题目内容
(本小题满分12分)
如图,菱形ABCD与矩形BDEF所在平面互相垂直,
.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240035089624905.png)
(1)求证:FC∥平面AED;
(2)若
,当二面角
为直二面角时,求k的值.
如图,菱形ABCD与矩形BDEF所在平面互相垂直,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003508947676.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240035089624905.png)
(1)求证:FC∥平面AED;
(2)若
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003508962569.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003508978548.png)
(1)根据面面平行的性质定理来分析得到证明,关键是证明平面FBC∥平面EDA
(2)![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003508994361.png)
(2)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003508994361.png)
试题分析:(1)证明:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509009587.png)
平面FBC∥平面EDA
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509025476.png)
故
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509040296.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509056278.png)
(2)取EF,BD的中点M,N. 由于AE=AF=CE=CF
所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509072538.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509087365.png)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509103310.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509118336.png)
连接AC,当
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509134312.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509150346.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003509165632.png)
即
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003508994361.png)
点评:解决立体几何中的平行和垂直的证明,需要熟练的运用线面平行和垂直 判定定理和性质定理阿丽解答。而对于角的求解,通常就是利用定义作出角,然后结合三角形来得到结论,属于中档题。
![](http://thumb.zyjl.cn/images/loading.gif)
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