Prove 4 points cyclic in an acute scalene triangle $ABC$ is an acute angled scalene triangle. $L,M,N$ are midpoints of the sides $BC,CA,AB$.
The perpendicular bisectors of $\overline{AB}$ and $\overline{CA}$ meet $\overline{AL}$ at point $D$ and point $E$. The rays $\overrightarrow{BD}$ and $\overrightarrow{CE}$ cut each other at point $F$ inside the triangle. 

Prove that $A,M,F,N$ are cyclic. 

Tried by taking a point on symmedian and also tried by applying Menelaus theorem. 

 A: 
Let $O=ME \cap ND$. By construction $O$ is the center of 
circumscribed circle of $\triangle ABC$
with circumradius $R=|AO|$.
Right triangles $ANO$ and $AOM$ share
common hypotenuse $AO$, 
which serves as the diameter 
of the circumscribed circle 
with the center $O_c$ in the middle of $AO$
for both of them,
and the radius of the circle is $\tfrac12|AO|=\tfrac12R$.
So, to prove that $A,M,F,N$ are cyclic,
suffice it to show that 
$|O_cF|=|O_cA|$
or $|O_cF|^2=|O_cA|^2$.
Considering the points as complex numbers, let
\begin{align}
P&=0,\quad
A=-4w,\quad
B=12w
,\\
C&=A+4(u+v\cdot i)=4(u-w+v\cdot i)
,\\
L&=\tfrac12\,(B+C)=2u+4w+2v\cdot i
,\\
M&=\tfrac12\,(A+C)=2u-4w+2v\cdot i
,\\
N&=\tfrac12\,(A+B)=4w
,\\
Q&=\tfrac12\,(A+M)=u-4w+v\cdot i
\end{align}
for some real numbers $u,v,w$.
The lines through points $z_1,z_2$
and $z_3,z_4$ intersect at a point
\begin{align}
z&=
f_{\times}(z_1,z_2,z_3,z_4)=
\frac{
(z_1-z_2)\,(\overline{z_3}\,z_4-\overline{z_4}\,z_3)
-(z_3-z_4)\,(\overline{z_1}\,z_2-\overline{z_2}\,z_1)
}{ 
(z_1-z_2)\,(\overline{z_3}-\overline{z_4})
-(z_3-z_4)\,(\overline{z_1}-\overline{z_2})
}
,
\end{align}
where $\overline{z}$ is the complex conjugate of $z$.
The center $O_c$ can be found as the point 
of intersection of 
perpendicular bisectors of $AM$ and $AN$.
The perpendicular bisector of $AN$
is defined by points $P$ and $P+i$,
the perpendicular bisector of $AM$
is defined by points $Q$ and $Q+(C-A)i$.
Given that,
\begin{align} 
Q_c&=f_{\times}(P,P+i,Q,Q+(C-A)\cdot i)
=\frac1v\,(v^2-4uw+u^2)\cdot i
.
\end{align}  
The other points can be found similarly,
\begin{align} 
D&=f_{\times}(N,N+i,A,L)
=4w+\frac{8vw}{u+4w}\cdot i
,\\
E&=
f_{\times}(M,M+O_c-Q,A,L)
\\
&=2\cdot\frac{u(u+4w)(u-2w)+v^2(u+2w)}{u^2+4uw+v^2}
+2\cdot\frac{v(u^2+v^2)}{u^2+4uw+v^2}\cdot i
,\\
F&=
f_{\times}(C,E,B,D)
\\
&=
4\cdot\frac{(8uw+3u^2+3v^2-16w^2)\,w}{(u+4w)^2+v^2}
+64\cdot\frac{v w^2}{(u+4w)^2+v^2}\cdot i
.
\end{align}
Now it can be verifies that
\begin{align} 
|O_cF|^2&=|O_cA|^2
=\frac1{v^2}(v^2+u^2)(v^2+(u-4w)^2)
.
\end{align}
The following maxima code can be used to verify the result:
_(z):=conjugate(z)$
xpoint(z1,z2,z3,z4):=
((z1-z2)*(_(z3)*z4-_(z4)*z3)-(z3-z4)*(_(z1)*z2-_(z2)*z1))
/((z1-z2)*(_(z3)-_(z4))-(z3-z4)*(_(z1)-_(z2)))$
declare([P,A,B,C,L,M,N,Q,Oc,D,E,F],complex)$
declare([u,v,w],real)$
P:0;
A:-4*w;
B:12*w;
C:A+4*(u+%i*v);
L:(B+C)/2;
M:(A+C)/2;
N:(A+B)/2;
Q:(A+M)/2;
Oc:factor(xpoint(P,P+%i,Q,Q+(C-A)*%i));
t:factor(xpoint(N,N+%i,A,L));
D:factor(realpart(t))+%i*imagpart(t);
t:factor(xpoint(M,M+Oc-Q,A,L));
E:realpart(t)+%i*imagpart(t);
t:factor(xpoint(C,E,B,D));
F:realpart(t)+%i*imagpart(t);
OcF2:factor(cabs(F-Oc)^2);
OcA2:factor(cabs(A-Oc)^2);
OcA2-OcF2;

