(1)证明:
A1在底面ABC上的射影在AC上⇒A1D⊥平面ABC⇒A1D⊥BC,∵AC⊥BC, ∴BC⊥平面A1C1CA…(3分)AC1⊂平面A1C1CA,∴BC⊥AC1,BA1⊥AC1,A1B∩BC=B,∴AC1⊥平面A1BC…(7分) (2)由(1)可知:A1C⊥AC1⇒ACC1A1是棱形;…(9分) ∵AC=2,点D为中点,AD⊥BC,∴△A1AC为正三角形,∴AD=…(11分) ∴V多面体B1C1ABC=VA1B1C1-ABC-VA-A1B1C1=VA1B1C1-ABC=××4×=…(13分) |