Fold △BCD into △BC'D along the straight line BD.

It can be seen that CD=6, BC'=BC= 10, BD=8,

That is BC'2 = C'D2+BD2,

Therefore, C.D. ⊥ BD ... (2 points)

∫ plane BC'D⊥ plane ABD, plane BC'D∩ plane ABD=BD, C'D? Planes in BC,

∴C'D⊥ Aircraft Abd ... (5 points)

(ii) According to (i), the plane ABD with known C'D⊥ and the CD⊥BD,

As shown in the figure, the spatial rectangular coordinate system D-XYZ is established with D as the origin. ... (6 points)

Then d (0 0,0,0), a (8 8,6,0), b (8 8,0,0), c' (0,0,6).

E is the midpoint of the AD line,

∴E(4,3,0),BD=(？ 8,0,0).

On BEC's plane, be = (? 4，3，0)，BC′=(？ 8,0,6),

Let the normal vector of plane BEC be n = (x, y, z),

∴BE？ n = 0BC′？ N = 0, that is? 4x+3y=0？ 8x+6z=0，

Let x=3, y=4 and z=4, so n = (3 3,4,4) ... (8 points).

Let the included angle between straight line BD and plane BEC be θ, then sin θ = | cos < n, BD > | = | n? BD||n|？ |BD|=34 14 1。 ? ... (9 points)

∴ The sine value of the angle formed by the straight line BD and the plane BEC' is 34 14 1.