在正方体ABCD-A1B1C1D1中,对角线A1C与平面BDC1交于点O,AC,BD交于点M

在正方体ABCDA1B1C1D1中,对角线A1C与平面BDC1交于点OACBD交于点M,求证:点C1OM共线.

答案

证明 如图所示,A1AC1C

A1AC1C确定平面A1C.

A1C平面A1COA1C

O平面A1C,而O=平面BDC1线A1CO平面BDC1

O在平面BDC1与平面A1C的交线上.

ACBDM

M平面BDC1,且M平面A1C

平面BDC1平面A1CC1M

OC1M,即C1OM三点共线.

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