连结AC、BD交于O,连结BM、DM、OM、OC'、A'C'、OC设正方体边长为2,由△ABM≌△ADM得BM=DM,又∵O是BD中点,∴MO⊥BD,由RT△AOM、RT△A'C'M、RT△OCC'分别得OM=√3,OC'=√6,MC'=3∴△OCM是RT△,MO⊥OC'∴OM⊥平面BDC',∴...