(1)∵CE= CC1= ,
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106044522-74258.gif) ∴VC—BDE=VE—BCD= S△BCD·CE = × ×1×1× = . (2)证明 连接AC、B1C. ∵AB=BC,∴BD⊥AC. ∵A1A⊥底面ABCD, ∴BD⊥A1A. ∵A1A∩AC=A, ∴BD⊥平面A1AC. ∴BD⊥A1C. ∵tan∠BB1C= = , tan∠CBE= = ,∴∠BB1C=∠CBE. ∵∠BB1C+∠BCB1=90°, ∴∠CBE+∠BCB1=90°,∴BE⊥B1C. ∵BE⊥A1B1,A1B1∩B1C=B1, ∴BE⊥平面A1B1C,∴BE⊥A1C. ∵BD∩BE=B,BE 平面BDE,BD 平面BDE, ∴A1C⊥平面BDE. |