Let $D'$ be symmetric to $D$ with respect to $AC$. If $D'$ coincides with $B$, then $N$ and $M$ are symmetric to $K$ and $L$, respectively, with respect to $AC$, and the problem clearly holds. Let $D' \neq B$. Since $\omega_2$ is symmetric to $\omega_1$ with respect to $AC$, If we denote by $M'$ and $N'$ the points symmetric to $M$ and $N$, respectively, with respect to $AC$ we will have that $B, D', M', L, N', K$ all lie on $\omega_1$ (fig. 24). Since $\angle AD'C = \angle ADC = \angle ABC$, points $A, B, D', C$ are concyclic. We get $\angle (CA, AD') = \angle (CB, BD') = \angle (LN', N'D')$, so $LN' \parallel AC$, and similarly $M'K \parallel AC$. Then $LM'KN'$ is a cyclic trapezoid, so it's isosceles, and $KL = M'N' = MN$. We also get $\angle (MN, AC) = \angle (AC, M'N') = \angle (N'L, M'N') = \angle (KL, N'L) = \angle (KL, AC)$, so $MN \parallel KL$. Since segments $KL$ and $MN$ are equal, $KLMN$ is a parallelogram, so $KM$ intersects $NL$ at its midpoint, and since $N'L \parallel AC$ and $N$ is symmetric to $N'$ with respect to $AC$, the midpoint of $NL$ lies on $AC$, as desired.