∴BD2+CD2=BC2,∴BD⊥CD，

∵CE⊥CD,∴CE∥BD，

And CE is not included in the plane ABD, BD? Aircraft ABD,

∴CE∥ plane ABD

(2) Solution: ∵ dihedral angle A-BD-C is 90, and AD⊥BD,

∴AD⊥ plane BDC, and BD⊥CD is known from (1).

With d as the origin, DB, DC and DA are the x, y and z axes respectively.

Establish spatial rectangular coordinate system, and CE⊥CD,

∴CE⊥ ACD, CE again? The ace of the plane,

∴ plane ACE⊥ plane ACD, let the midpoint of AC be f,

If DF is connected, then DF⊥AC, and DF=2, DF⊥ACE,

According to (1), BD = CD = AD = 22,

B(2，22，0)，C(0，22，0)，

A(0，0，22)，F(0，2，2)，

Normal vector df of plane ACE = (0,2,2),

Similarly, take the normal vector n of ADE = (2, 1, 0),

cos=225= 10 10。

The dihedral angle C-AE-D is arccos 10 10.