Problem: Isosceles triangle $ABC$, with $AB=AC$, is inscribed in circle $\omega$.
Let $D$ be an arbitrary point inside $BC$ such that $BD \neq DC$.
Ray $AD$ intersects $\omega$ again at $E$ (other than $A$).
Point $F$ (other than $E$) is chosen on $\omega$ such that $\angle DFE = 90\dg$.
Line $FE$ intersects rays $AB$ and $AC$ at points $X$ and $Y$, respectively.
Prove that $\angle XDE = \angle EDY$.

\end{enumerate}

Solution: We present three solutions.

\paragraph{Angle chasing solution.}
Note that $(BDA)$ and $(CDA)$ are congruent,
since $BA=CA$ and $\angle BDA + \angle CDA = 180\dg$.
So these two circles are reflections around line $ED$.
Moreover, $(DEF)$ is obviously also symmetric around line $ED$.
\begin{center}
\begin{asy}
size(10cm);

pair E = dir(100);
pair B = dir(160);
pair C = dir(0);
pair A = dir(260);
draw(unitcircle, gray);

draw(E--B--C--cycle, dashed+gray);
pair D = extension(E, A, B, C);
pair W = -E;
pair F = foot(E, W, D);
pair X = extension(A, B, E, F);
pair Y = extension(A, C, E, F);
draw(E--A, blue);
draw(X--Y--A--cycle, dashed+gray);
draw(X--D--Y, red);

pair X_prime = foot(E, D, X);
pair Y_prime = foot(E, D, Y);
filldraw(circumcircle(E, X_prime, Y_prime), opacity(0.1)+cyan, blue);
filldraw(circumcircle(B, D, A), opacity(0.1)+cyan, blue);
filldraw(circumcircle(C, D, A), opacity(0.1)+cyan, blue);

dot("$E$", E, dir(E));
dot("$B$", B, dir(160));
dot("$C$", C, dir(340));
dot("$A$", A, dir(A));
dot("$D$", D, dir(315));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot(X_prime);
dot(Y_prime);

clip(scale(1.7)*unitcircle);
dot("$Y$", D+2*dir(Y-D), dir(Y-D));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(14cm);
E = dir 100
B 160 = dir 160
C 340 = dir 0
A = dir 260
unitcircle / gray
E--B--C--cycle / dashed gray
D 315 = extension E A B C
W := -E
F = foot E W D
X = extension A B E F
Y = extension A C E F
E--A blue
X--Y--A--cycle / dashed gray
X--D--Y / red
X' 70 .= foot E D X
Y' 130 .= foot E D Y
circumcircle E X' Y' / 0.1 cyan / blue
circumcircle B D A / 0.1 cyan / blue
circumcircle C D A / 0.1 cyan / blue
*/
\end{asy}
\end{center}
Hence, the radical axis of $(BDA)$ and $(DEF)$,
and the radical axis of $(CDA)$ and $(DEF)$,
should be symmetric about line $DE$.
But these radical axii are exactly lines $XD$ and $YD$, so we're done.

\begin{remark*}
  [Motivation]
  The main idea is that you can replace $DX$ and $DY$ with the radical axii,
  letting $X'$ and $Y'$ be the second intersections of the blue circles.
  Then for the problem to be true, you'd need $X'$ and $Y'$ to be reflections.
  That's equivalent to $(BDA)$ and $(CDA)$ being congruent;
  you check it and it's indeed true.
\end{remark*}

\paragraph{Harmonic solution (mine).}
Let $T$ be the point on line $\ol{XFEY}$ such that $\angle EDT = 90\dg$,
and let $\ol{AT}$ meet $\omega$ again at $K$.
Then
\[ TD^2 = TF \cdot TE = TK \cdot TA \implies \angle DKT = 90\dg \]
so line $DK$ passes through the antipode $M$ of $A$.
\begin{center}
\begin{asy}
size(14cm);

pair E = dir(110);
pair B = dir(160);
pair C = dir(20);
pair A = dir(270);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);

filldraw(E--B--C--cycle, opacity(0.1)+yellow, blue);
pair D = extension(E, A, B, C);
pair W = -E;
pair F = foot(E, W, D);
pair X = extension(A, B, E, F);
pair Y = extension(A, C, E, F);
draw(E--A, blue);
draw(X--Y--A--cycle, blue);
draw(X--D--Y, deepgreen);
pair M = dir(90);
pair T = extension(E, F, D, D+M-E);
draw(X--T, blue);
draw(D--T, red);
draw(E--M, red);
pair K = extension(D, M, T, A);
draw(T--A, red);
draw(K--M, red);

dot("$E$", E, dir(E));
dot("$B$", B, dir(200));
dot("$C$", C, dir(340));
dot("$A$", A, dir(A));
dot("$D$", D, dir(315));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$M$", M, dir(35));
dot("$T$", T, dir(T));
dot("$K$", K, dir(K));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(14cm);
E = dir 110
B 200 = dir 160
C 340 = dir 20
A = dir 270
unitcircle / 0.1 lightcyan / blue
E--B--C--cycle / 0.1 yellow / blue
D 315 = extension E A B C
W := -E
F = foot E W D
X = extension A B E F
Y = extension A C E F
E--A blue
X--Y--A--cycle / blue
X--D--Y / deepgreen
M 35 = dir 90
T = extension E F D D+M-E
X--T / blue
D--T / red
E--M / red
K = extension D M T A
T--A / red
K--M / red
*/
\end{asy}
\end{center}

Thus,
\[ -1 = (AM;CB)_\omega \overset{D}{=} (EK;BC)_\omega \overset{A}{=} (TE;XY) \]
and since $\angle EDT = 90\dg$ we're done.

\begin{remark*}
  [Motivation]
  The idea is to kill the points $X$ and $Y$
  by reinterpreting the desired condition as $(TD;XY)=-1$
  and then projecting through $A$ onto $\omega$.
  This eliminates points $X$ and $Y$ altogether
  and reduces the problem to showing that $\ol{TA}$ passes
  through the harmonic conjugate of $E$ with respect to $BC$ on $\omega$.

  The labels on the diagram are slightly misleading in that
  $\triangle EBC$ should probably be thought of as the ``reference'' triangle.
\end{remark*}

\paragraph{Pascal solution (Zuming Feng).}
Extend ray $FD$ to the antipode $T$ of $E$ on $\omega$. Then,
\begin{itemize}
\ii By Pascal's theorem on $EFTABC$,
the points $X$, $D$, and $P \coloneq \ol{EC} \cap \ol{AT}$ are collinear.
\ii Similarly by Pascal's theorem on $EFTACB$, the points
the points $Y$, $D$, and $Q \coloneq \ol{EB} \cap \ol{AT}$ are collinear.
\end{itemize}

\begin{center}
\begin{asy}
size(14cm);
pair E = dir(100);
pair B = dir(160);
pair C = dir(0);
pair A = dir(260);
filldraw(unitcircle, opacity(0.1)+yellow, blue);

draw(E--B--C--cycle, dashed+gray);
pair D = extension(E, A, B, C);
pair W = -E;
pair F = foot(E, W, D);
pair X = extension(A, B, E, F);
pair Y = extension(A, C, E, F);
draw(E--A, blue);
draw(X--Y--A--cycle, dashed+gray);
draw(X--D--Y, red);

pair T = -E;
pair P = extension(C, E, T, A);
pair Q = extension(B, E, T, A);

draw(P--D--Q, red);
draw(P--Q--E--cycle, deepgreen);
draw(F--T, blue);
filldraw(circumcircle(E, D, F), opacity(0.1)+cyan, blue);

dot("$E$", E, dir(E));
dot("$B$", B, dir(160));
dot("$C$", C, dir(0));
dot("$A$", A, dir(A));
dot("$D$", D, dir(315));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$T$", T, dir(T));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(14cm);
E = dir 100
B 160 = dir 160
C 0 = dir 0
A = dir 260
unitcircle / 0.1 yellow / blue
E--B--C--cycle / dashed gray
D 315 = extension E A B C
W := -E
F = foot E W D
X = extension A B E F
Y = extension A C E F
E--A blue
X--Y--A--cycle / dashed gray
X--D--Y / red
T = -E
P = extension C E T A
Q = extension B E T A
P--D--Q / red
P--Q--E--cycle / deepgreen
F--T / blue
circumcircle E D F / 0.1 cyan / blue
*/

\end{asy}
\end{center}

Now it suffices to prove $\ol{ED}$ bisects $\angle QDP$.
However, $\ol{ED}$ is the angle bisector of $\angle QEP = \angle BEC$,
but also $\ol{EA} \perp \ol{QP}$.
Thus triangle $QEP$ is isosceles with $QE=PE$, and $\ol{EA}$ cuts it in half.
Since $D$ is on $\ol{EA}$, the result follows now.