Problem on Triangles

Question

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?

Answer 1

The points $A$, $F$, $E$, and $D$ are concyclic because $A$ and $F$ subtends equal angles with respect to the segment $ED$ that is $\angle{EAD}=\angle{EFD}$. Moreover $EF$ is parallel to $AB$ because $\triangle ABC$ and $\triangle FEC$ are similar. Hence $$40^{\circ}=\angle{EAB}=\angle{AEF}=\angle{ADF}$$ where in the last equality we used the fact that $A$, $F$, $E$, and $D$ are concyclic and therefore $E$ and $D$ subtends equal angles with respect to the segment $AF$.

Answer 2

$40^{\circ}$ because $AEDF$ is cyclic and $EF||AB$.