UA-UMO 2023 Problem 96

A point $O$ lies inside $\triangle ABC$ so that $\angle BOC=90-\angle BAC$ . Let $BO, CO$ meet $AC, AB$ at $K, L$ . Points $K_1, L_1$ lie on the segments $CL, BK$ so that $K_1B=K_1K$ and $L_1C=L_1L$ . If $M$ is the midpoint of $BC$ , then prove that $\angle K_1ML_1=90^{o}$ .

Proposed by Anton Trygub