In convex $\square ABCD$, $\angle BAD=\angle CDA$
The midpoints of $AB$,$CD$,$DA$ are $L$,$M$,$N$ respectively.
$\overline{AC}$ meet $\overline{BD}$ at point $E$.
Let $w$ be a circle that passes through $E$ and is tangent to $\overset{\longleftrightarrow}{AD}$ at point $A$.
Let $\overline{NE}$ cut $w$ at point $F$, which is not $E$.
Show that $\angle LFE=\angle MFE$
