Let $L$ be the reflection of $N$ across line $BC.$ Since $\angle LBC=\angle MBA=60^{\circ},$ we deduce that $\angle MBL=\angle ABC$ and because of $MB=MA$ and $BL=BC,$ it follows that $\triangle ABC \cong \triangle MBL$ by SAS criterion $\Longrightarrow$ $ML=AC=AK.$ Similar reasoning yields $KL=AM$ $\Longrightarrow$ $AMLK$ is a parallelogram, thus $AL$ and $MK$ bisect each other at $U.$ If $H_a,M_a$ denote the foot of the A-altitude and midpoint of $BC,$ then perpendicular $\ell_a$ from $U$ to $BC$ is the midline of the trapezoid $ALM_aH_a$ $\Longrightarrow$ $\ell_a$ passes through the 9-point center of $\triangle ABC.$ Likewise, perpendiculars from midpoints of $MN,NK$ to $CA,AB$ pass through the 9-point center of $\triangle ABC.$ P.S. The result is still true for similar isosceles triangles AMB,BNC,CKA and the proof is similar.