Problem: Let $\Omega$ and $\Gamma$ be circles with centres $M$ and $N$, respectively,
such that the radius of $\Omega$ is less than the radius of $\Gamma$.
Suppose $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$.
Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$,
so that $C$, $M$, $N$, $D$ lie on $MN$ in that order.
Let $P$ be the circumcenter of triangle $ACD$.
Line $AP$ meets $\Omega$ again at $E \neq A$ and meets $\Gamma$ again at $F \neq A$.
Let $H$ be the orthocenter of triangle $PMN$.

Prove that the line through $H$ parallel to $AP$ is tangent
to the circumcircle of triangle $BEF$.

Solution: Throughout the solution, we define
\begin{align*}
  \alpha &\coloneq \dang DCA = \dang BCD \implies \dang PAD = \dang CAB = 90\dg - \alpha \\
  \beta &\coloneq \dang ADC = \dang CDB \implies \dang CAP = \dang BAD = 90\dg - \beta.
\end{align*}
Ignore the points $H$, $M$, $N$ for now and focus on the remaining ones.
\begin{claim*}
  We have $\ol{CE} \parallel \ol{AD}$ and $\ol{DF} \parallel \ol{AC}$.
\end{claim*}
\begin{proof}
  $\dang AEC = \dang ABC = \dang CAB = 90\dg - \alpha$.
\end{proof}
Hence, if we let $A' \coloneq \ol{CE} \cap \ol{DF}$, we have a parallelogram $ACA'D$.
Note in particular that $\ol{BA'} \parallel \ol{CD}$.

\begin{center}
\begin{asy}
size(15cm);

pair A = dir(117);
pair C = dir(150);
pair D = dir(30);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
pair B = -A+2*foot(A, C, D);
pair M = circumcenter(A, B, C);
pair N = circumcenter(A, B, D);
pair P = origin;
draw(circumcircle(A, C, B), gray);
draw(circumcircle(A, D, B), gray);
draw(C--D, blue);
draw(A--B, blue);
pair H = orthocenter(P, M, N);
pair A_prime = C+D-A;
pair T = M+N-A;
pair E = extension(A, P, M, T);
pair F = extension(A, P, N, T);
draw(A--F, lightred);
draw(C--A_prime, lightred);
draw(F--D, lightred);
draw(M--H--N, lightred);
draw(circumcircle(A_prime, E, F), gray);

draw(A_prime--B, blue);
draw(circumcircle(B, E, F), purple);
draw(M--T, dashed);
draw(N--F, dashed);
draw(C--A--D, lightred);
markangle("$\alpha$", n=1, radius=18.0, D,C,A, deepgreen);
markangle("$\beta$", n=2, radius=30.0, A,D,C, deepgreen);
markangle("$\alpha$", n=1, radius=18.0, C,D,F, deepgreen);
markangle("$\beta$", n=2, radius=30.0, E,C,D, deepgreen);
markangle("$90^{\circ}-\alpha$", n=3, radius=14.0, C,A,B, deepcyan);
markangle("$90^{\circ}-\alpha$", n=3, radius=14.0, E,A,D, deepcyan);

dot("$A$", A, dir(A));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$B$", B, dir(225));
dot("$M$", M, dir(125));
dot("$N$", N, dir(N));
dot("$P$", P, dir(250));
dot("$H$", H, dir(270));
dot("$A'$", A_prime, dir(315));
dot("$T$", T, dir(230));
dot("$E$", E, dir(260));
dot("$F$", F, dir(315));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(15cm);

A = dir 117
C = dir 150
D = dir 30
unitcircle / 0.1 lightcyan / blue
B 225 = -A+2*foot A C D
M 125 = circumcenter A B C
N = circumcenter A B D
P 250 = origin
circumcircle A C B / gray
circumcircle A D B / gray
C--D / blue
A--B / blue
H 270 = orthocenter P M N
A' 315 = C+D-A
T 230 = M+N-A
E 260 = extension A P M T
F 315 = extension A P N T
A--F / lightred
C--A' / lightred
F--D / lightred
M--H--N / lightred
circumcircle A' E F / gray

A'--B / blue
circumcircle B E F / purple
M--T / dashed
N--F / dashed
C--A--D / lightred
!markangle("$\alpha$", n=1, radius=18.0, D,C,A, deepgreen);
!markangle("$\beta$", n=2, radius=30.0, A,D,C, deepgreen);
!markangle("$\alpha$", n=1, radius=18.0, C,D,F, deepgreen);
!markangle("$\beta$", n=2, radius=30.0, E,C,D, deepgreen);
!markangle("$90^{\circ}-\alpha$", n=3, radius=14.0, C,A,B, deepcyan);
!markangle("$90^{\circ}-\alpha$", n=3, radius=14.0, E,A,D, deepcyan);
*/
\end{asy}
\end{center}


Next, let $T$ denote the circumcenter of $\triangle A'EF$.
(This will be the tangency point later in the problem.)
\begin{claim*}
  Point $T$ also lies on $\ol{BA'}$ and is also the arc midpoint of $\widehat{EF}$ on $(BEF)$.
\end{claim*}
\begin{proof}
  We compute the angles of $\triangle A'EF$:
  \begin{align*}
    \dang FEA' &= \dang AEC = \dang ABC = \dang CAB = 90\dg - \alpha \\
    \dang A'FE &= \dang DFA = \dang DBA = \dang BAD = 90\dg - \beta \\
    \dang EA'F &= \alpha + \beta.
  \end{align*}
  Then, since $T$ is the circumcenter, it follows that:
  \[ \dang EA'T = \dang 90\dg - \dang A'FE = \beta = \dang A'CD = \dang CA'B. \]
  This shows that $T$ lies on $\ol{BA'}$.

  Also, we have $\dang ETF = 2 \dang EA'F = 2(\alpha+\beta)$ and
  \begin{align*}
    \dang EBF &= \dang EBA + \dang ABF = \dang ECA + \dang ADF \\
    &= \dang A'CA + \dang ADA' = (\alpha+\beta) + (\alpha+\beta) = 2(\alpha+\beta)
  \end{align*}
   which proves that $T$ also lies on $\ol{AB'}$.
\end{proof}

We then bring $M$ and $N$ into the picture as follows:
\begin{claim*}
  Point $T$ lies on both lines $ME$ and $NF$.
\end{claim*}
\begin{proof}
  To show that $F$, $T$, $N$ are collinear,
  note that $\triangle FEA' \sim \triangle FAD$ via a homothety at $F$.
  This homothety maps $T$ to $N$.
\end{proof}

We now deal with point $H$ using two claims.
\begin{claim*}
  We have $\ol{MH} \parallel \ol{AD}$ and $\ol{NH} \parallel \ol{AC}$.
\end{claim*}
\begin{proof}
  Note that $\ol{MH} \perp \ol{PN}$,
  but $\ol{PN}$ is the perpendicular bisector of $\ol{AD}$,
  so in fact $\ol{MH} \parallel \ol{AD}$.
  Similarly, $\ol{NH} \parallel \ol{AC}$.
\end{proof}
\begin{claim*}
  Lines $\ol{MH}$ and $\ol{NH}$ bisect $\angle NMT$ and $\angle MNT$.
  In fact, point $H$ is the incenter of $\triangle TMN$,
  and $\dang NTH = \dang HTM = 90\dg-(\alpha+\beta)$.
\end{claim*}
\begin{proof}
  Hence, $\dang HMN = \dang A'CD = \dang ADC = \beta$.
  But $\dang TMN = \dang CME = 2 \dang CAE = -2(90\dg-2\beta) = 2\beta$.
  That proves $\ol{MH}$ bisects $\angle NMT$; the other one is similar.

  To show that $H$ is an incenter (rather than an excenter)
  and get the last angle equality, we need to temporarily undirect our angles.
  Assume WLOG that $\triangle ACD$ is directed counterclockwise.
  The problem condition that $C$ and $D$ are the farther intersections of line $MN$
  mean that $\angle NHM = \angle CAD > 90\dg$.
  We are also promised $C$, $M$, $N$, $D$ are collinear in that order.
  Hence the reflections of line $MN$ over lines $MH$ and $NH$, which meet at $T$,
  should meet at a point for which $T$ lies on the same side as $H$.
  In other words, $\triangle MTN$ is oriented counterclockwise and contains $H$.

  Working with undirected $\alpha = \angle DCA$ and $\beta = \angle ADC$ with $\alpha + \beta < 90\dg$,
  \[ \angle NTH = \angle HTM = \half \angle NTM = \half(180\dg-2(\alpha+\beta))
    = 90\dg - (\alpha + \beta). \]
  This matches the claim and finishes the result.
\end{proof}

Now
\[ \dang NFA = 90\dg - \dang ADF = 90\dg-(\alpha+\beta) = \dang NTH \]
so $\ol{HT} \parallel \ol{AP}$.
And since $TE = TF$, we have the tangency requested too now, as desired.

\begin{remark*}
  There are many other ways to describe the point $T$.
  For example, $AMTN$ is a parallelogram and $MBTN$ is an isosceles trapezoid.
  In coordination, we joked that it was impossible to write a false conjecture.
\end{remark*}