请将下面的推理过程补充完整,并在括号内填入恰当的理由.
(1)已知,如图,AB⊥BC于点B,CD⊥BC于点C,∠1=∠2,BE与CF平行吗?
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/26/21924213-01a8-497f-a247-e27cb6dd6936.png?resizew=240)
解:BE
CF
证明:∵AB⊥BC,CD⊥BC(已知)
∴∠ABC=∠BCD=90° ( )
∴∠1+∠3=90°,∠2+∠4=90°
又∵∠1=∠2 ( )
∴ = ( )
∴BE
CF ( )
(2)已知:如图,直线AB
CD,直线EF与直线AB,CD分别交于点G,H,GM平分∠FGB,∠3=60°,求∠1的度数.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/26/5f9e5add-49cf-4d89-9918-b3b76bcb5bbf.png?resizew=197)
解:∵EF与CD交于点H,(已知)
∠4=∠3= °( )
∵AB
CD,(已知)
∴∠4+∠FGB=180°( )
∴∠FGB= °
∵GB平分∠FGB,(已知)
∴∠1=
∠BGF= °.( )
(1)已知,如图,AB⊥BC于点B,CD⊥BC于点C,∠1=∠2,BE与CF平行吗?
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/26/21924213-01a8-497f-a247-e27cb6dd6936.png?resizew=240)
解:BE
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2a9bfa68259d7a331be323b2038d628a.png)
证明:∵AB⊥BC,CD⊥BC(已知)
∴∠ABC=∠BCD=90° ( )
∴∠1+∠3=90°,∠2+∠4=90°
又∵∠1=∠2 ( )
∴ = ( )
∴BE
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2a9bfa68259d7a331be323b2038d628a.png)
(2)已知:如图,直线AB
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2a9bfa68259d7a331be323b2038d628a.png)
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/26/5f9e5add-49cf-4d89-9918-b3b76bcb5bbf.png?resizew=197)
解:∵EF与CD交于点H,(已知)
∠4=∠3= °( )
∵AB
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2a9bfa68259d7a331be323b2038d628a.png)
∴∠4+∠FGB=180°( )
∴∠FGB= °
∵GB平分∠FGB,(已知)
∴∠1=
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f89eef3148f2d4d09379767b4af69132.png)
更新时间:2022-04-10 09:44:47
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【知识点】 根据平行线判定与性质证明
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