Let ABCABC be an acute-angled triangle with ABACAB \neq AC, and let OO be its circumcentre. The line AOAO intersects the circumcircle ω\omega of ABCABC a second time in point DD, and the line BCBC in point EE. The circumcircle of CDECDE intersects the line CACA a second time in point PP. The line PEPE intersects the line ABAB in point QQ. The line through OO parallel to PEPE intersects the altitude of the triangle ABCABC that passes through AA in point FF.

Prove that FP=FQFP = FQ.