Problem: Let $\Omega$ be the circumcircle of the triangle $ABC$.
The circle $\omega$ is tangent to the sides $AC$ and $BC$,
and it is internally tangent to the circle $\Omega$ at the point $P$.
A line parallel to $AB$ intersecting the
interior of triangle $ABC$ is tangent to $\omega$ at $Q$.
Prove that $\angle ACP = \angle QCB$.

Solution: First, let us extend $\ol{AQ}$ to meet $\ol{BC}$ at $Q_1$.
By homothety, we see that $Q_1$ is just the
contact point of the $A$-excircle with $\ol{BC}$.

\begin{center}
\begin{asy}
  size(7cm);
  pair A = Drawing("A", dir(100), dir(100));
  pair B = Drawing("B", dir(210), dir(190));
  pair C = Drawing("C", dir(330), dir(350));
  pair I = incenter(A,B,C);
  pair Ia = 2*dir(-90)-I;
  pair Ib = 2*dir(35)-I;
  pair Ic = 2*dir(155)-I;
  draw(unitcircle);
  pair E = dir(90);
  pair T = Drawing("P", 2*foot(origin,I,E)-E, dir(I-E));
  pair K = extension(A,B,T,midpoint(I--Ic));
  pair L = extension(A,C,T,midpoint(I--Ib));
  path w = Drawing(circumcircle(T,K,L));

  Drawing("I_A", Ia, dir(-90));
  pair Q1 = Drawing("Q_1", foot(Ia, B, C), dir(45));
  pair Q = Drawing("Q", IP(A--Q1, w), dir(45));
  draw(arc(Ia, abs(Ia-Q1), -20, 200));
  draw(B--C);
  draw(A--foot(Ia, A, B));
  draw(A--foot(Ia, A, C));
  draw(T--A--Q1);
  draw( (Q-0.8)--(Q+0.8) );
\end{asy}
\end{center}

Now let us perform an inversion around $A$
with radius $\sqrt{AB \cdot AC}$ followed by
a reflection around the angle bisector;
call this map $\Psi$.
Note that $\Psi$ fixes $B$ and $C$.
Moreover it swaps $\ol{BC}$ and $(ABC)$.
Hence, this map swaps the $A$-excircle
with the $A$-mixtilinear incircle $\omega$.
Hence $\Psi$ swaps $P$ and $Q_1$.
It follows that $\ol{AP}$ and $\ol{AQ_1}$
are isogonal with respect to $\angle BAC$,
meaning $\angle BAP = \angle CAQ_1$.
Since $\angle CAQ = \angle CAQ_1$ we are done.