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.