62nd Ukrainian National Mathematical Olympiad

Inside the triangle $ABC$ there exists a point $O$ such that $\angle BOC=90^{\circ }-\angle BAC$. The rays $BO$ and $CO$ intersect the sides $AC$ and $AB$ at the points $K$ and $L$ respectively. The points $K_1$ and $L_1$ are selected on the segments $LC$ and $BK$ respectively, so that $B{K}_1=K_{1}K$ and $C{L}_1=L_{1}L$. Let $M$ be the middle of the side $BC$. Prove that $\angle K_{1}M{L}_1$ is a right angle.

(Anton Trygub)

Fig. 18