题目内容
(本题满分12分) 如图,正方形
所在平面与平面四边形
所在平面互相垂直,△
是等腰直角三角形![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851588739.gif)
(1)求证:
;
(2)设线段
的中点为
,在直线
上是否存在一点
,使得
?若存在,请指出点
的位置,并证明你的结论;若不存在,请说明理由;
(3)求二面角
正切值的大小。
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231508517911207.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851307301.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851479406.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851494270.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851588739.gif)
(1)求证:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851603548.gif)
(2)设线段
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851619242.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851635202.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851697234.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851713327.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851728585.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851713327.gif)
(3)求二面角
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851759330.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231508517911207.gif)
(1)略
(2)略
(3)二面角
正切值为![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851900266.gif)
(2)略
(3)二面角
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851759330.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851900266.gif)
解:(Ⅰ)因为平面ABEF⊥平面ABCD,BC
平面ABCD,BC⊥AB,平面ABEF∩平面ABCD=AB,
所以BC⊥平面ABEF.
所以BC⊥EF. ……………………………………2分
因为⊿ABE为等腰直角三角形,AB=AE,
所以∠AEB=45°,
又因为∠AEF=45,
所以∠FEB=90°,即EF⊥BE. …………………3分
因为BC
平面ABCD, BE
平面BCE,
BC∩BE=B
所以
…………………………4分(II)取BE的中点N,连结CN,MN,则MN![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852025444.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852071299.gif)
PC
∴PMNC为平行四边形,所以PM∥CN. ………6分
∵CN在平面BCE内,PM不在平面BCE内,PM∥平面BCE ………8分
(III)由EA⊥AB,平面ABEF⊥平面ABCD,易知EA⊥平面ABCD.
作FG⊥AB,交BA的延长线于G,则FG∥EA.从而FG⊥平面ABCD,
作GH⊥BD于H,连结FH,则由三垂线定理知BD⊥FH.
∴ ∠FHG为二面角F-BD-A的平面角. …………………10分
∵ FA=FE,∠AEF=45°,∠AEF=90°, ∠FAG=45°.
设AB=1,则AE=1,AF=
,则![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852118372.gif)
在Rt⊿BGH中, ∠GBH=45°,BG=AB+AG=1+
=
,
,
在Rt⊿FGH中,
,
∴ 二面角
正切值为
………………12分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851915135.gif)
所以BC⊥平面ABEF.
所以BC⊥EF. ……………………………………2分
因为⊿ABE为等腰直角三角形,AB=AE,
所以∠AEB=45°,
又因为∠AEF=45,
所以∠FEB=90°,即EF⊥BE. …………………3分
因为BC
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851915135.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851915135.gif)
BC∩BE=B
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851603548.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852025444.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852071299.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852087443.jpg)
∴PMNC为平行四边形,所以PM∥CN. ………6分
∵CN在平面BCE内,PM不在平面BCE内,PM∥平面BCE ………8分
(III)由EA⊥AB,平面ABEF⊥平面ABCD,易知EA⊥平面ABCD.
作FG⊥AB,交BA的延长线于G,则FG∥EA.从而FG⊥平面ABCD,
作GH⊥BD于H,连结FH,则由三垂线定理知BD⊥FH.
∴ ∠FHG为二面角F-BD-A的平面角. …………………10分
∵ FA=FE,∠AEF=45°,∠AEF=90°, ∠FAG=45°.
设AB=1,则AE=1,AF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852103156.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852118372.gif)
在Rt⊿BGH中, ∠GBH=45°,BG=AB+AG=1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852149109.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852212113.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852227573.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852243644.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852290332.jpg)
在Rt⊿FGH中,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150852321389.gif)
∴ 二面角
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851759330.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823150851900266.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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