DG in G after D is parallel to AB and AC extension lines.

According to the parallel line theorem, AB/BC=DG/CD.

And AB/BC=DE/DC ∴DE=DG.

∴∠DEG=∠G Because AB is parallel to DG, ∴∠ G =∠ A.

And ∠AEF=∠DEG, so ∠A=∠AEF ∴AF=EF.

Solution 2:

Theorem: a/sinA=b/sinB=c/sinC=2R(R is the radius of △ABC circumscribed circle).

get

AB/sin∠ABC=BC/sin∠A

DE/sin∠ECD = CD/sin∠CED = CD/sin∠AEF

So ab/CB = sin ∠ ABC/sin ∠ A.

DE/CD=sin∠ECD/sin∠AEF

Because ∠ABC and ∠ECD are complementary, the sine values of complementary angles are equal.

So sin∠ABC=sin∠ECD.

Because AB/BC=DE/DC.

So sin∠ABC/sin∠A=sin∠ECD/sin∠AEF.

sin∠ABC=sin∠ECD。

So sin∠A=sin∠AEF.

Because ∠ A+∠ AEF < 180 degrees

So ∠A=∠AEF ∴AF=EF.

I am exhausted from typing. ...

I hope it can be adopted as the best solution. ....