(1)取AC1 的中点M,则MD是三角形ABC1的中位线,可得 MD∥BC1,从而得到 BC1∥面A1DC.
(2)由AA1⊥面ABC,可得 AA1⊥BC,又 AC⊥BC,BC⊥面A1AC,从而得到面A1BC⊥面A1AC.
【解析】
(1)证明:取AC1 的中点M,∵D是AB中点,∴MD是三角形ABC1的中位线,∴MD∥BC1.
而MD在平面A1DC中,BC1 不在平面A1DC中,∴BC1∥面A1DC.
(2)∵AA1⊥面ABC,∴AA1⊥BC,又 AC⊥BC,AC∩AA1=A,∴BC⊥面A1AC.
∵BC⊂面A1BC,∴面A1BC⊥面A1AC.