E and f are the midpoint of AA 1 and DD 1, respectively.

∴b 1c 1∥a 1d 1,ef∥a 1d 1，

∴B 1C 1∥EF，

∵B 1C 1 is not included in the plane EFC, EF? Aircraft EFC,

∴B 1C 1∥ Aircraft Company ..

(2) Proof: Let AB= 1, then AA 1=2AB=2, D 1F=DF= 1,

∴c 1f=cf=2,∴c 1f2+cf2=cc 12，

∴C 1F⊥CF, ∵ a1d1c1ef ∑ a1d/kloc.

∴EF⊥ plane CDD 1C 1,

∵C 1F？ ∴ c1f ⊥ efcdd1c1plane

∴C 1F⊥ Aircraft Company ..

(Ⅲ) Solution: There is a midpoint p on the side BB 1 so that the plane ADP⊥ the plane EFC. ..

The proof is as follows: ∵ In the cuboid ABCD-a1b1c1d1,f is the midpoint of DD 1 and p is the midpoint of BB 1.

∴C 1F∥AP, ∵C 1F⊥ plane EFC, ∴AP⊥ plane EFC,

∵AP？ Aircraft ADP, aircraft ADP, aircraft EFC,

At this time, bpbb 1 = 12.