In △ABD, m, n is the midpoint of edges AD, BD, ∴MN∥AB and MN = 1/2ab = 1 = ef.

In quadrilateral MEFN, EF∥AB,EF=AB∴ quadrilateral MEFN is a parallelogram.

∴ME∥FN

∫me∨fn, me∨ADF, fn ∈ (symbol error) ADF, ∴EM∥ADF.

(2) Prove AB=BF=2, AD=DF=BC=√ 13, just prove it yourself, and then three lines can be regarded as one line.

The angle formed by the midpoint g and B.D of AF is a dihedral angle.

It is easy to get AF=2 (it can be found by parallel lines with point E as AF, not to mention).

BG=√3, DG=2√3, COS ∠ BGD = 1/2 (cosine theorem)

As shown in the figure, the dihedral angle is π/3.

(3) translate AF to cross AB, and in A', F'

Let the length of BP be x, then (2) shows that △ABF is a regular triangle ∴ BF ′ = x, Ba ′ = √ 3x/3, and A ′ f ′ = 2 √ 3x/3.

∠CBE=90？ (from EB⊥ ABCD) ∴CP=√ 13-x? ，A′C =√9+(√3x/3+2)？

The angle formed is 30? Solve x with cosine theorem, there is a solution, and there is no solution.