Problem

Source: Mexico National Olympiad Mock Exam 2019 P6

Tags: geometry, symmedian, Circumcenter, perpendicular bisector



Let ABC be a scalene triangle with circumcenter O, and let D and E be points inside angle BAC such that A lies on line DE, and ADB=CBA and AEC=BCA. Let M be the midpoint of BC and K be a point such that OK is perpendicular to AO and BAK=MAC. Finally, let P be the intersection of the perpendicular bisectors of BD and CE. Show that KO=KP. Proposed by Victor Domínguez