MEMO 2025 T6

Let $ABC$ be an acute triangle with an interior point $D$ such that $\angle BDC = 180^\circ - \angle BAC$. The lines $BD$ and $AC$ intersect at the point $E$, and the lines $CD$ and $AB$ intersect at the point $F$. The points $P \neq E$ and $Q \neq F$ lie on the line $EF$ so that $BP = BE$ and $CQ = CF$. Assume that the segments $AP$ and $AQ$ intersect the circumcircle $\omega$ of $ABC$ at the points $R \neq A$ and $S \neq A$, respectively. Prove that the lines $RF$ and $SE$ intersect on $\omega$.