(1)证明:在△ABC中,由余弦定理得BC=4,∴△ABC为直角三角形,∴AC⊥BC.
又∵CC 1⊥面ABC,∴CC 1⊥AC,CC 1∩BC=C,∴AC⊥面BCC 1∴AC⊥BC 1.
(2)证明:设B 1C交BC 1于点E,则E为BC 1的中点,连接DE,则DE为△ABC 1的中位线,
则在△ABC 1中,DE ∥ AC 1,又DE⊂面CDB 1,则AC 1∥ 面B 1CD.
(3)在△ABC中过C作CF⊥AB垂足为F,
由面ABB 1A 1⊥面ABC知,CF⊥面ABB 1A 1,∴ V A 1 - B 1 CD = V C- A 1 D B 1 .
而 S △D A 1 B 1 =
1
2 A 1 B 1 •A A 1 =5×4×
1
2 =10 ,
CF=
AC•BC
AB =
3×4
5 =
12
5 ,
∴
V A 1 - B 1 CD =
1
3 ×10×
12
5 =8 .