(Ⅰ)证明:∵∠ACB=90°,∴BC⊥AC. ∵三棱柱ABC-A1B1C1中,CC1⊥平面ABC, ∴BC⊥CC1, ∵AC∩CC1=C, ∴BC⊥平面ACC1A1. ∵A1D⊂平面ACC1A1,∴BC⊥A1D, 而BC∥B1C1,则B1C1⊥A1D. 在Rt△ACC1与Rt△DC1A1中,==,∴△ACC1~△DC1A1, ∴∠AC1C=∠DA1C1, ∴∠AC1C+∠C1DA1=90°.即A1D⊥AC1. ∵B1C1∩AC1=C1, ∴A1D⊥平面AB1C1; (Ⅱ)如图,设A1D∩AC1=H,过A1作AB1的垂线,垂足为G,连GH, ∵A1D⊥平面AB1C1,∴AB1⊥A1D,∴AB1⊥平面A1GH, ∴∠A1GH为二面角A1-AB1-C1的平面角. 在Rt△AA1B1中,AA1=,A1B1=2, ∴AB1=, ∴由等面积,可得A1G=; 在Rt△AA1C1中,AA1=,A1C1=, ∴AC1=3,∴由等面积,可得A1H=. ∴在Rt△A1GH中,sin∠A1GH=, ∴cos∠A1GH=, ∴二面角B-AB1-C1的余弦值为-. |