∫△a 1bc 1, where A 1D=DC 1 and A 1O=OB.

∴OD∥BC 1

∵OD？ Aircraft AB 1D, BC 1? Aircraft AB 1D,

∴BC 1∥ Aircraft AB1d; ;

(2) In the ABC-A1b1,AA 1⊥ plane A 1C 1,

∵B 1D？ ∴ b1d ⊥ aa1a1b1plane,

∵B 1D is the center line of a regular triangle a1b1c, and we can get B 1D⊥A 1C 1.

Combining aa1∩ a1= a1,we get the plane of b1d ⊥ aa1c1c.

∵A 1C？ ∴ b1d ⊥ a1CAA1c aircraft,

∵ab=2aa 1,∴a 1daa 1=aa 1ac=22

∠∠da 1A =∠a 1AC = Rt∠

∴△DA 1A∽△A 1AC, you can get ∠ ada1= ∠ ca1a = 90-∠ da1c.

Therefore ∠ ada 1+∠ da 1c = 90, so A 1C⊥AD.

∵B 1D and AD are straight lines intersecting on the AB 1D plane.

∴A 1C⊥ Aircraft AB 1D ..