(1)证明:在题图a中,EF∥AB,AB⊥AD, ∴EF⊥AD,在题图b中,CE⊥EF,又平面CDFE⊥平面ABEF,且平面CDFE∩平面ABEF=EF, CE⊥平面ABEF,AB⊂平面ABEF,∴CE⊥AB,又∵AB⊥BE,BE∩CE=E,∴AB⊥平面BCE; (2)解:∵平面CDFE⊥平面ABEF,且平面CDFE∩平面ABEF=EF,AF⊥FE,AF⊂平面ABEF,∴AF⊥平面CDEF,∴AF为三棱锥A CDE的高,且AF=1,又∵AB=CE=2,∴S△CDE=×2×2=2, ∴VC ADE=·S△CDE·AF=×2×1=. |