Assume that rays $AB$ and $DC$ intersect. Let's draw a line $l$ tangent to $\omega_1$ at $L'$ which is parallel to $AD$ and for which $\omega_1$ and $O$ are on opposite sides of the line. Let $\omega$ be tangent to $BC, AD$ at $K^*, L^*$ respectively. Let $M_1, M_2$ be the midpoints of $BC, AD,$ respectively. Let $I$ be the center of $\omega$. Then, observe that homothety implies that $O, L, L'$ are collinear. Therefore, $O, L', K, L$ are collinear, hence implying that line $KL$ is perpendicular to the angle bisector of the angle formed by $BC$ and $l$. As $l || AD,$ $KL$ is perpendicular to the angle bisector of the angle formed by $BC, AD.$ Notice that since $M_1$ is the midpoint of $K^*K$ and $I$ is the midpoint of the diameter of $\omega$ containing $K^*,$ we have that $IM_1$ bisects $OK^*.$ Similarly, $IM_2$ bisects $OL^*,$ and so it only suffices to show that $M_1M_2 || OL.$ However, since $KL, K^*L^*$ are both perpendicular to the angle formed by $BC, AD,$ we know that $KK^*L^*L$ is an isosceles trapezoid. Therefore, it clearly follows that $M_1M_2 || KL,$ as desired.