$\angle BKC=\angle AKC+\angle AKB=\angle AMN+\angle ANB=180^{\circ}-30^{\circ}=150^{\circ}.$ Then it follows that $A$ is the circumcenter of $\triangle BKC$ $\Longrightarrow$ triangle $\triangle AKC$ is isosceles with apex $A.$ But because of $\angle AKN=\angle ABN=\angle ACN=60^{\circ},$ we deduce that $AN$ is the perpendicular bisector of $\overline{KC},$ i.e. $AN \perp KC$ $\Longrightarrow$ $\angle KAN=90^{\circ}-\angle AMN.$ Therefore, $AK$ passes through the circumcenter of $\triangle AMN.$