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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.