Problem

Source: Sharygin Correspondence Round 2024 P10

Tags: geometry



Let $\omega$ be the circumcircle of triangle $ABC$. A point $T$ on the line $BC$ is such that $AT$ touches $\omega$. The bisector of angle $BAC$ meets $BC$ and $\omega$ at points $L$ and $A_0$ respectively. The line $TA_0$ meets $\omega$ at point $P$. The point $K$ lies on the segment $BC$ in such a way that $BL = CK$. Prove that $\angle BAP = \angle CAK$.