Given an acute-angled triangle $ABC$, $AA_1$ and $CC_1$ are its angle bisectors, $I$ is its incenter, $M$ and $N$ are the midpoints of $AI$ and $CI$. Points $K$ and $L$ in the interior of triangles $AC_1I$ and $CA_1I$ respectively are such that $\angle AKI = \angle CLI = \angle AIC$, $\angle AKM = \angle ICA$, $\angle CLN = \angle IAC$. Prove that the circumradii of triangles $KIL$ and $ABC$ are equal. Proposed by Anton Trygub