已知:如图,
,∠1=∠2.求证:
.
证明:∵
,(已知)
∴∠ABC=∠BCD.(_________________)
∵∠1=∠2,(已知)
∴∠ABC-∠ = -∠2,(_________________)
即∠EBC=_______.
∴
.(_________________)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b0fff774b4b0087a6f304ce930d359be.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cce38d8a8a7043586aad206f8153d0bc.png)
证明:∵
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b0fff774b4b0087a6f304ce930d359be.png)
∴∠ABC=∠BCD.(_________________)
∵∠1=∠2,(已知)
∴∠ABC-∠ = -∠2,(_________________)
即∠EBC=_______.
∴
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cce38d8a8a7043586aad206f8153d0bc.png)
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/8/b9c98f21-ef89-4bcb-a63d-439ed246547f.png?resizew=191)
更新时间:2022-09-07 15:55:33
|
【知识点】 根据平行线判定与性质证明
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解答题-证明题
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【推荐1】如图,
为
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与过点C的切线互相垂直,垂足为D.连接
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解答题-问答题
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较易
(0.85)
【推荐2】如图,已知点E在直线DC上,射线EF平分∠AED,过E点作EB⊥EF,G为射线EC上一点,连接BG,且∠EBG+∠BEG=90°.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/7/31/408eab73-fec9-44a2-ad59-acf1d9a013cd.png?resizew=166)
(1)若∠AED=72°,求∠EBG的度数;
(2)若∠EBG=∠A,试判断AB与EF的位置关系,并说明理由.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/7/31/408eab73-fec9-44a2-ad59-acf1d9a013cd.png?resizew=166)
(1)若∠AED=72°,求∠EBG的度数;
(2)若∠EBG=∠A,试判断AB与EF的位置关系,并说明理由.
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