International Mathematical Olympiad

Solution 1. If $A A_{1} = C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ and $A_{1} B C_{1}$ are tangent to each other. So $A A_{1}$ and $C C_{1}$ must be different. Since the points $A$ and $A_{1}$ can be interchanged with $C$ and $C_{1}$, respectively, we may assume $A A_{1} < C C_{1}$.
Let $R$ be the radical center of the circles $A E B C$ and $A_{1} E B C_{1}$, and the circumcircle of the symmetric trapezoid $A C C_{1} A_{1}$; that is the common point of the pairwise radical axes $A C$, $A_{1} C_{1}$, and $B E$. By the symmetry of $A C$ and $A_{1} C_{1}$, the point $R$ lies on the common perpendicular bisector of $A A_{1}$ and $C C_{1}$, which is the external bisector of $\angle A D C$.
Let $F$ be the second intersection of the line $D R$ and the circle $A C D$. From the power of $R$ with respect to the circles $\omega$ and $A C F D$ we have $R B \cdot R E = R A \cdot R C = R D \cdot D F$, so the points $B, E, D$ and $F$ are concyclic.
The line $R D F$ is the external bisector of $\angle A D C$, so the point $F$ bisects the $\operatorname{arc} \overline{C D A}$. By $A B = B C$, on circle $\omega$, the point $B$ is the midpoint of $\operatorname{arc} \overline{A E C}$; let $M$ be the point diametrically opposite to $B$, that is the midpoint of the opposite $\operatorname{arc} \widetilde{C A}$ of $\omega$. Notice that the points $B, F$ and $M$ lie on the perpendicular bisector of $A C$, so they are collinear.

Finally, let $X$ be the second intersection point of $\omega$ and the line $D E$. Since $B M$ is a diameter in $\omega$, we have $\angle B X M = 90^{\circ}$. Moreover,
$$\angle E X M = 180^{\circ} - \angle M B E = 180^{\circ} - \angle F B E = \angle E D F,$$
so $M X$ and $F D$ are parallel. Since $B X$ is perpendicular to $M X$ and $B B_{1}$ is perpendicular to $F D$, this shows that $X$ lies on line $B B_{1}$.

Solution 2. Define point $M$ as the point opposite to $B$ on circle $\omega$, and point $R$ as the intersection of lines $A C$, $A_{1} C_{1}$ and $B E$, and show that $R$ lies on the external bisector of $\angle A D C$, like in the first solution.
Since $B$ is the midpoint of the $\operatorname{arc} \overline{A E C}$, the line $B E R$ is the external bisector of $\angle C E A$. Now we show that the internal angle bisectors of $\angle A D C$ and $\angle C E A$ meet on the segment $A C$. Let the angle bisector of $\angle A D C$ meet $A C$ at $S$, and let the angle bisector of $\angle C E A$, which is line $E M$, meet $A C$ at $S^{\prime}$. By applying the angle bisector theorem to both internal and external bisectors of $\angle A D C$ and $\angle C E A$,
$$A S : C S = A D : C D = A R : C R = A E : C E = A S^{\prime} : C S^{\prime},$$
so indeed $S = S^{\prime}$.
By $\angle R D S = \angle S E R = 90^{\circ}$ the points $R, S, D$ and $E$ are concyclic.

Now let the lines $B B_{1}$ and $D E$ meet at point $X$. Notice that $\angle E X B = \angle E D S$ because both $B B_{1}$ and $D S$ are perpendicular to the line $D R$, we have that $\angle E D S = \angle E R S$ in circle $S R D E$, and $\angle E R S = \angle E M B$ because $S R \perp B M$ and $E R \perp M E$. Therefore, $\angle E X B = \angle E M B$, so indeed, the point $X$ lies on $\omega$.