Question 55
In triangle $ABC$ points $P, Q$ lie on the exterior angle bisector of vertex $A$ such that $P, B$ lie on the same side of line $AC$ and $Q$ is on the other side. Perpendicular line from $P$ to $AB$ cuts the perpendicular line from $Q$ to $AC$ at $X$. Points $P', Q'$ are respectively chosen on $PB, QC$ such that $QX = Q'X$ and $PX = P'X$. Let $T$ be the midpoint of arc $BC$ (the one that doesn't contain $A$) of the circumcircle of triangle $ABC$. Prove that $P', Q', T$ are collinear if and only if
$$
\widehat{PBA} + \widehat{QCA} = 90^\circ
$$
$$
\widehat{PBA} + \widehat{QCA} = 90^\circ
$$