Question 184
In $\triangle ABC$, $AB = AC$. A line parallel to $BC$ meets sides $AB$ and $AC$ at $D$ and $E$, respectively. The angle bisector of $\angle BAC$ meets the circumcircles of $\triangle ABC$ and $\triangle ADE$ at points $X$ and $Y$, respectively. Let $F$ and $G$ be the midpoints of $BY$ and $XY$, respectively. Let $T$ be the intersection of lines $CY$ and $DF$. Prove that the circumcenter of $\triangle FGT$ lies on line $XY$.