Autumn tournament

Let $A^{\prime }$ be the antipode of $A$ and let $AO\cap (BOC)=R$. Since $NO\parallel P{A}^{\prime }$, by Reim's theorem we need $AP{A}^{\prime }F$ being cyclic, or $DP\cdot DF=DA\cdot D{A}^{\prime }=DB\cdot DC=DO\cdot DR$, so we need $OPRF$ being cyclic, or that $\angle OFP=\angle ORP$. By shooting lemma, $RDEF$ is cyclic, so $\angle OFP=\angle ERD$, so we need $AR$ being the angle bisector of $\angle PRE$. Since $AR$ bisects $\angle BRC$, it is sufficient to show that $\angle BRP=\angle CRE$.

To use that $EQDO$ is cyclic, let $EQ\cap (BOC)=S$; Reim's implies $RS\parallel BC$, which together with the length condition gives that $RSQP$ is an isosceles trapezoid. Hence, $\angle CRE=\angle CSQ=\angle BRP$, which finishes the problem.