By angle chase wededuce that $BQPD$ is cyclic then $\frac{AP}{AQ}=\frac{AD}{AB}$; more since $AC$ is the $A$-median of the triangle $ABD$ then $AR$ is the $A$ symmedian of the triangle $APQ$ thus $\frac{RP}{RQ}=\frac{AP^2}{AQ^2}=\frac{AD^2}{AB^2}=\frac{BC^2}{DC^2}$ RH HAS

By parallel of $ABCD$ we have $\angle DEB=\angle ABE=\angle ADF=\angle BFD$, so $\angle BPD=\angle BQD$, which implies $BQPD$ are concyclic. Let $AC\cap EF=G, AC\cap(ABD)=H$. By Reim's theorem, we have $PQ//EF$. Since $AC$ pass through the midpoint of $BD$, $\dfrac{PR}{RQ}=\dfrac{FG}{GE}$. By ratio lemma and Ceva's theorem for chords, $$\dfrac{FG}{GE}=\dfrac{FA}{AE}\cdot\dfrac{FH}{HE}=\dfrac{FH}{HE}=\dfrac{FB}{BA}\cdot\dfrac{AD}{DE}=\left(\dfrac{BC}{CD}\right)^2$$

Let $O$ be the intersection of the parallel to $DC$ that passes through $Q$ with $AC$. Then \[ \frac{PR}{RQ} = \frac{PA}{OQ}\]and \[\frac{OQ}{QA} = \frac{DC}{AD} = \frac{CD}{BC}\]Since $\triangle PDA \sim \triangle BFC$ \[\frac{PA}{DA} = \frac{BC}{FC}\]then $BC^{2} = PA \cdot FC$. Analogously, $\triangle DCE \sim \triangle QAB$ \[CD^{2} = QA \cdot CE\]Hence $\triangle CFE \sim \triangle CBD$ \[\frac{CF}{CE} = \frac{BC}{CD}\]Therefore \[(\frac{BC}{CD})^{2} = \frac{PA \cdot FC}{QA \cdot CE} = \frac{PR \cdot OQ \cdot CF}{RQ \cdot QA \cdot CE} = \frac{PR}{RQ} \]$\blacksquare$