Belarusian Mathematical Olympiad

Answer: $90^{\circ }$.

Let points $A$, $B$, and $C$ lie in the same half-plane with respect to the line $LN$. Let $E$ and $D$ be the midpoints of the segments $AL$ and $AN$, respectively (see the Fig.). By condition, the triangle $BKL$ is equilateral and $K$ is the midpoint of the side $AB$, so $BK=KA=KL$. Therefore, the triangle $BLA$ is a right-angled triangle (the median $LK$ is half as long as the side $AB$) and $\angle BLA=90^{\circ }$. By condition, $\angle KBL=60^{\circ }$, so $\angle LAB=30^{\circ }$. By condition, the triangle $ANM$ is equilateral, then $\angle NAM=60^{\circ }$. Therefore, $$\begin{array}{rl}\angle LAN& =360^{\circ }-\angle LAK-\angle NAM-\angle KAM=\\ & =360^{\circ }-30^{\circ }-60^{\circ }-\angle KAM=270^{\circ }-\angle KAM.\end{array}(1)$$ Since $FD$ and $FE$ are the midlines of the triangle $LAN$, we have $FD\parallel LA$ and $FE\parallel NA$, so $EFDA$ is a parallelogram. Then $$\angle FDA=\angle FEA=180^{\circ }-\angle LAN\stackrel{(1)}{=}\angle KAM-90^{\circ }.(2)$$ Since $D$ is the midpoint of the side $AN$ of the equilateral triangle $ANM$, we have $\angle MDA=90^{\circ }$. Now from (2) it follows that: $$\angle FDM=\angle FDA+\angle MDA=\angle KAM-90^{\circ }+90^{\circ }=\angle KAM.(3)$$

In the similar way, we easily find that $$\angle FEK=\angle KAM.(4)$$

Since $EFDA$ is a parallelogram, we have $$FD=EA=[\angle KEA=90^{\circ },\angle EAK=30^{\circ }]=KA\frac{\sqrt{3}}{2}.$$ Also $DM=[\angle MDA=90^{\circ },\angle MAD=60^{\circ }]=MA\frac{\sqrt{3}}{2}$. Therefore, $FD:DM=KA:MA$, so, taking into account (3), we see that the triangles $FDM$ and $KAM$ are similar, hence $\angle MFD=\angle MKA$, $\angle FMD=\angle KMA$. Thus, $$\angle KMF=\angle KMA+\angle AMF=\angle FMD+\angle AMF=\angle AMD=30^{\circ }.$$ In the same way, we get $△KEF\sim △FAM$, since $$KE:EF=KE:AD=\frac{1}{2}KA:\frac{1}{2}MA=KA:MA,\angle KEF=\angle KAM.$$ So $\angle EKF=\angle AKM$, and then $$\angle FKM=\angle FKA+\angle AKM=\angle FKA+\angle EKF=\angle EKA=60^{\circ }.$$ $$\text{Thus, }\angle KFM=180^{\circ }-\angle KMF-\angle FKM=180^{\circ }-30^{\circ }-60^{\circ }=90^{\circ }.$$