Problem: Let $H$ be the orthocenter of an acute triangle $ABC$,
let $F$ be the foot of the altitude from $C$ to $AB$,
and let $P$ be the reflection of $H$ across $BC$.
Suppose that the circumcircle of triangle $AFP$ intersects line $BC$
at two distinct points $X$ and $Y$.
Prove that $CX = CY$.

Solution: Let $Q$ be the antipode of $B$.
\begin{claim*}
  $AHQC$ is a parallelogram, and $APCQ$ is an isosceles trapezoid.
\end{claim*}
\begin{proof}
  As $\ol{AH} \perp \ol{BC} \perp \ol{CQ}$ and $\ol{CF} \perp \ol{AB} \perp \ol{AQ}$.
\end{proof}

\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = -B*C/A;
pair H = orthocenter(A, B, C);
pair F = foot(C, A, B);
pair Q = -B;
filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
draw(unitcircle, lightblue);
filldraw(A--H--C--Q--cycle, opacity(0.1)+lightred, red);
draw(H--P, blue);
pair O = origin;
draw(B--Q, dotted+blue);
draw(F--H, blue);
pair M = midpoint(Q--C);
pair N = midpoint(A--F);
draw(M--N, red);
draw(A--M--P, deepgreen+dashed);
draw(M--F, deepgreen+dashed);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$H$", H, dir(220));
dot("$F$", F, dir(F));
dot("$Q$", Q, dir(Q));
dot("$O$", O, dir(270));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
A = dir 110
B = dir 210
C = dir 330
P = -B*C/A
H 220 = orthocenter A B C
F = foot C A B
Q = -B
A--B--C--cycle / 0.1 lightblue / blue
unitcircle / lightblue
A--H--C--Q--cycle / 0.1 lightred / red
H--P / blue
O 270 = origin
B--Q / dotted blue
F--H / blue
M = midpoint Q--C
N = midpoint A--F
M--N / red
A--M--P / deepgreen dashed
M--F / deepgreen dashed
*/
\end{asy}
\end{center}

Let $M$ be the midpoint of $\ol{QC}$.
\begin{claim*}
  Point $M$ is the circumcenter of $\triangle AFP$.
\end{claim*}
\begin{proof}
  It's clear that $MA = MP$ from the isosceles trapezoid.
  As for $MA = MF$, let $N$ denote the midpoint of $\ol{AF}$;
  then $\ol{MN}$ is a midline of the parallelogram, so $\ol{MN} \perp \ol{AF}$.
\end{proof}

Since $\ol{CM} \perp \ol{BC}$ and $M$ is the center of $(AFP)$, it follows $CX = CY$.