Problem: Triangle $ABC$ has circumcircle $\Gamma$ and obeys $\angle BCA > 90^{\circ}$.
There is a point $P$ in the interior of the line segment $AB$
such that $PB = PC$ and the length of $PA$ equals the radius of $\Gamma$.
The perpendicular bisector of $\ol{PB}$ intersects $\Gamma$ at the points $D$ and $E$.
Prove $P$ is the incenter of triangle $CDE$.

Solution: Let $O$ be the center of $\Gamma$ and $M$
the arc midpoint of $\ol{DE}$.
\begin{claim*}
  Quadrilateral $APMO$ is a rhombus.
\end{claim*}
\begin{proof}
  Since $PA = MO$ and both are perpendicular to $\ol{DE}$,
  it follows $APMO$ is a parallelogram.
  In fact though, because $AO = MO$, we get the rhombus.
\end{proof}
\begin{center}
\begin{asy}
size(6cm);
pair B = dir(0);
pair C = dir(180);
pair D = dir(37);
pair A = D*D;
pair E = dir(60)*A;
pair F = dir(-60)*A;
pair O = origin;
pair J = A+O-D;
filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);

filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, heavycyan);
draw(A--D, red);
draw(O--J, red);
filldraw(C--E--F--cycle, opacity(0.1)+green, heavygreen);
filldraw(A--E--O--F--cycle, opacity(0.1)+lightblue, blue);
draw(E--F, blue);
draw(A--O, blue);

dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$A$", A, dir(A));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$O$", O, dir(-90));
dot("$J$", J, dir(J));
label("IMO 2002/2", dir(-90), dir(-90));
\end{asy}
\qquad
\begin{asy}
size(6cm);
pair C = dir(180);
pair K = dir(37);
pair M = K*K;
pair D = dir(60)*M;
pair E = dir(-60)*M;
pair O = origin;
pair P = M+O-K;
filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);

filldraw(C--D--E--cycle, opacity(0.1)+green, heavygreen);
filldraw(M--D--O--E--cycle, opacity(0.1)+lightblue, blue);
draw(D--E, blue);
draw(M--O, blue);
draw(C--P, deepcyan);
draw(P--M, blue+dotted);
pair A = P+O-M;
pair B = -A+2*foot(O, P, A);
draw(A--B--O, orange);

dot("$C$", C, dir(C));
dot("$M$", M, dir(M));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$O$", O, dir(-90));
dot("$P$", P, dir(P));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
label("EGMO 2020/5", dir(-90), dir(-90));

/* TSQ Source:

C = dir 180
K := dir 37
M = K*K
D = dir(60)*M
E = dir(-60)*M
O = origin R-90
P = M+O-K
unitcircle 0.1 lightcyan / heavycyan

C--D--E--cycle 0.1 green / heavygreen
M--D--O--E--cycle 0.1 lightblue / blue
D--E blue
M--O blue
C--P deepcyan
P--M blue dotted
A = P+O-M
B = -A+2*foot O P A
A--B--O orange

*/
\end{asy}
\end{center}

Since $PC = PB$, and $PA = PM$,
it follows that $C$, $P$, $M$ are collinear now.
\begin{claim*}
  We have $MP = MD = ME$ equal to the radius of $\Gamma$.
\end{claim*}
\begin{proof}
  Note $PBMO$ is an isosceles trapezoid.
  Since $\ol{DE}$ is the perpendicular bisector of $\ol{PB}$,
  it is the perpendicular bisector of $\ol{OM}$ too.
  Hence $MDOE$ is a rhombus (with $60\dg$ internal angles), the end.
\end{proof}

Since $MP = MD = ME$ we are done by Fact 5.

\begin{remark*}
  The same figure appears in IMO 2002/2,
  drawn on the left for contrast.
\end{remark*}