题目内容
如图所示,在正方体
中,点
分别是棱
的中点,
求证:点
共面.
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求证:点

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证明见答案
∵
三点不共线,∴
三点确定一平面
.
又由题意可知
与
共面且不平行,故分别延长
,
相交于
,则
直线
平面
.
∴
,同理设直线
则
平面
,
又∵点
均属于平面
,又
中点,且
,
∴
,∴∠
.∴同理∠
.
又
,∴点
共线于
平面
,从而点
.
∴点
四点共面.

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又由题意可知
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
∴
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
又∵点
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
∴
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又
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
∴点
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