Take AE midpoint o and cONnect OM and on.

Yi Zheng OM∨DE, ON∨CE

∴ surface omn surface CDE

∴MN∥ surface CDE

(2) It is easy to prove AE⊥BE.

∵AD⊥BE,∴BE⊥ face ADE

Connecting OD, OB, OC

∫od is contained in the face ADE,∴BE⊥OD.

∵AD=DE=2,∴OD⊥AE

∴OD⊥ surface

It is easy to prove OD=√2, OB=OC=√ 10 of cosine theorem and BD=CD=2√3 of Pythagorean theorem.

The cosine theorem is ∠ ced = 120, ∴ s △ CDE =1/2 * ce * de * sin ∠ ced = ∠ 3.

S△BCE= 1/2*BC*CE=2。

Let the distance from B to CDE be H, and h*S△CDE=OD*S△BCE can be obtained by volume method.

The solution is h=2√2/√3.

Let the included angle between BD and CDE be β, then sinβ=h/BD=√2/3.