题目内容

已知在如图所示的正方体ABCD—A1B1C1D1中,AC∩BD=O,A1C1∩B1D1=O1,求证:OO1⊥平面AC.

证法一:连结O1A和O1C,在正方体ABCD—A1B1C1D1中,∵AA1⊥A1B1,AA1⊥A1D1,且A1B1∩A1D1=A1,

∴A1A⊥平面A1C1.∵A1C1平面A1C1,∴A1A⊥A1C1.同理,CC1⊥A1C1.

∵A1O1=O1C1,AA1=CC1,

∴△AA1O1≌△CC1O1,O1A=O1C.

∵O为AC中点,∴O1O⊥AC.

同理,O1O⊥BD.又AC∩BD=O,∴OO1⊥平面AC.

证法二:∵ABCD—A1B1C1D1为正方体,∴AA1⊥AB,AA1⊥AD.

∵AB∩AD=A,∴AA1⊥平面AC.

∵AA1BB1,BB1CC1,∴AA1CC1,四边形AA1C1C为平行四边形.

∵O、O1分别为AC、A1C1的中点,∴OO1∥AA1,OO1⊥平面AC.

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