Question 57
Let the internal angle bisector of $\angle BAC$ of $\triangle ABC$ meet side $BC$ at $D$. Let $\Gamma$ be the circle through $A$ tangent to $BC$ at $D$. Suppose $\Gamma$ meets sides $AB$ and $AC$ at $E$ and $F$ again, respectively. Lines $BF$ and $CE$ meet $\Gamma$ again at $P$ and $Q$, respectively. Let $AP$ and $AQ$ intersect side $BC$ at $X$ and $Y$, respectively. Prove that $XY = \frac{1}{2}BC$.