Let ABC be an acute triangle for which AB=AC, and let O be its circumcenter. Line AO meets the circumcircle of ABC again in D, and the line BC in E. The circumcircle of CDE meets the line CA again in P. The lines PE and AB intersect in Q. Line passing through O parallel to the line PE intersects the A-altitude of ABC in F.Prove that FP=FQ. geometrycircumcirclegeometry proposedLaw of Cosines