BALTIC-WAY 2018 Problem 14

A quadrilateral $A B C D$ is circumscribed about a circle $\omega$. The intersection point of $\omega$ and the diagonal $A C$, closest to $A$, is $E$. The point $F$ is diametrically opposite to the point $E$ on the circle $\omega$. The tangent to $\omega$ at the point $F$ intersects lines $A B$ and $B C$ in points $A_{1}$ and $C_{1}$, and lines $A D$ and $C D$ in points $A_{2}$ and $C_{2}$, respectively. Prove that $A_{1} C_{1}=A_{2} C_{2}$.