Problem 1747

Let $I$ be the incenter of $\triangle ABC$. We have $\angle IFG = \angle CBG = 90^\circ - \frac{1}{2}\angle A = \angle CIE$, that is $FG \parallel IE$. Similarly $GE \parallel IF$. We prove that the intersection point $W$ of $BY$ and $CX$ lies on $\Omega$: consider the broken line $BWCFGE$; by Pascal's theorem, $W$ lying on $\Omega$ is equivalent to $X, Y, I$ being collinear, and this holds because
$$A(X,Y; I, G) = (B,C; D, G) = -1 = G(E,F; I, \infty_{EF}) = G(X,Y; I, A).$$
Let $U$ be the second intersection point of $BY$ and $\odot(BEX)$; then from $\angle XUB = \angle XEB = \angle XGY$ we know that $U$ lies on $\odot(GXY)$. Similarly, the second intersection point $V$ of $CX$ and $\odot(CFY)$ also lies on $\odot(GXY)$. Since $W$ lies on $\Omega$, $\angle XUB = \angle GEB = \angle GWB$, that is $XU \parallel GW$. Similarly, $YV \parallel GW$. Therefore $XYVU$ is an isosceles trapezoid.

Since the internal angle bisector $WD$ of $\angle BWC$ is the perpendicular bisector $WS$ of $\overline{XU}$, and also the perpendicular bisector $WT$ of $\overline{YV}$, therefore $D, S, T$ are collinear, which completes the proof. $\Box$

Solution 2. Let $I$ be the incenter of $\triangle ABC$. Then likewise $IEGF$ is a parallelogram and $X, Y, I$ are collinear. Let $X^*, Y^*, A^*$ be the points diametrically opposite $X, Y, A$ with respect to $\odot(BEX), \odot(CFY), \Omega$ respectively. Then from $A^*B \perp AB, DE \perp GE$ we know $X^* = A^*B \cap DE$, and similarly $Y^* = CA^* \cap FD$. Note that
$$\angle X^*XE = \angle X^*BE = \angle A^*BE = \angle A^*GE = \angle(AI, XE),$$
that is $XX^* \parallel AI$. Similarly $YY^* \parallel AI$. Therefore, combining $X, Y, I$ collinear with the fact that the reflection of $I$ about $D$ is the $A$-excenter $J$ of $\triangle ABC$, we obtain that the original proposition is equivalent to $X^*, Y^*, J$ being collinear.

Let $E^*, F^*$ be the points diametrically opposite $E, F$ with respect to $\Omega$ respectively, and let $Z$ be the second intersection point of $A^*J$ and $\Omega$. Then
$$\begin{aligned} D(X^*, Y^*; J, A^*) &= (E, F; A, A^*) = (E^*, F^*; A^*, A) \\ &\stackrel{J}{=} (B, C; Z, D) = A^*(X^*, Y^*; J, D), \end{aligned}$$
that is $X^*, Y^*, J$ are collinear. Thus the original proposition holds. $\square$