Estonian Mathematical Olympiad

A reflection across the centre of the segment $BD$ sends the triangle $BDP$ into the triangle $DBA$. Thus the circumcircle of $BDP$ goes to the circumcircle of $DBA$. Analogously we see that a reflection across the centre of $CD$ sends the circumcircle of $CDQ$ into the circumcircle of $DCA$.



The reflection of a circle across the midpoint of one of its chords is equivalent to a reflection across the line corresponding to the chord. Thus the reflections of the circumcircles of $BDP$ and $CDQ$ across $BC$ are the circumcircles of $ABD$ and $ACD$, respectively. The first two circles intersect in points $A$ and $D$, whereas the latter two circles intersect in $R$ and $D$. Thus $A$ and $R$ are reflections of each other across $BC$. Therefore the points $P$, $Q$, $R$ are all located on the opposite side of $BC$ compared to $A$, but at the same distance from $BC$ as $A$. Thus $P$, $Q$, $R$ lie on one line parallel to $BC$.

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Alternative solution.

We will use directed angles. Angle chasing yields $$\angle QRD=\angle QCD=\angle ADC=\angle ADB=\angle PBD=\angle PRD.$$ The equality $\angle QRD=\angle PRD$ means that $P,R,Q$ are collinear.