5%A6%82%E5%9B%BE%EF%BC%8C%E2%96%B3ABC%E7%9A%84%E4%B8%89%E6%9D%A1%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9O%EF%BC%8C%E8%BF%87%E7%82%B9O%E4%BD%9COE%E2%8A%A5BC%E4%BA%8E%E7%82%B9E%EF%BC%8C%E6%B1%82%E8%AF%81%EF%BC%9A%E2%88%A0BOD%3D%E2%88%A0COE%EF%BC%8E)
![](http://hiphotos.baidu.com/zhidao/pic/item/0eb30f2442a7d933c21c0dc1ae4bd11372f001cf.jpg)
1
2∠ABC+∠ACB,
∴∠AOF=180°-(∠DAC+∠AF0)
=180°-[
1
2∠BAC+
1
2∠ABC+∠ACB]
=180°-[
1
2(∠BAC+∠ABC)+∠ACB]
=180°-[
1
2(180°-∠ACB)+∠ACB]
=180°-[90°+
1
2∠ACB]
=90°-
1
2∠ACB,
∴∠BOD=∠AOF=90°-
1
2∠ACB,
又∵在直角△OCE中,∠COE=90°-∠OCD=90°-
1
2∠ACB,
∴∠BOD=∠COE.
如图,△ABC的三条角平分线交于点O,过点O作OE⊥BC于点E,求证:∠BOD=∠COE.
如图所示·,三角形ABC的三条角平分线相交于O点,过O点做OE垂直于BC于E,求证:∠BOD=∠COE
已知,如图,三角形ABC的三个内角平分线交于o点,过o作oe垂直bc于点e,求证三角形bod全等于三角形coe
△ABC中AD平分∠BAC交BC于点D,∠ABC、∠ACB的平分线交AD于点O,过点O作OE⊥BC于点E,试探究∠BOD
AD平分∠BAC交BC于D,∠ABC、∠ACB的平分线交AD于O,过O点作OE⊥BC于E.求证:∠BOD=∠EOC.
如下图,△ABC中,三个角的平分线交于O点,OE⊥BC于E,试猜想∠BOD和∠COE的关系,并说明理由.
如图,已知△ABC中,AD平分∠BAC交BC于D,∠ABC∠ACB的平分线交AD于O,过O作OE⊥BC于点E.证明:∠B
如图,已知△ABC中,AD平分∠BAC交BC于D,∠ABC∠ACB的平分线交AD于O,过O作OE⊥BC于点E
已知△ABC中,AD平分∠BAC交BC于D,∠ABC、∠ACB的平分线交AD于O,过O点作OE丄BC于E,试判断∠BOD
如图,在△ABC中,∠ABC和∠ACB的平分线相交于点O,过点O作EF∥BC交AB于E,交AC于F,过点O作OD⊥AC于
如图,△ABC的三条内角平分线相交于点O,过点O作OE⊥BC于E点,
如图6,在△ABC中,∠ABC和∠ACB的平分线相交于点O,过O点作EF//BC,交AB于E,交AC于F,从点O作OD⊥