摘要:19.证明:连结AB.A1D.在正方形中.A1B=A1D.O是BD中点.∴A1O⊥BD, 连结OM.A1M.A1C1.设AB=a.则AA1=a.MC=a=MC1.OA=OC=a.AC=a. ∴A1O2=A1A2+AO2=a2+a2=a2.OM2=OC2+MC2=a2.A1M2=A1C12+MC12=2a2+a2=a2, ∴A1M2=A1O2+OM2.∴A1O⊥OM.∴AO1⊥平面MBD.

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