解答：证明：（Ⅰ）连接AC，则AC∥EF，∵AA1∥CC1且AA1=CC1，∴四边形AA1C1C是平行四边形，∴A1C1∥AC∴EF∥A1C1，∴EF∥面A1C1B．（Ⅱ）连接B1D1，则B1D1⊥A1C1．∵DD1⊥平面A1B1C1D1，∴DD1⊥A1C1，∴A1C1⊥平面DD1B1，∴A1C1⊥B1D同理可证，A1B⊥B1D，∴B1D⊥平面A1C1B．
