摘要:11.如图所示.三棱柱ABC-A1B1C1.D是BC上一点.且A1B∥平面AC1D.D1是B1C1的中点.求证:平面A1BD1∥平面AC1D. [证明] 连结A1C交AC1于点E. ∵四边形A1ACC1是平行四边形. ∴E是A1C的中点.连结ED. ∵A1B∥平面AC1D. 平面A1BC∩平面AC1D=ED. ∴A1B∥ED. ∵E是A1C的中点.∴D是BC的中点. 又∵D1是B1C1的中点. ∴在三棱柱ABC-A1B1C1中.BD1∥C1D.A1D1∥AD. 又A1D1∩BD1=D1.AD∩C1D=D. ∴平面A1BD1∥平面AC1D.

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