盐城style

(Ⅰ)证明:∵平面ABCD⊥平面ABEF,CB⊥AB,平面ABCD∩平面ABEF=AB

∴CB⊥平面ABEF,∵AF?平面ABEF,∴AF⊥CB

又AF⊥BF,且BF∩BC=B,BF、BC?平面CBF

∴AF⊥平面CBF

(Ⅱ)解:设DF的中点为N,则MN

.
1
2
CD,又AO
.
1
2
CD,则MN
.
AO,

∴MNAO为平行四边形

∴OM∥AN

又AN?平面DAF,OM?平面DAF

∴OM∥平面DAF

(III)∵AF=1,AF⊥BF,AB=2

∴∠FAB=60°

过点E作EH⊥AB于H,则∠EBH=60°,

∴EH=

3
2
,EF=AB-2HB=1,

故S△BEF=

1
2
×1×
3
2
=
3
4

∵CB⊥平面ABEF

∴三棱锥C-BEF的高为CB=1

∴VC-BEF=

1
3
×S△BEF×BC=
1
3
×
3
4
×1=
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