Problem: Let $ABC$ be a scalene triangle,
and let $D$ be a point on side $BC$ satisfying $\angle BAD=\angle DAC$.
Suppose that $X$ and $Y$ are points inside $ABC$
such that triangles $ABX$ and $ACY$ are similar and quadrilaterals $ACDX$ and $ABDY$ are cyclic.
Let lines $BX$ and $CY$ meet at $S$ and lines $BY$ and $CX$ meet at $T$.
Prove that lines $DS$ and $AT$ are parallel.

Solution: \paragraph{Solution by characterizing $\boldsymbol X$ and $\boldsymbol Y$.}
We first state an important property of $X$ and $Y$.

\begin{claim*}
  Points $X$ and $Y$ are isogonal conjugates with respect to $\triangle ABC$.
\end{claim*}

\begin{center}
	\begin{asy}
    unitsize(120);
    pair A = dir(125);
    pair C = dir(-35);
    pair B = dir(-145);
    pair D = extension(A,incenter(A,B,C),B,C);
    pair gammaB = extension(0, B+C, B, -A);
    pair gammaC = extension(0, B+C, C, -A);
    pair omegaB = circumcenter(A,B,D);
    pair omegaC = circumcenter(A,C,D);
    pair Y = 2*foot(B,omegaB,gammaC) - B;
    pair X = 2*foot(C,omegaC,gammaB) - C;
    pair E = 2*foot(gammaB, A, C) - C;
    pair F = extension(B,E,A,D);
    pair S = extension(B,X,C,Y);
    pair T = extension(B,Y,C,X);
    pair Cp = 2*foot(omegaC,B,X)-X;
    pair Bp = 2*foot(omegaB,C,Y)-Y;

    draw(A--B--C--A, red);
    draw(A--D, red);
    draw(circumcircle(A,B,D),blue);
    draw(circumcircle(A,C,D),blue);
    draw(B--E,dotted);
    draw(circumcircle(A,E,F),dotted);
    draw(B--Y,dashed+heavygreen);
    draw(C--X,dashed+heavygreen);
    draw(B--Cp,dashed+heavygreen);
    draw(C--Bp,dashed+heavygreen);

    dot("$A$", A, dir(A));
    dot("$B$", B, dir(B));
    dot("$C$", C, dir(C));
    dot("$D$", D, dir(-90));
    dot("$X$", X, dir(X)*dir(10));
    dot("$Y$", Y, 1.5*dir(20));
    dot("$E$", E, dir(E));
    dot("$F$", F, 1.5*dir(150));
    dot("$S$", S, 1.5*dir(100));
    dot("$T$", T, dir(T));
    dot("$B'$", Bp, dir(Bp));
    dot("$C'$", Cp, dir(Cp));
  \end{asy}
\end{center}

Here are two proofs of the claim.

\begin{proof}[First proof of claim by Maxim Li]
  We prove an equivalent statement that $S$ and $T$ are isogonal conjugates
  with respect to $\triangle ABC$.

  First, we note that $\angle YBC = \angle YAD = \angle XAD = \angle XCB$, so $BT=TC$.
  Now, let $T'$ be the isogonal conjugate of $S$ w.r.t. $\triangle ABC$.
  Since $\angle ABS = \angle ACS$, it follows that $T'B=T'C$.
  Thus, both $T$ and $T'$ lies on perpendicular bisector of $BC$.

  Moreover, since $AX$ and $AY$ are isogonal with respect to $\angle BAC$,
  by DDIT on point $A$ and $BXCY$, we get that
  $AS$ and $AT$ are isogonal with respect to $\angle BAC$.
  Hence, $A, T, T'$ are collinear.
  Since $\triangle ABC$ is scalene, combining this with the previous
  paragraph gives $T=T'$, or $S$ and $T$ are isogonal conjugates.
\end{proof}

\begin{proof}[A second different proof of the claim.]
  Let $\gamma_B$ be the circle through $B$ and $C$ tangent to $AB$,
  and define $\gamma_C$ similarly.
  We claim that $X \in \gamma_B$ and $Y \in \gamma_C$.

  Let $E$ be the second intersection of $\gamma_B$ with $AC$,
  and let $F$ be the intersection of $BE$ and $AD$.
  Consider the transformation $\tau$ that is the
  composition of a reflection across $AD$ and a homothety at $A$ with ratio $\frac{AB}{AC}$.
  Note that $\tau$ maps $Y \mapsto X$, $B \mapsto E$, and $D \mapsto F$.
  Thus, $X$ lies on $\tau(\odot(ABD)) = \odot(AEF)$,
  so $X$ is the second intersection of $\odot(AEF)$ and $\odot(ADC)$,
  i.e.\ it is the Miquel point of quadrilateral $CDFE$.
  Thus, $X$ lies on $\odot(BEC) = \gamma_B$, as claimed.

  Finally, the main claim follows from $\angle BCX = \angle ABX = \angle ACY$.
\end{proof}

Here are two ways to finish after the claim.

\subparagraph{Finish with isosceles trapezoid.}
We compute
\[\angle AXB = 360\dg-\angle BXC-\angle CXA = 360\dg-\angle BEC-\angle CDA
  = 180\dg-\frac{\angle A}{2}.\]
Thus, if $BX$ meets $\odot(ACD)$ again at $C'$, then
\[ \angle ADC' = \angle AXC' = \frac{\angle A}2 = \angle CAD, \]
so $ADCC'$ is an isosceles trapezoid.
Similarly, if $CY$ meets $\odot(ABD)$ again at $B'$, then $ADBB'$ is an isosceles trapezoid.
Hence, $BCC'B'$ is an isosceles trapezoid whose diagonals meet at $S$, so we have that $SA=SD$.

However, by the claim, $S$ and $T$ are isogonal conjugates in $ABC$. Thus,
\[ \angle ADS = \angle SAD = \angle DAT, \]
which completes the proof.

\subparagraph{Finish with angle chasing (Pitchayut Saengrungkongka).}
By the claim, let $\angle DAX = \angle DAY = \angle ABX = \angle ACY = \theta$.
Then,
\begin{align*}
  \angle XSY &= \angle A + 2\theta \\
  \angle XDY &= \angle BDY + \angle CDX - 180\dg = 180\dg - (\angle BAY + \angle CAX) \\
  &= 180\dg - (\angle A + 2\theta),
\end{align*}
so $XSYD$ is cyclic. Moreover,
\[ \angle XTY = 180\dg - \angle YBC - \angle XCB
  = 180\dg - 2\theta = 180\dg - \angle XAY, \]
so $AXTY$ is also cyclic. Finally,
\begin{align*}
  \dang(AT,AX)
  &= \dang TYX = \dang BYD + \dang DYX \\
  &= \dang BAD + \dang DSX = \dang BXA + \dang DSX \\
  &= \dang(DS,AX).
\end{align*}

\paragraph{Solution 2 (Ruben Carpenter).}
Let $M$ be the midpoint of $BC$, and $S'$, $T'$ the reflections of $S$, $T$ over $M$.
In what follows, DDIT is short for the dual of Desargues' involution theorem.

\begin{claim*}
  $S'$ lies on $AT$.
\end{claim*}
\begin{proof}
  From $\triangle ABX \overset{-}{\sim} \triangle AYC$,
  reflection along the angle bisector of $\angle ABC$ is an involution on the pencil of
  lines through $A$ with pairs $(AB, AC)$, $(AX, AY)$ and $(A\infty_{BS}, A\infty_{CS})$.
  By DDIT from $A$ onto complete quadrilateral $XYSTBC$, $AS$ is sent to $AT$.
  By DDIT from $A$ onto $BC\infty_{BX}\infty_{CY}SS'$, $(AS, AS')$
  is also an involutive pair, so $AS' \equiv AT$.
\end{proof}

\begin{claim*}
  $T'$ lies on $DS$.
\end{claim*}
\begin{proof}
	Using the circles $(ACDX)$, $(ABDY)$ we obtain
	\[\angle BDT=\angle DAY=\angle DAC-\angle YAC = \angle BAD-\angle BAX=\angle XAD=\angle TCB,\]
	so $T$ lies on the perpendicular bisector of $BC$. Furthermore
	\[\angle CDY=\angle BAY=\angle XAC=\angle XDB,\]
	so by DDIT from $D$ onto $XSYTBC$ shows $\angle CDT = \angle SDB$.
  The conclusion follows.
\end{proof}

Finally, since $STS'T'$ is a parallelogram we immediately have $DS \parallel AT$.

\begin{remark*}
  This problem was discovered by taking a degenerate case of IMO 2018/6 in which
  three of the vertices of the quadrilateral are collinear. Nevertheless, this
  origin is very obscured in the problem statement and does not seem to help with
  the solution at all. The original statement asked to show that the midpoints of
  $AD$, $BC$, and $XY$ are collinear. However, it was found that this is a
  straightforward consequence of a high-powered result about isogonal conjugates,
  so the statement was changed.
\end{remark*}