Belarusian Mathematical Olympiad

Answer: $60^{\circ }$.

Let $E$ and $F$ be the midpoints of the segments $BM$ and $CM$, respectively. Since $\angle AMB=\angle CMD=60^{\circ }$, we have $$\begin{array}{rl}\angle BMC& =360^{\circ }-\angle AMB-\angle CMD-\angle LMN=360^{\circ }-60^{\circ }-60^{\circ }-\angle LMN=\\ & =240^{\circ }-\angle LMN.\end{array}$$ Since $KF$ is the midline in the triangle $CBM$, we have $KF\parallel BM$, therefore, $\angle KFC=\angle BMC$. Then $$\begin{array}{rl}\angle KFM& =180^{\circ }-\angle KFC=180^{\circ }-\angle BMC=\\ & =180^{\circ }-(240^{\circ }-\angle LMN)=\angle LMN-60^{\circ }.\end{array}$$ By condition, $CM=DM$ and $\angle CMD=60^{\circ }$, so the triangle $CMD$ is equilateral. Since $FN$ is the midline in the triangle $CMD$, we see that the triangle $FMN$ is equilateral too and $FM=NM=FN$, $\angle MFN=60^{\circ }$. Therefore, $$\angle KFN=\angle KFM+\angle MFN=[\angle MFN=60^{\circ }]=\angle LMN-60^{\circ }+60^{\circ }=\angle LMN.$$ In the same way, one can show that $\angle KEL=\angle LMN$ and $EM=ML=EL$. Thus, $△KEL=△NFK=△NML$ (by two sides and the angle between them), so $KL=KN=LN$. Therefore, the triangle $KLN$ is equilateral and $\angle LKN=60^{\circ }$.