XVI Silk Road Math Competition

It is enough to show that points $E_2$ and $F_2$ are symmetrical with respect to the middle of the chord $E_1{F}_1$. 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 $D{E}_2$ and $C{F}_2$ as $K$, and let $\angle (AD,AC)=\alpha$. Then lines $E_2{F}_2$, $K{E}_2$ and $K{F}_2$ form an isosceles triangle, since $\angle (E_{2}K,E_2{F}_2)=\angle (F_2{E}_2,F_{2}K)=\alpha$. 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 $\angle (KM,F_2{E}_2)=\angle (KM,KD)+\angle (KD,F_2{E}_2)=90^{\circ }$. Then, we have that $KM$ passes through the center of circle $\omega$ and is perpendicular to the chord $E_1{F}_1$, as well as splits segment $E_2{F}_2$ 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 $E_1{F}_1$. Proved.