And because AB, AC and AD are perpendicular to each other, that is, AB⊥AC and AB⊥AD, so DE⊥AC, DE⊥AD,

Because AC, AD∈ plane ACD, and AC∩AD=A, DE⊥ plane ACD,

Because de ∈ plane ECD, plane ACD⊥ plane ECD.

(2) Connect BF as shown in the figure.

Because AB, AC and AD are vertical, BC=CD=DB=√2,

So △ABC, △ACD and △ABD are congruent isosceles right triangles, and AB=AC=AD=DE= 1.

Therefore, the volume of triangular pyramid A-BCD =△ABC area× ad×1/3 =/kloc-0 /×1÷ 2×/kloc-0 /×1/3 =1/6,

Because there is AE//BD in the parallelogram ABDE, and BD∈ plane BCD, AE? Plane BCD,

So AE// plane BCD, point A and point E have the same distance from plane BCD,

That is, the triangular pyramid A-BCD and the triangular pyramid E-BCD are triangular pyramids with equal base and equal height, both of which are 1/6.

From the question (1), it is concluded that "DE⊥ plane ACD" and CD∈ plane ACD know DE⊥CD,

Therefore, in the right triangle CDE, according to CD=√2 and DE= 1, the area of the triangle CDE is √2/2.

Therefore, the volume of triangular pyramid E-BCD = 1/6=△CDE area× height×1/3 = √ 2/2× height×1/3.

Calculation height =√2/2, that is, the distance from point B to plane ECD is √2/2.