题目内容
如图,在四棱锥
中,底面
是正方 形,侧棱
底面
,
,点
是
的中点,作
交
于点

(1)求证:
∥平面
(2)求证:
平面

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(1)求证:

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(2)求证:


略
方法一:

(1)
两两垂直,以O为原点,射线
为非负
轴,建立空间直角坐标系
,
,
,
,
,
,
,
2分
可求得平面
的法向量为
又
平面

∥平面
4分
(2)

又


平面
6分
解:方法二:

(1) 连接BE,BD,AC,设AC交BD于G,
则G为AC的中点
在
中,E为PC的中点,
则PA∥EG,
面BED,
面BED
所以
∥平面
3分
(2)
PD⊥面ABCD
PD⊥BC
BC⊥CD

BC⊥面PCD
面PCD
BC⊥DE
PD=CD,E为PC中点,
DE⊥PC
DE⊥面PBC
DE⊥PB,又因为PB⊥EF

平面
6分

(1)

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可求得平面


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(2)
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解:方法二:

(1) 连接BE,BD,AC,设AC交BD于G,
则G为AC的中点
在

则PA∥EG,


所以
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(2)
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