摘要:10.在直平行六面体AC1中.四边形ABCD是菱形.∠DAB=60°.AC∩BD=O.AB=AA1. (1)求证:C1O∥平面AB1D1, (2)求证:平面AB1D1⊥平面 ACC1A1. 证明:(1)连接A1C1交B1D1于O1.连接AO1. 在平行四边形AA1C1C中.C1O1∥AO.C1O1=AO. ∴四边形AOC1O1为平行四边形. ∴C1O∥AO1. ∵C1O⊄平面AB1D1.AO1⊂平面AB1D1. ∴C1O∥平面AB1D1. (2)在直平行六面体AC1中.A1A⊥平面A1B1C1D1. ∴A1A⊥B1D1. ∵四边形A1B1C1D1为菱形.∴B1D1⊥A1C1. ∵A1C1∩AA1=A1.A1C1⊂平面ACC1A1.AA1⊂平面ACC1A1. ∴B1D1⊥平面ACC1A1. ∵B1D1⊂平面AB1D1.∴平面AB1D1⊥平面ACC1A1.

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