(1) Take the midpoint g of BC and connect FG and AG, then FG ∨= BD/2 ∨= AE.

∴ quadrilateral AEFG is a parallelogram, ∴EF∥AG.

Surface ABC,∴EF∥AG. contains AG.

Geometric method

(2) Pythagorean theorems are CE = 3 √ 5, CD = 6 √ 2 and DE = 3 √ 5.

∴EF⊥CD

ef = ag = acsin 60 =3√3,∴s△cde= 1/2*ef*cd=9√6

∵BD∥AE,∴BD∥ ACE,∴D distance to ace =B distance to ace.

Let BH⊥AC become the ace pilot of H, AE ⊥ ABC, AE ⊥ BH, BH ∴.

∴ BH = bcsin60 = 3 √ 3 is the distance from B to ACE.

S△ACE= 1/2*AC*AE=9, and the distance from A to surface CDE obtained by volume method is d=S△ACE*BH/S△CDE=3/√2.

Vector method

(2) Take the midpoint O of AB, connect OC, and pass O for l⊥ plane ABC.

With OA, OC, L as the axis, then A (3 3,0,0), C (0 0,3 √ 3,0), E (3 3,0,3), D (-3,0,6).

CE→=(3，-3√3，3)，CD→=(-3，-3√3，6)

Let the normal vector of CDE be n→=(x, y, 2), then

3x-3√3y+6=0

-3x-3√3y+ 12=0

X = 1，y = √ 3，∴ n→ = ( 1，√ 3，2)。

ac→=(-3,3√3,0),∴d=|ac→n→|/| n→| = |-3+3√3 *√3+0 |/√( 1+3+4)= 3/√2