证明:(Ⅰ)连接AD1,在长方体ABCD-A1B1C1D1中, ABD1C1,则四边形ABC1D1是平行四边形, ∴AD1∥BC1, 又∵E,F分别是AD,DD1的中点 ∴AD1∥EF, ∴EF∥BC1,又EF⊄面A1BC1,BC1⊂面A1BC1, ∴EF∥平面A1BC1(3分) (II)在平面CC1D1D中作D1Q⊥C1D交CC1于Q, 过Q作QP∥CB交BC1于点P,则A1P⊥C1D.(7分) 因为A1D1⊥平面CC1D1D,C1D⊂平面CC1D1D, ∴C1D⊥A1D1,而QP∥CB,CB∥A1D1,∴QP∥A1D1, 又∵A1D1∩D1Q=D1,∴C1D⊥平面A1PQC1, 且A1P⊂平面A1PQC1,∴A1P⊥C1D.(10分) ∵△D1C1Q∽Rt△C1CD, ∴=,∴C1Q= 又∵PQ∥BC, ∴PQ=BC=1. ∵四边形A1PQD1为直角梯形,且高D1Q=, ∴A1P==.(14分)
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