Problem

Source: Kyiv City MO 2024 Round 1, Problem 9.2/10.2

Tags: geometry, angle bisector



Let $BL, AD$ be the bisector and the altitude correspondingly of an acute triangle ABC. They intersect at point $T$. It turned out that the altitude $LK$ of $\triangle ALB$ is divided in half by the line $AD$. Prove that $KT \perp BL$. Proposed by Mariia Rozhkova