It is enough to show that points $E_2$ and $F_2$ are symmetrical with respect to the middle of the chord $E_1F_1$. In order to ignore cases of mutual arrangement of points, lines and circles, we use oriented angles. Let $M$ be the center of circle $\omega$. Denote the intersection point of lines $DE_2$ and $CF_2$ as $K$, and let $\angle(AD, AC) = \alpha$. Then lines $E_2F_2$, $KE_2$ and $KF_2$ form an isosceles triangle, since $\angle(E_2K, E_2F_2) = \angle(F_2E_2, F_2K) = \alpha$. Then, the exterior angle at vertex $K$ in this triangle equals $2\alpha$. Since points $K$ and $M$ lie on one side of the line $CD$ and $\angle CMD = 2\alpha$, then points $C$, $M$, $D$ and $K$ lie on one circle. Since triangle $CMD$ is isosceles, we have: $\angle(KM, KD) = \angle(CM, CD) = 90^\circ - \alpha$ or $\angle(KM, F_2E_2) = \angle(KM, KD) + \angle(KD, F_2E_2) = 90^\circ$. Then, we have that $KM$ passes through the center of circle $\omega$ and is perpendicular to the chord $E_1F_1$, as well as splits segment $E_2F_2$ in half. Hence, pairs of points $E_2, F_2$ and $E_1, F_1$ are symmetrical with respect to the middle of the chord $E_1F_1$. Proved.