Problem

Source: Sharygin Correspondence Round 2024 P10

Tags: geometry



Let ω be the circumcircle of triangle ABC. A point T on the line BC is such that AT touches ω. The bisector of angle BAC meets BC and ω at points L and A0 respectively. The line TA0 meets ω at point P. The point K lies on the segment BC in such a way that BL=CK. Prove that BAP=CAK.