BMO Shortlist, 2022

Since $BP$ is tangent to the circumcircle and $BR \parallel AC$, we have $\angle PBZ = \angle BAC = \angle TBR$. It follows that the right-angled triangles $RTB$ and $PZB$ are similar and therefore $\frac{PZ}{RT} = \frac{PB}{RB}$.

Analogously the triangles $REB$ and $PDB$ are also similar and therefore $\frac{PD}{RE} = \frac{PB}{RB}$.

Since $P$, $R$ belong on the bisector of $A$, we have that $PD = PL$ and $RT = RM$ so from the results of the previous two paragraphs we get that $\frac{PL}{RE} = \frac{PZ}{RM}$.

The quadrilaterals $ZPLC$ and $ERMC$ are cyclic, therefore $\angle ZPL = 180^\circ - \angle ECL = \angle ERM$. Together with the previous result we get that the triangles $ZPL$ and $MRE$ are similar. Using this together with properties of cyclic quadrilaterals and the fact that $BC \parallel XY$ we get that
$$\angle XYC = \angle ZCP = \angle ZLP = \angle MER = \angle MCR$$
Since $BR \parallel AC \parallel QP$ and $QX \parallel BC$ we get
$$\frac{AP}{PR} = \frac{AQ}{QB} = \frac{AX}{XC}$$
which implies that $XP \parallel CR$. Thus $\angle PXC = \angle RCM = \angle XYC$. So the result follows.