Question 84
Quadrilateral $ABCD$ is circumscribed about a circle $\omega$. $E$ is the intersection point of $\omega$ and the diagonal $AC$, which is nearest to $A$. Point $F$ is diametrically opposite to point $E$ in the circle $\omega$. The line which is tangent to $\omega$ in the point $F$ intersects lines $AB$ and $BC$ in points $A_1$ and $C_1$, and lines $AD$ and $CD$ in points $A_2$ and $C_2$ respectively. Prove that $A_1C_1 = A_2C_2$.