By some straightforward angle chasing we have $\angle AQP = \angle BCP = \angle BDC = \angle ABD$ so $\overline{BD}$ and $\overline{PQ}$ are antiparallel in $\angle A$. Thus, since $AR$ is the $A$-symmedian in $\triangle APQ$, it must be the $A$-median in $\triangle ABD;$ but $C$ lies on the $A$-median too since $ABCD$ is a parallelogram, so we are done.

Let $RQ \cap BC=E,RP \cap DC=F$ We have $\angle DQC=\angle BCP=\angle BDC=\angle DBA$ so $Q,D,B,P$ are cyclic. $\angle PBE=\angle PQE $ so $E,Q,B,P $ are cyclic. $\angle FPQ= \angle FDQ $ so $F,P,D,Q$ are cyclic. $\implies E,Q,D,B,P,F$ are cyclic. By Pascal at $EQDFPB,$ we get $R,A,C$ are collinear.