如图,在四边形ABCD中,连接BD,点E,F分别在AB和CD上,连接CE,AF,CE与AF分别交B于点N,M.已知∠AMD=∠BNC.
![](https://img.xkw.com/dksih/QBM/2018/6/13/1966467593756672/1967356499492864/STEM/e3488dbc21bb4659be9f9e09992c4aad.png?resizew=121)
(1)若∠ECD=60°,求∠AFC的度数;
(2)若∠ECD=∠BAF,试判断∠ABD与∠BDC之间的数量关系,并说明理由.
![](https://img.xkw.com/dksih/QBM/2018/6/13/1966467593756672/1967356499492864/STEM/e3488dbc21bb4659be9f9e09992c4aad.png?resizew=121)
(1)若∠ECD=60°,求∠AFC的度数;
(2)若∠ECD=∠BAF,试判断∠ABD与∠BDC之间的数量关系,并说明理由.
更新时间:2018-06-14 23:39:57
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【知识点】 根据平行线判定与性质求角度解读
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【推荐1】如图,点E、F、G分别在直线
、
、
上,
交
于点H.已知
,
.
与
平行吗?说明理由;
(2)若
,求
的度数.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9d78abbad68bbbf12af10cd40ef4c353.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f52a58fbaf4fea03567e88a9f0f6e37e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/03902478df1a55bc99703210bccab910.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/85c4bdfb0db1e31e8459df1d15f9ab55.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/85c4bdfb0db1e31e8459df1d15f9ab55.png)
(2)若
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7b11dbd664bd252b1e3c15b421c806a0.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d0944ff7ee6bd93f215be6d566b42481.png)
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【推荐2】如图,EF∥AD,∠1=∠2,∠BAC=70°.将求∠AGD的过程填写完整.
∴∠2=______.(两直线平行,同位角相等;)
又∵∠1=∠2,( )
∴∠1=∠3.( )
∴AB∥DG.( )
∴∠BAC+______=180°( )
又∵∠BAC=70°,( )
∴∠AGD=______.
∴∠2=______.(两直线平行,同位角相等;)
又∵∠1=∠2,( )
∴∠1=∠3.( )
∴AB∥DG.( )
∴∠BAC+______=180°( )
又∵∠BAC=70°,( )
∴∠AGD=______.
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