The points $A$, $H$, $M$, and $K$ lie on a circle $\omega$ with diameter $AK$ (see Fig. 4). Since $\angle MAH = 30^\circ$, we have $HM = AK/2$. On the other hand, $BC = 2HM$.

First solution. Since $\angle ANK = \angle AMK = 90^\circ$, the points $A$, $H$, $M$, and $K$ lie on a circle $\omega$ with diameter $AK$ (see Fig. 4). By the condition, the chord $HM$ of this circle subtends the angle $\angle MAN = 30^\circ$, so $HM = 2R \sin 30^\circ = AK/2$. On the other hand, $HM$ is the median in the right triangle $BHC$, so $BC = 2HM = AK$, which is what was required to prove.

Second solution. Complete triangle $ABC$ to parallelograms $ABRC$ and $AQBC$ (see Fig. 5); then $BP = AC = BQ$ and $KH \perp PQ$. Moreover, point $M$ is the midpoint of $AP$ (as the intersection point of the diagonals of parallelogram $ABRC$). Therefore, lines $MK$ and $HK$ are the perpendicular bisectors of segments $AP$ and $PQ$. Consequently, point $K$ is the center of the circle $\Omega$ circumscribed about triangle $APQ$. Furthermore, $\angle APQ = \angle PAC = 30^\circ$, so the chord $AQ$ of circle $\Omega$ is equal to the radius $AK$. Finally, from parallelogram $AQBC$ we get $BC = AQ = AK$.