Let $D$ be the foot of the perpendicular from $C$ to $AB$ and $H$ be the orthocenter of $\triangle ABC$. Note first that $\omega_1$ and $\omega_2$ both intersect $AB$ at $D$. By Power of a Point, $LH \cdot HN = CH \cdot HD = KH \cdot HM$, implying that $KLMN$ is a cyclic quadrilateral. Noting that $AC$ and $BC$ are perpendicular bisectors of the diagonals of $KLMN$, we conclude that the center of its circumcircle is $C$. Observe now that since $\angle ANC = \angle ALC = 90^\circ$, we have that $AN$ and $AL$ are tangent to the circumcircle of $KLMN$. Thus, $H$ is on the polar of $A$. Similarly, $H$ is on the polar of $B$, and so by Brocard's theorem, $KN$ and $LM$ meet on the polar of $H$, which is $AB$.