Let lines $AB$ and $CD$ intersect at $E$. We have that $\angle BKC = 60^\circ$. Let us prove that $BK > \frac{1}{2}BC$. Indeed, one of the angles $\angle KBC$ or $\angle KCB$ is at least $60^\circ$. WLOG, this angle is $\angle KBC$ (Fig. 35). Then in $\triangle BMK$ $\angle KBM \ge 60^\circ = \angle BKC > \angle BKM$. So $MK > BM$, hence on the segment $BK$ there is point $O'$, such that $MK \cdot MO' = BM^2 = MC^2$. Then circumcircles of $\triangle BO'K$ and $\triangle CO'K$ are tangent to $BC$. Hence, $\angle BKM = \angle O'BC$, $\angle CKM = \angle O'CB$. So $\angle BO'C = 180^\circ - \angle O'BC - \angle O'CB = 180^\circ - \angle BKM - \angle MKC = 180^\circ - 60^\circ = 120^\circ$

Fig. 35

Then point $O'$ is on circumcircle of $\triangle EBC$. Let line $BO'$ intersect $CE$ at $D'$, $CO'$ intersect $BE$ at $A'$. From the fact that $BECO'$ is cyclic we conclude that (Fig. 36)
$$\angle BO'E = \angle BCE = 60^\circ = \angle EBC = \angle EO'C.$$
Consider $\triangle O'EB$ and $\triangle EBD'$. Two pairs of angles are equal. Hence $\angle BEO' = \angle ED'B$. Then $\angle BEO' = \angle BCO' = \angle CKO' = \angle ED'B$. So point $D'$ is on circumcircle of $\triangle KO'C$. Similarly, point $A'$ is on the circle $\triangle BKO'$. Then $\angle A'KB = \angle A'O'B = 60^\circ = \angle CO'D' = \angle CKD'$. It follows that points $A'$, $K$ and $D'$ are on the same line.

Point $O$ lies on segment $MK$. If this point is inside $MO'$, then $A$ lies inside $A'B$, $D'$ inside $CD'$. Then $K$ lies strictly outside $\triangle EAD$, but by the problem $K$ is inside $AD$, a contradiction. Similarly, if point $O$ is outside $MO'$, then $A$, $D$ are outside $EA'$ and $ED'$, so $K$ is strictly inside $\triangle EAD$, again a contradiction. Hence $O = O'$. Then $A = A'$ and $D = D'$. But this is already proved that $\angle A'KB = \angle CKD' = 60^\circ$, which is needed, because $A = A'$ and $D = D'$.

Let us draw external bisector of $\angle BKC$. Let it intersect lines $BA$ and $CD$ at $A'$ and $D'$ (Fig. 36). We draw on sides $\triangle BKC$ outside equilateral triangles $BPC$, $BRK$ and $CQK$. Obviously $R$ and $Q$ are on $A'D'$, and $P$ is an intersection of $AB$ and $CD$.

Consider $\triangle BKC$. We saw, that

$$\angle CBR = \angle CBK + 60^\circ = 180^\circ - (120^\circ - \angle CBK) = \angle KBA'$$

It implies that $BR$ and $BA'$ are isogonal, also $KR$ is external bisector. So $A'$ and $R$ are isogonal, hence $CA'$ and $CR$ are isogonal. Similarly we have that segments $BD'$ and $BQ$ are isogonal. It is easy to see that $P$ - the point of intersection of tangents to circumcircle of $\triangle BKC$, so $\angle PBC = \angle BKC = \angle PCB = 60^\circ$. For this $KP$ is symmedian of $\triangle BKC$, so $KP$ and $KM$ are isogonal. As it is known $KP$, $BQ$ and $CR$ intersect in the same point, namely at Fermat point, so as lines $KM$, $BD'$ and $CA'$ intersect in the same point. Let this be point $O'$. Assume that $MO > MO'$ (Fig. 37). Then points $A'$ and $D'$ are on sides $BA$ and $CD$, but in this case $AD$ and $A'D'$ don't intersect each other. Similarly if $MO < MO'$. Hence $MO = MO' \Rightarrow A = A', D = D' \Rightarrow \angle AKB = \angle DKC$.

Fig. 37