Problem: Let $ABCD$ be a quadrilateral inscribed
in circle $\omega$ with $\ol{AC} \perp \ol{BD}$.
Let $E$ and $F$ be the reflections of $D$ over
$\ol{BA}$ and $\ol{BC}$, respectively, and let $P$ be
the intersection of $\ol{BD}$ and $\ol{EF}$.
Suppose that the circumcircles of $EPD$ and $FPD$
meet $\omega$ at $Q$ and $R$ different from $D$.
Show that $EQ = FR$.

Solution: Most of this problem is about realizing where
the points $P$, $Q$, $R$ are.

\paragraph{First solution (Evan Chen).}
Let $X$, $Y$, be the feet from $D$ to $\ol{BA}$, $\ol{BC}$,
and let $Z = \ol{BD} \cap \ol{AC}$.
By Simson theorem, the points $X$, $Y$, $Z$ are collinear.
Consequently, the point $P$ is the reflection of $D$ over $Z$,
and so we conclude $P$ is the orthocenter of $\triangle ABC$.

\begin{center}
\begin{asy}
size(10cm);
pair B = dir(100);
pair A = dir(210);
pair C = dir(330);
pair D = -A*C/B;
pair X = foot(D, B, A);
pair Y = foot(D, B, C);
pair Z = extension(A, C, B, D);
pair E = 2*X-D;
pair F = 2*Y-D;
pair P = A+B+C;
pair Q = -A*B/C;
pair R = -B*C/A;

filldraw(unitcircle, opacity(0.1)+lightred, red);
draw(A--B--C--D--cycle, red);
draw(A--C, red);
draw(B--P, red);
draw(A--X, red);
draw(X--Y, orange);
draw(E--F, orange);
draw(F--D--E, orange);

filldraw(circumcircle(P, F, D), opacity(0.1)+yellow, orange);
filldraw(circumcircle(P, E, D), opacity(0.1)+yellow, orange);
draw(A--R, red+dotted);
draw(C--Q, red+dotted);
draw(R--F, lightcyan);
draw(P--D, lightcyan);
draw(Q--E, lightcyan);


dot("$B$", B, dir(B));
dot("$A$", A, dir(180));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(80));
dot("$Z$", Z, dir(315));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$P$", P, dir(135));
dot("$Q$", Q, dir(110));
dot("$R$", R, dir(80));

/* TSQ Source:

!size(10cm);
B = dir 100
A = dir 210 R180
C = dir 330
D = -A*C/B
X = foot D B A
Y = foot D B C R80
Z = extension A C B D R315
E = 2*X-D
F = 2*Y-D
P = A+B+C R135
Q = -A*B/C R110
R = -B*C/A R80

unitcircle 0.1 lightred / red
A--B--C--D--cycle red
A--C red
B--P red
A--X red
X--Y orange
E--F orange
F--D--E orange

circumcircle P F D 0.1 yellow / orange
circumcircle P E D 0.1 yellow / orange
A--R red dotted
C--Q red dotted
R--F lightcyan
P--D lightcyan
Q--E lightcyan

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

Suppose now we extend ray $CP$ to meet $\omega$ again at $Q'$.
Then $\ol{BA}$ is the perpendicular bisector
of both $\ol{PQ'}$ and $\ol{DE}$;
consequently, $PQ'ED$ is an isosceles trapezoid.
In particular, it is cyclic, and so $Q' = Q$.
In the same way $R$ is the second intersection
of ray $\ol{AP}$ with $\omega$.

Now, because of the two isosceles trapezoids we have found, we conclude
\[ EQ = PD = FR \]
as desired.

\begin{remark*}
  Alternatively, after identifying $P$,
  one can note $\ol{BQE}$ and $\ol{BRF}$ are collinear.
  Since $BE = BD = BF$,
  upon noticing $BQ = BP = BR$ we are also done.
\end{remark*}

\paragraph{Second solution (Danielle Wang).}
Here is a solution which does not identify the point $P$ at all.
We know that $BE = BD = BF$, by construction.

\begin{center}
\begin{asy}
size(12cm);
pair B = dir(100);
pair A = dir(210);
pair C = dir(330);
pair D = -A*C/B;
pair X = foot(D, B, A);
pair Y = foot(D, B, C);
pair Z = extension(A, C, B, D);
pair E = 2*X-D;
pair F = 2*Y-D;
pair P = A+B+C;
pair Qp = -A*B/C;
pair Rp = -B*C/A;

filldraw(unitcircle, opacity(0.1)+lightred, red);
draw(A--B--C--D--cycle, red);
draw(A--C, red);
draw(B--D, heavycyan);
draw(A--X, red);
draw(E--F, orange);
draw(F--D--E, orange);

filldraw(circumcircle(P, F, D), opacity(0.1)+yellow, dotted);
filldraw(circumcircle(P, E, D), opacity(0.1)+yellow, dotted);
draw(E--B--F, heavycyan);
draw(Qp--D--Rp, orange);
draw(arc(B,abs(B-D),190,350), heavycyan);

dot("$B$", B, dir(B));
dot("$A$", A, dir(180));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot(X);
dot(Y);
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$P$", P, dir(135));
dot("$Q'$", Qp, dir(110));
dot("$R'$", Rp, dir(80));

/* TSQ Source:

!size(12cm);
B = dir 100
A = dir 210 R180
C = dir 330
D = -A*C/B
X .= foot D B A
Y .= foot D B C R80
Z := extension A C B D R315
E = 2*X-D
F = 2*Y-D
P = A+B+C R135
Q' = -A*B/C R110
R' = -B*C/A R80

unitcircle 0.1 lightred / red
A--B--C--D--cycle red
A--C red
B--D heavycyan
A--X red
E--F orange
F--D--E orange

circumcircle P F D 0.1 yellow / dotted
circumcircle P E D 0.1 yellow / dotted
E--B--F heavycyan
Qp--D--Rp orange
!draw(arc(B,abs(B-D),190,350), heavycyan);

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

\begin{claim*}
  The points $B$, $Q$, $E$ are collinear.
  Similarly the points $B$, $R$, $F$ are collinear.
\end{claim*}
\begin{proof}
  Work directed modulo $180\dg$.
  Let $Q'$ be the intersection of $\ol{BE}$ with $(ABCD)$.
  Let $\alpha = \dang DEB = \dang BDE$
  and $\beta = \dang BFD = \dang FDB$.

  Observe that $BE = BD = BF$, so $B$ is the circumcenter
  of $\triangle DEF$. Thus, $\dang DEP = \dang DEF = 90\dg - \beta$.
  Then
  \begin{align*}
    \dang DPE &= \dang DEP + \dang PDE = (90\dg-\beta) + \alpha \\
    &= \alpha - \beta + 90\dg \\
    \dang DQ'B &= \dang DCB = \dang DCA + \dang ACB \\
    &= \dang DBA - (90\dg - \dang DBC) = -(90\dg-\alpha) - (90\dg - (90\dg - \beta)) \\
    &= \alpha - \beta + 90\dg.
  \end{align*}
  Thus $Q'$ lies on the desired circle, so $Q' = Q$.
\end{proof}

Now, by power of a point we have $BQ \cdot BE = BP \cdot BD = BR \cdot BF$,
so $BQ = BP = BR$. Hence $EQ = PD = FR$.