69th Belarusian Mathematical Olympiad

By the Archimedes's lemma, $BL$ is the bisector of $\angle ABC$, so $M_1\in BC$ and $N_1\in AB$. Denote $M{M}_1\cap LB=T$ and $M{M}_2\cap LA=P$. Since $\angle MTL=\angle MPL=90^{\circ }$, the quadrilateral $MTLP$ is cyclic with the diameter $ML$. Hence $$\angle M{M}_1{M}_2=\angle MTP=\angle MLP=\angle MBL=\angle TB{M}_1=90^{\circ }-\angle T{M}_{1}B.$$ Thus $\angle M_2{M}_{1}B=90^{\circ }$ and, similarly $\angle N_2{N}_{1}B=90^{\circ }$, which implies that the quadrilateral $B{N}_{1}K{M}_1$ is cyclic. Homotety with center $B$, mapping $\Omega$ to $\Gamma$, maps $△MBN$ to $△ABC$, hence these triangles are similar. Since the lines $MN$ and $M_1{N}_1$ are antiparallel with respect to $\angle ABC$, we can write (in oriented angles): $$ $$\begin{array}{rl}\angle (BK,AC)& =\angle (BK,N_{1}K)+\angle (N_{1}K,AB)+\angle (AB,AC)=\\ & =\angle (B{M}_1,M_1{N}_1)+90^{\circ }+\angle (MB,MN)=90^{\circ }.\end{array}$$