Problem: Let $\Gamma $ be the circumcircle of triangle $ABC$.
A circle $\Omega$ is tangent to the line segment $AB$
and is tangent to $\Gamma$ at a point
lying on the same side of the line $AB$ as $C$.
The angle bisector of $\angle BCA$ intersects $\Omega$
at two different points $P$ and $Q$. Prove that $\angle ABP = \angle QBC$.

Solution: If we let $M$ denote the midpoint of arc $\widehat{AB}$
then the inversion at $M$ with radius $MA = MB$ fixes $\Omega$,
so it swaps $P$ and $Q$, thus
\[ \dang MPB = \dang QBM. \]
\begin{center}
\begin{asy}
pair A = dir(210);
pair B = dir(330);
pair C = dir(110);
pair T = dir(20);
pair M = dir(270);
pair S = extension(T, M, A, B);
pair O = extension(origin, T, S, S+M);
pair P = IP(CP(O, T), C--M);
pair Q = OP(CP(O, T), C--M);
filldraw(unitcircle, opacity(0.1)+lightcyan, deepcyan);
draw(C--M, deepcyan);
draw(A--B--C--cycle, deepcyan);
filldraw(CP(O, T), opacity(0.1)+lightgreen, deepgreen);

draw(A--P--B--Q--cycle, lightred);
draw(M--T, blue+dotted);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$T$", T, dir(T));
dot("$M$", M, dir(M));
dot("$S$", S, dir(S));
dot("$P$", P, dir(140));
dot("$Q$", Q, dir(Q));

/* TSQ Source:

A = dir 210
B = dir 330
C = dir 110
T = dir 20
M = dir 270
S = extension T M A B
O := extension origin T S S+M
P = IP CP O T C--M R140
Q = OP CP O T C--M
unitcircle 0.1 lightcyan / deepcyan
C--M deepcyan
A--B--C--cycle deepcyan
CP O T 0.1 lightgreen / deepgreen

A--P--B--Q--cycle lightred
M--T blue dotted

*/
\end{asy}
\end{center}
But
\begin{align*}
  \dang MPB &= \dang MCB + \dang CBP \\
  \dang QBM &= \dang ABM + \dang QBA
\end{align*}
implying the desired isogonality,
since $\dang ABM = \dang ACM  = \dang MCB$.