Let $H$ be the orthocentre of $\triangle ABC$ and let $P$ be a point on the circumcircle of $\triangle ABC$, distinct from $A,B,C$. Let $E$ and $F$ be the feet of altitudes from $H$ onto $AC$ and $AB$ respectively. Let $PAQB$ and $PARC$ be parallelograms. Suppose $QA$ meets $RH$ at $X$ and $RA$ meets $QH$ at $Y$. Prove that $XE$ is parallel to $YF$.