Problem 1711

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,
$$\begin{aligned} \angle LAN &= 360^\circ - \angle LAK - \angle NAM - \angle KAM = \\ &= 360^\circ - 30^\circ - 60^\circ - \angle KAM = 270^\circ - \angle KAM. \end{aligned} \quad (1)$$
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
$$\angle FDA = \angle FEA = 180^\circ - \angle LAN \stackrel{(1)}{=} \angle KAM - 90^\circ. \quad (2)$$
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:
$$\angle FDM = \angle FDA + \angle MDA = \angle KAM - 90^\circ + 90^\circ = \angle KAM. \quad (3)$$

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

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 $\triangle KEF \sim \triangle FAM$, since
$$KE : EF = KE : AD = \frac{1}{2}KA : \frac{1}{2}MA = KA : MA, \quad \angle KEF = \angle KAM.$$
So $\angle EKF = \angle AKM$, and then
$$\angle FKM = \angle FKA + \angle AKM = \angle FKA + \angle EKF = \angle EKA = 60^\circ.$$
$$\text{Thus, } \angle KFM = 180^\circ - \angle KMF - \angle FKM = 180^\circ - 30^\circ - 60^\circ = 90^\circ.$$