TITLE: Problem on Triangles
QUESTION [5 upvotes]: I was solving this question. I solved it. Then I forgot how I solved it.

An acute angled triangle $ABC$ has $AD$ as altitude. $E$ is the midpoint of $BC$. $F$ is the midpoint of $AC$. $\angle{EAB}=40°$ and $\angle{EAD}=\angle{EFD}$. Find $\angle{ADF}$.

This is the diagram I made.

What I cannot understand is how I proved $\angle{AEF}=\angle{ADF}$, or if that is even true, and how to actually get the solution.
Can anyone help?

REPLY [3 votes]: $40^{\circ}$ because $AEDF$ is cyclic and $EF||AB$.