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 geometrysymmedianCircumcenterperpendicular bisector