Let ABCABC be an acute triangle with an interior point DD such that ∠BDC=180∘−∠BAC\angle BDC = 180^{\circ} - \angle BAC. The lines BDBD and ACAC intersect at the point EE, and the lines CDCD and ABAB intersect at the point FF. The points P≠EP \neq E and Q≠FQ \neq F lie on the line EFEF so that BP=BEBP = BE and CQ=CFCQ = CF. Assume that the segments APAP and AQAQ intersect the circumcircle ω\omega of ABCABC at the points R≠AR \neq A and S≠AS \neq A, respectively. Prove that the lines RFRF and SESE intersect on ω\omega.