已知:如图,∠1+∠2=180°,∠3=100°,OK平分∠DOH.
(1)证明AB∥CD;
(2)∠KOH的度数是多少?
(1)证明AB∥CD;
(2)∠KOH的度数是多少?
![](https://img.xkw.com/dksih/QBM/editorImg/2023/11/2/e1deeaa3-0cd4-4c3a-944b-21abcbfff9b6.jpg?resizew=141)
更新时间:2020-05-20 23:48:43
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解答题-问答题
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【推荐1】直线
相交于点
平分
.
,则直接写出
的大小__________;
(2)如图1,若
平分
,试说明E,O,F三点在同一直线上;
(3)如图2,若
平分
,则求出
的大小.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b5c4cd264c97c1f261229925cc5a6761.png)
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(2)如图1,若
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6cad4595d5352b2884568a59d8d766a4.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d22df2977de56cc69be0c1e847653d7a.png)
(3)如图2,若
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/307bd991211ec79b47a4be52933bb8e7.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b23ed5ddd713fac0501ddb0c2ba007f9.png)
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解答题-问答题
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适中
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【推荐2】如图,∠AOC:∠COD:∠BOD=2:3:4,且A,O,B三点在一条直线上,OE,OF分别平分∠AOC和∠BOD,OG平分∠EOF,求∠GOF的度数.将下列解题过程补充完整.
解:因为,∠AOC:∠COD:∠BOD=2:3:4,
所以∠AOC= ,∠COD= ,∠BOD= ,
因为OE,OF分别平分∠AOC和∠BOD,
所以∠AOE= ,∠BOF= ,
所以∠EOF= ,
又因为 ,所以∠GOF=60°.
解:因为,∠AOC:∠COD:∠BOD=2:3:4,
所以∠AOC= ,∠COD= ,∠BOD= ,
因为OE,OF分别平分∠AOC和∠BOD,
所以∠AOE= ,∠BOF= ,
所以∠EOF= ,
又因为 ,所以∠GOF=60°.
![](https://img.xkw.com/dksih/QBM/2018/12/4/2089638163030016/2090847322185728/STEM/a3563f2776ca4c47a3529fe533d8e0e5.png?resizew=217)
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解答题-证明题
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适中
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【推荐1】用反证法证明:已知直线a、b被直线c所截,∠1+∠2≠180°.求证:a与b不平行.
![](https://img.xkw.com/dksih/QBM/2015/12/24/1573945266880512/1573945272451072/STEM/5702f5325e144944b1af878ed041e9da.png)
证明:假设_________________________,则:∠1+∠2=180°(___________________________)
这与____________________矛盾,故假设不成立.所以a与b不平行.
![](https://img.xkw.com/dksih/QBM/2015/12/24/1573945266880512/1573945272451072/STEM/5702f5325e144944b1af878ed041e9da.png)
证明:假设_________________________,则:∠1+∠2=180°(___________________________)
这与____________________矛盾,故假设不成立.所以a与b不平行.
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解答题-问答题
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【推荐2】已知直线
,
被直线
所截,交点分别为点
,
,且
.
(1)如图1,请说明
与
的位置关系;
(2)如图2,点
,
分别在射线
,
上,点
,
分别在射线
,
上,连接
,
,并分别延长交于点
,且
,求
的度数;
(3)如图3,在(2)的条件下,连接
.若
平分
,且
平分
,
,求
的度数.
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(1)如图1,请说明
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f52a58fbaf4fea03567e88a9f0f6e37e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9d78abbad68bbbf12af10cd40ef4c353.png)
(2)如图2,点
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dad2a36927223bd70f426ba06aea4b45.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/acc290b44635265137fdf13146b6a6d9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8b46c607b3deac746c0ef3389ad8f65c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1460aa3d83df61f6c411b34412135451.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/73465a1f9aa03481295bf6bd3c6903ac.png)
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(3)如图3,在(2)的条件下,连接
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解答题-证明题
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适中
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【推荐3】已知:如图,DE⊥AC,垂足为点E,∠AGF=∠ABC,∠BFG+∠BDE=180°,
求证:BF⊥AC.
![](https://img.xkw.com/dksih/QBM/editorImg/2023/2/5/1b9adb4c-646a-4beb-8285-7db12392aecd.png?resizew=160)
请完成下面的证明的过程,并在括号内注明理由.
证明:∵∠AGF=∠ABC(已知)
∴FG∥ ( )
∴∠BFG=∠FBC( )
∵∠BFG+∠BDE=180°(已知)
∴∠FBC+∠BDE=180°( )
∴BF∥DE( )
∴∠BFA= (两直线平行,同位角相等)
∵DE⊥AC(已知)
∴∠DEA=90°( )
∴∠BFA=90°(等量代换)
∴BF⊥AC(垂直的定义)
求证:BF⊥AC.
![](https://img.xkw.com/dksih/QBM/editorImg/2023/2/5/1b9adb4c-646a-4beb-8285-7db12392aecd.png?resizew=160)
请完成下面的证明的过程,并在括号内注明理由.
证明:∵∠AGF=∠ABC(已知)
∴FG∥ ( )
∴∠BFG=∠FBC( )
∵∠BFG+∠BDE=180°(已知)
∴∠FBC+∠BDE=180°( )
∴BF∥DE( )
∴∠BFA= (两直线平行,同位角相等)
∵DE⊥AC(已知)
∴∠DEA=90°( )
∴∠BFA=90°(等量代换)
∴BF⊥AC(垂直的定义)
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【推荐1】如图,有一张长方形纸条
,
,在线段
,
上分别取点G,H,将四边形
沿直线
折叠,点C,D的对应点为
,
,将四边形
沿直线
折叠,点A,B的对应点为
,
,设
.
、
在直线
的上方,当
且满足
时,求
的度数.
(2)在(1)的条件下,猜想直线
和
的位置关系,并证明
(3)在点G,H运动的过程中,若
,请直接用含有
的式子表示
的度数
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/411b38a18046fea8e9fab1f9f9b80a5f.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f84af857b7c2db51841c8359ec0f0a65.png)
(2)在(1)的条件下,猜想直线
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(3)在点G,H运动的过程中,若
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解答题-问答题
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【推荐2】如图,四边形ABCD中,∠BAD=106°,∠BCD=64°,点M,N分别在AB,BC上,得
FMN,若MF∥AD,FN∥DC.
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求:(1)∠F的度数;
(2)∠D的度数.
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求:(1)∠F的度数;
(2)∠D的度数.
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