【答案】
分析:(Ⅰ)利用线面垂直,证明面面垂直,先证明A
1A⊥面ABC,再证明面A
1AC⊥面ABC;
(Ⅱ)取BC的中点E,证明四边形CEB
1C
1为平行四边形,可得B
1E∥C
1C,从而可得B
1E∥面A
1C
1C,再证明AE∥面A
1C
1C,利用面面平行的判定,可得面B
1AE∥面A
1C
1C,从而可得AB
1∥面A
1C
1C.
解答:![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025122941621447943/SYS201310251229416214479018_DA/images0.png)
证明:(Ⅰ)∵四边形ABB
1A
1为正方形,∴A
1A=AB=AC=1,A
1A⊥AB
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025122941621447943/SYS201310251229416214479018_DA/0.png)
…(2分)
∵A
1C=A
1B,∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025122941621447943/SYS201310251229416214479018_DA/1.png)
,∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025122941621447943/SYS201310251229416214479018_DA/2.png)
∴A
1A⊥AC…(4分)
∵AB∩AC=A,∴A
1A⊥面ABC
又∵A
1A?面A
1AC,∴面A
1AC⊥面ABC…(6分)
(Ⅱ)取BC的中点E,连接AE,C
1E,B
1E
∵B
1C
1∥BC,B
1C
1=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025122941621447943/SYS201310251229416214479018_DA/3.png)
,∴B
1C
1∥EC,B
1C
1=EC
∴四边形CEB
1C
1为平行四边形,∴B
1E∥C
1C
∵C
1C?面A
1C
1C,B
1E?面A
1C
1C,∴B
1E∥面A
1C
1C…(8分)
∵B
1C
1∥BC,B
1C
1=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025122941621447943/SYS201310251229416214479018_DA/4.png)
,∴B
1C
1∥BE,B
1C
1=BE
∴四边形BB
1C
1E为平行四边形,∴B
1B∥C
1E,且B
1B=C
1E
又∵ABB
1A
1是正方形,∴A
1A∥C
1E,且A
1A=C
1E
∴AEC
1A
1为平行四边形,∴AE∥A
1C
1,
∵A
1C
1?面A
1C
1C,AE?面A
1C
1C,∴AE∥面A
1C
1C…(10分)
∵AE∩B
1E=E,∴面B
1AE∥面A
1C
1C
∵AB
1?面B
1AE,∴AB
1∥面A
1C
1C…(12分)
点评:本题考查面面垂直,考查线面平行,解题的关键是掌握面面垂直的判定方法,正确运用面面平行判断线面平行,属于中档题.