Question 55
Let $ABC$ be an acute triangle with $AB \neq AC$ and circumcircle $\Gamma$. The angle bisector of $BAC$ intersects $BC$ and $\Gamma$ at $D$ and $E$ respectively. Circle with diameter $DE$ intersects $\Gamma$ again at $F \neq E$. Point $P$ is on $AF$ such that $PB = PC$ and $X$ and $Y$ are feet of perpendiculars from $P$ to $AB$ and $AC$ respectively. Let $H$ and $H'$ be the orthocenters of $ABC$ and $AXY$ respectively. $AH$ meets $\Gamma$ again at $Q$. If $AH'$ and $HH'$ intersect the circle with diameter $AH$ again at points $S$ and $T$, respectively, prove that the lines $AT$, $HS$ and $FQ$ are concurrent.