Thinking: (limited to the first and second volumes of the eighth grade of Beijing Normal University) This problem cannot be directly proved by parallel lines, because there is no angle formed by MN and AE and AF and BC and AE and AF. Here, only a diagonal line can be proved by triangles, such as ∠AMN=∠AEF. So it can be proved by the similarity of the triangle where MN and EF are located. That is to say, em: am = fn: an comes first. Add ∠EAF=∠MAN, and we can prove △AMN∽△AEF.

Auxiliary line: Extension line BG and AD extension line intersect at point J, and extension line BH and AD extension line intersect at point K. 。

As shown in the figure:

Turn left | turn right

Simple steps:

1, prove △NBF∽△NKA, and get the value of NF: Na;

2. Prove △NBE∽△MJA and get the value of MF: MA.

The two are equal. So am: AE = an: af, the included angle ∠EAF=∠MAN, then two triangles are similar, then the corresponding angles ∠AMN=∠AEF are equal, and finally MF and EF are parallel.

Maybe there's a better way. Look downstairs.