证明:(Ⅰ)∵A 1B 1=A 1C 1,点D是棱B 1C 1的中点.
∴A 1D⊥B 1C 1.
由直三棱柱ABC-A 1B 1C 1,可得BB 1⊥B 1C.
∵BB 1∩B 1C 1=B 1.
∴A 1D⊥平面BB 1C 1C.
(Ⅱ)∵A 1B 1=A 1C 1=2,∠B 1A 1C 1=90°,
∴ B 1 C 1 =2
2 .
∵点D是棱B 1C 1的中点,∴ A 1 D=
2 .
∵A 1A ∥ 平面BB 1C 1C,∴点A与A 1到平面BB 1C 1C的距离相等,
∴ V B 1 -AD = V A- B 1 CD =
1
3 ×
1
2 ×
2 ×2×
2 =
2
3 .