## Solution

As $A D=A C, \triangle C D A$ is isosceles. If $\text{A D C= A C D=}$ and $\text{B C E=}$, then $\beta=120^{\circ}-\alpha$. The quadrilateral $A B E D$ is cyclic, so $\text{A B E=180 -}$. Then $\text{C B E=}$ $120^{\circ}-\alpha$ so $\text{C B E=}$. Thus $\triangle C B E$ is isosceles, so $A E$ is the perpendicular bisector of $B C$, so it bisects $\text{B A C}$. 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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