Problem: Let $ABC$ be a triangle with an obtuse angle at $A$.
Let $E$ and $F$ be the intersections of the external bisector of angle $A$
with the altitudes of $ABC$ through $B$ and $C$ respectively.
Let $M$ and $N$ be the points on the segments $EC$ and $FB$ respectively
such that $\angle EMA = \angle BCA$ and $\angle ANF = \angle ABC$.
Prove that the points $E$, $F$, $N$, $M$ lie on a circle.

Solution: Call the altitudes $\ol{BZ}$ and $\ol{CY}$,
and $H$ the orthocenter.
Let $W$ be the midpoint of $\ol{BC}$.
Then according to
\textbf{\href{https://aops.com/community/c6h89098p519896}{IMO Shortlist 2005 G5}},
the line $AW$ is concurrent
with $(HYZ)$, $(HEF)$, $(HBC)$ at a point $Q$.

\begin{center}
\begin{asy}
import graph; size(9.98895cm);
pen ffxfqq = rgb(1.,0.49803,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.);
pair H = (0.2,3.2), B = (-1.,0.2), C = (2.2,0.2), A = (0.2,1.), Y = (-0.55862,1.30344), Z = (1.21538,1.67692), F = (1.57146,1.14279), Q = (-0.68,2.76), M = (0.87710,0.51931);
draw(H--B--C--cycle, linewidth(0.6) + zzttqq);
draw(circle((0.34465,1.81068), 1.39682), linewidth(0.6) + ffxfqq);
draw(circle((0.2,2.1), 1.1), linewidth(0.6) + yqqqqq);
draw(circle((0.6,1.3), 1.94164), linewidth(0.6) + yqqqqq);
draw(H--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--H, linewidth(0.6) + zzttqq);
draw(B--Z, linewidth(0.6));
draw(C--Y, linewidth(0.6));
draw(B--F, linewidth(0.6) + blue);
draw(C--(-0.71824,0.90439), linewidth(0.6) + blue);
draw((0.6,0.2)--Q, linewidth(0.6) + green);
draw(circle((-1.,1.5), 1.3), linewidth(0.6) + qqwuqq);
draw((-0.71824,0.90439)--F, linewidth(0.6) + yqqqqq);
dot("$H$", H, dir((-5.899, 6.902)));
dot("$B$", B, dir((-8.675, -9.665)));
dot("$C$", C, dir((5.858, -4.103)));
dot("$A$", A, dir((1.801, 3.594)));
dot("$Y$", Y, dir((-13.028, -3.649)));
dot("$Z$", Z, dir((6.784, -1.212)));
dot("$E$", (-0.71824,0.90439), dir((-10.327, -2.673)));
dot("$F$", F, dir((7.966, -3.412)));
dot("$Q$", Q, dir((-8.162, 8.978)));
dot("$N$", (-0.15969,0.50808), dir((1.837, -9.672)));
dot("$W$", (0.6,0.2), dir((-1.409, -11.804)));
dot("$M$", M, dir((-1.312, -9.939)));
\end{asy}
\end{center}

Since $WA \cdot WQ = WB^2$,
it follows that $(AQB)$ is tangent to $\ol{BC}$,
ergo $N \in (AQB)$.
Then
\[ \dang QNF = \dang QNB = \dang QAB = \dang QAZ = \dang QHZ = \dang QHF \]
and hence $N$ lies on $(HQEF)$.
Similarly, so does $M$.