Let the intersection point d be the midpoint of the straight line B 1C, and e be the midpoint of communication.

Then in the triangle AB 1C, ED is the center line and AB 1 is the opposite side of the center line.

So ab1/CE, and because ce is in the plane BEC 1,

So ab1/faces BEC 1

(2) The volume method is simple.

ab 1//face bec 1 from( 1)

So the distance from B to BEC 1 is the expected value (set to H).

BE=√3，EC 1=√5，BC 1=√8

Therefore, the triangle BEC 1 is a right triangle.

S(BEC 1)=√ 15/2

AEB =√3/2，

V(C 1-AEB)=V(A-BEC 1)

1/3 * S(AEB)* cc 1 = 1/3 * S(bec 1)* h

h=2√5/5

That is, the distance from straight line AB 1 to curved surface BEC 1 is 2√5/5.