Problem

Source: Iran MO 3rd round 2017 finals - Geometry P3

Tags: geometry, circumcircle



In triangle $ABC$ points $P$ and $Q$ lies on the external bisector of $\angle A$ such that $B$ and $P$ lies on the same side of $AC$. Perpendicular from $P$ to $AB$ and $Q$ to $AC$ intersect at $X$. Points $P'$ and $Q'$ lies on $PB$ and $QC$ such that $PX=P'X$ and $QX=Q'X$. Point $T$ is the midpoint of arc $BC$ (does not contain $A$) of the circumcircle of $ABC$. Prove that $P',Q'$ and $T$ are collinear if and only if $\angle PBA+\angle QCA=90^{\circ}$.