Altitudes $AH_A, BH_B, CH_C$ of triangle $ABC$ intersect at $H$, and let $M$ be the midpoint of the side $AC$. The bisector $BL$ of $\triangle ABC$ intersects $H_AH_C$ at point $K$. The line through $L$ parallel to $HM$ intersects $BH_B$ in point $T$. Prove that $TK = TL$. Proposed by Anton Trygub
Problem
Source: Ukrainian Mathematical Olympiad 2024. Day 1, Problem 10.3
Tags: geometry, orthocenter