Equal Angle Related To Midpoints In Quadrilateral 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 $\angle LFE=\angle MFE$


 A: Consider the most extreme case: if $ABCD$ were to be square.

If that were the case, $L$ and $M$ would lie on the same line, the midpoint of the chord $AE$ is $J$, and with $AD\perp AB$, this happens to show that $M=w$. If $ABCD$ is a square, then $E, L$ and $M$ will be co-linear. Since $AM=AE$, it goes to show that $F=E$, and by pure consequence $\angle LEF$ and $\angle MEF$ are both $180^\circ$, thus equal.

Suppose $ABCD$ is a rectangle.
W.l.o.g., although this holds true too for all rectangles, assume that $AD=BC=s$ and $CD=AB=2s$. Inevitably, the center of circle $w$ will lie on $AB$ since $AB\perp AD$ 
If you take some time to calculate, you'll get the following lengths and angles:
$$\begin{align}
Aw&=\frac{3\sqrt{20}}{20}s\\
FA&=\frac{\sqrt{30}}{10}s\\
NF&=\frac{\sqrt5}{10}s\\
FE&=s-NF=\frac{10-\sqrt5}{10}s\\
\end{align}$$
For angles $\angle LFE$ and $\angle MFE$, since $L, E$ and $M$ are co-linear and thus create right triangles with $F$, you get
$$\tan LFE=\frac{EL}{EF}\Rightarrow LFE=\tan^{-1}\frac{EL}{EF}$$
$$\tan MFE=\frac{EM}{EF}\Rightarrow MFE=\tan^{-1}\frac{EM}{EF}$$
Since they both angles share $FE$ and $EL=EM=\frac12s$, we are able to prove that indeed:
$$\angle LFE=\angle MFE$$
