Question 29
You are given a triangle $ABC$, $\omega$ is its circumscribed circle, $I$ is its incenter. Let $K$ be any point on the arc $AC$ of $\omega$, not containing point $B$. Point $P$ is symmetrical to the point $I$ with respect to the point $K$. Point $T$ on the arc $AC$ of the circle $\omega$, which contains point $B$, is such that $\angle KCT = \angle PCI$. Prove that the bisectors of angles $AKC$ and $ATC$ meet at the line $CI$.