The vertices $A$ and $B$ of an equilateral $\triangle A B C$ lie on a circle $k$ of radius 1, and the vertex $C$ is inside $k$. The point $D \neq B$ lies on $k, A D=A B$ and the line $D C$ intersects $k$ for the second time in point $E$. Find the length of the segment $C E$.
As $A D=A C, \triangle C D A$ is isosceles. If $\angle A D C=\angle A C D=\alpha$ and $\angle B C E=\beta$, then $\beta=120^{\circ}-\alpha$. The quadrilateral $A B E D$ is cyclic, so $\angle A B E=180^{\circ}-\alpha$. Then $\angle C B E=120^{\circ}-\alpha$ so $\angle C B E=\beta$. Thus $\triangle C B E$ is isosceles, so $A E$ is the perpendicular bisector of $B C$, so it bisects $\angle B A C$. Now the arc $B E$ is intercepted by a $30^{\circ}$ inscribed angle, so it measures $60^{\circ}$. Then $B E$ equals the radius of $k$, namely 1. Hence $C E=B E=1$.

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