(Ⅰ)连结A1B与AB1交于E,连结DE,则E为A1B的中点,故DE为△A1BC1的中位线,∴BC1∥DE. 又DE⊂平面AB1D,BC1⊄平面AB1D,∴BC1∥平面AB1D.(6分) (Ⅱ)过点D作DH⊥A1B1,∵正三棱柱ABC-A1B1C1,∴AA1⊥平面A1B1C1,AA1⊥DH,AA1∩A1B1=A1, ∴DH⊥平面ABB1A1.DH为三棱锥D-ABB1的高.(8分) ∵S△ABB1=•AB•BB1=MH=A1B1=,(10分) 且 DH=A1Dtan=, ∵VB-AB1D=VD-ABB1=××=.(12分) |