Problem: Given a triangle $ABC$, let $P$ and $Q$ be points
on segments $\ol{AB}$ and $\ol{AC}$, respectively, such that $AP=AQ$.
Let $S$ and $R$ be distinct points on segment $\ol{BC}$
such that $S$ lies between $B$ and $R$,
$\angle BPS=\angle PRS$, and $\angle CQR=\angle QSR$.
Prove that $P$, $Q$, $R$, $S$ are concyclic.

Solution: Assume for contradiction that $(PRS)$ and $(QRS)$ are distinct.
Then $\ol{RS}$ is the radical axis of these two circles.

\begin{center}
\begin{asy}
size(12cm);
pair P = dir(155);
pair Q = dir(55);
pair S = dir(245);
pair R = dir(295);
pair A = 2*P*Q/(P+Q);
pair B = extension(A, P, R, S);
pair C = extension(A, Q, R, S);
draw(unitcircle, dashed);
markangle(B,P,S,radius=18,n=1,deepgreen,StickIntervalMarker(1,deepgreen));
markangle(P,R,S,radius=18,n=1,deepgreen,StickIntervalMarker(1,deepgreen));
markangle(R,Q,C,radius=12,n=2,purple);
markangle(R,S,Q,radius=12,n=2,purple);

filldraw(A--B--C--cycle, opacity(0.1)+gray, brown);
draw(A--P, red+1.4);
draw(A--Q, red+1.4);
draw(S--P--R--Q--cycle, black);

dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$S$", S, dir(S));
dot("$R$", R, dir(R));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(12cm);
P = dir 155
Q = dir 55
S = dir 245
R = dir 295
A = 2*P*Q/(P+Q)
B = extension A P R S
C = extension A Q R S
unitcircle / dashed
!markangle(B,P,S,radius=18,n=1,deepgreen,StickIntervalMarker(1,deepgreen));
!markangle(P,R,S,radius=18,n=1,deepgreen,StickIntervalMarker(1,deepgreen));
!markangle(R,Q,C,radius=12,n=2,purple);
!markangle(R,S,Q,radius=12,n=2,purple);

A--B--C--cycle / 0.1 gray / brown
A--P / red 1.4
A--Q / red 1.4
S--P--R--Q--cycle / black
*/
\end{asy}
\end{center}

However, the given angle conditions give two tangencies:
\begin{itemize}
  \item the given $\angle BPS = \angle PRS$ means $\ol{AP}$ is tangent to $(PRS)$;
  \item the given $\angle CQR = \angle QSR$ means $\ol{AQ}$ is tangent to $(QRS)$.
\end{itemize}
Since $AP = AQ$, it follows the point $A$ has equal power $AP^2 = AQ^2$ to both circles.

Hence $A$ should lie on the radical axis $\ol{RS}$,
which is impossible since $A$ does not lie on line $BC$.
This contradiction finishes the problem.