(1)证明：如图所示：∵正方体ABCD~A1B1C1D1中，B1E=BF∴BD=AB1==>BF/BD=EB1/AB1过F作FG//AD交AB于G ，连接EG∴BF/BD=BG/AB==> BG/AB=EB1/AB1∴EG//BB1==>BB1//面EFG∵BB1⊥底面ABCD∴面EFG⊥底面ABCD又面BB1C1C⊥底面ABCD∴面EFG//面BB1C1C∴EF//面BB1C1C
连接B1C
