Problem: Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$.
Line $\ell$ is tangent to $\omega_1$ at $P$
and to $\omega_2$ at $Q$ so that $A$ is closer to $\ell$ than $B$.
Let $X$ and $Y$ be points on major arcs $\arc{PA}$
(on $\omega_1$) and $\arc{AQ}$ (on $\omega_2$), respectively,
such that $AX/PX = AY/QY = c$.
Extend segments $PA$ and $QA$ through $A$ to $R$ and $S$,
respectively, such that $AR = AS = c\cdot PQ$.
Given that the circumcenter of triangle $ARS$ lies on line $XY$,
prove that $\angle XPA = \angle AQY$.

Solution: We begin as follows:
\begin{claim*}
  There is a spiral similarity centered at $X$ mapping $AR$ to $PQ$.
  Similarly there is a spiral similarity centered at $Y$ mapping $SA$ to $PQ$.
\end{claim*}
\begin{proof}
  Since $\dang XAR = \dang XAP = \dang XPQ$,
  and $AR/AX = PQ/PX$ is given.
\end{proof}
Now the composition of the two spiral similarities
\[ AR \xmapsto{X} PQ \xmapsto{Y} SA \]
must be a rotation, since $AR = AS$.
The center of this rotation must coincide with the circumcenter $O$ of $\triangle ARS$,
which is known to lie on line $XY$.

\begin{center}
\begin{asy}
import graph; size(10cm);
pen uququq = rgb(0.25098,0.25098,0.25098); pen zzttqq = rgb(0.6,0.2,0.);
draw((13.97096,-7.13004)--(20.23013,-5.53182)--(27.19232,-12.23004)--cycle, linewidth(0.6) + zzttqq);
draw(circle((25.,-3.), 9.48683), linewidth(0.6) + uququq);
draw(circle((13.,-3.), 4.24264), linewidth(0.6) + uququq);

draw((13.97096,-7.13004)--(16.,0.), linewidth(0.6) + red);
draw((13.97096,-7.13004)--(11.14589,0.81605), linewidth(0.6) + red);
draw((11.14589,0.81605)--(25.35640,-1.57297), linewidth(0.6) + blue);
draw((27.19232,-12.23004)--(20.85410,5.53296), linewidth(0.6) + red);
draw((27.19232,-12.23004)--(16.,0.), linewidth(0.6) + red);
draw((20.85410,5.53296)--(9.74299,-7.13207), linewidth(0.6) + blue);
draw((13.97096,-7.13004)--(20.23013,-5.53182), linewidth(0.6) + zzttqq);
draw((20.23013,-5.53182)--(27.19232,-12.23004), linewidth(0.6) + zzttqq);
draw((27.19232,-12.23004)--(13.97096,-7.13004), linewidth(0.6) + zzttqq);
draw((20.23013,-5.53182)--(19.26755,-9.17733), linewidth(0.6));
draw((11.14589,0.81605)--(20.85410,5.53296), linewidth(0.6));
dot("$A$", (16.,0.), dir(100));
dot("$Q$", (20.85410,5.53296), dir(120));
dot("$P$", (11.14589,0.81605), dir(120));
dot("$X$", (13.97096,-7.13004), dir(280));
dot("$Y$", (27.19232,-12.23004), dir(280));
dot("$R$", (25.35640,-1.57297), dir((8.043, 16.110)));
dot("$S$", (9.74299,-7.13207), dir(225));
dot("$O$", (19.26755,-9.17733), dir(45));
dot("$O'$", (20.23013,-5.53182), dir(135));
\end{asy}
\end{center}

As $O$ is a fixed-point of the composed map above,
we may let $O'$ be the image of $O$ under the rotation at $X$,
so that
\[ \triangle XPA \overset{+}{\sim} \triangle XO'O, \qquad
  \triangle YQA \overset{+}{\sim} \triangle YO'O. \]
Because
\[ \frac{XO}{XO'} = \frac{XA}{XP} = c \frac{YQ}{YA} = \frac{YO}{YO'} \]
it follows $\ol{O'O}$ bisects $\angle XO'Y$.
Finally, we have
\[ \dang XPA = \dang XO'O = \dang OO'Y = \dang AQY. \]

\begin{remark*}
  Indeed, this also shows $\ol{XP} \parallel \ol{YQ}$; so the positive homothety
  from $\omega_1$ to $\omega_2$ maps $P$ to $Q$ and $X$ to $Y$.
\end{remark*}