Let $ S$ be center of circle $ k$. Let $ \mathcal{C}$ be circle with center at $ A$ and $ r=AB$. Then $ C,D\in \mathcal{C}$. Let $ \angle CBD = \alpha$. Then we have: $ \angle CAD = 2\alpha$ and $ \angle ABD = 60-\alpha$. Beacuse triangles $ ABD$ and $ ACD$ are isoscale we have $ \angle ADB = 60-\alpha$ and $ \angle ACD =\angle ADC = 90-\alpha$. Then $ \angle BSE = 2\cdot \angle BDC = 60$. So $ \overline{BE} =1$. Now $ \angle BCE = 180- 60-(90-\alpha) = 30+\alpha$ and $ \angle BEC = 120-2\alpha$ since $ ABED$ is cyclic. Finaly $ \angle CBE =30+\alpha$ wich means $ \overline{CE} = \overline{BE} =1$.

Here is another solution. Let $ S$ be the center of $ k$. Since $ ABDE$ is cyclic $ \angle CBE=\angle ABE-60=180-\angle ADE-60=180-\angle ACD-60=\angle BCE$ so $ \triangle BCE$ is isosceles with $ EB=EC$. Extend $ AC$ to meet $ k$ at $ G$. Then $ \angle CGE=180-\angle ABE=180-60-\angle CBE=180-60-\angle BCE=\angle GCE$. so $ \triangle CGE$ is isosceles, with $ EC=EG$, so $ B,C,G$ lie on a circle centered at $ E$. Now $ \angle BSG=2\angle BAG=120$ and $ \angle BCG=180-\angle ACB=120$, so $ BSCG$ is cyclic, so $ S$ also lies on this circle centered at $ E$, so $ CE=SE=1$.

A simpler solution based upon Number1: A is the excenter of the triangle BCD; since angle <BAC = 60, it follows that <BDC = 30 = <BDE = < BAE, hence BE = 1 ( radius of the excircle BADE ) and AE is the angle bisector of <BAC, thence the perpendicular bisector of BC and CE = BE. Best regards, sunken rock

ABED being cyclic and AB = AD, it follows that EA is the angle bisector of <BED. It is well known that the middle of the arc BC of the circle (BEC) is on EA and since that point and A lie on the perpendicular bisector of BC; it follows that AE is the perpendicu;ar bisector of BC, and CE = BE ( 1 ) and <BAE = 30, which gives BE = 1 ( the radius of circle BADE ), hence CE = 1. Best regards, sunken rock

Let $ O$ be the center of $ k$ and for any $ P,Q$ on $ k$ define the centerangle $ \angle POQ = \widehat{PQ}$. Let $ A',B'$ be the intersection of $ AC,BC$ with $ k$ different from $ A,B$ respectively. From the power of a point: $ CA \cdot CA' = CB \cdot CB' \Rightarrow CA' = CB'$ Let $ \widehat{BA} = \widehat{AD} = v$ $ \angle CAB = \frac {\widehat{BA} + \widehat{B'A'}}{2} \iff \widehat{B'A'} = 120^\circ - v$ $ \angle ABB' = \frac {\widehat{AD} + \widehat{DB'}}{2} \iff \widehat{DB'} = 120 \circ - v$ $ \widehat{DB'} = \widehat{B'A} \iff DB' = B'A \Rightarrow \triangle DB'C$ is isosceles. $ \angle DB'C = \frac {\widehat{BA} + \widehat{AD}}{2} = v \Rightarrow \angle CDB' = 90 - \frac {v}{2} = \angle BCE$. $ \angle CDB' = \angle EDB' = \frac {\widehat{B'A'} + \widehat{A'E}}{2} \iff \widehat{A'E} = 60 ^\circ$ $ \triangle OEA'$ equilateral $ \Rightarrow EA' = OE = 1$ $ \angle CEA = 120^\circ - v, \angle EA'C = \angle ECA' = 30 ^\circ + \frac {v}{2} \Rightarrow CE = A'E = 1$. QED

Let $\angle{ADC}=a$ and $O$ denote the center of circle $k$. Since $AD=AC$, $\angle{BCE}=180-\angle{BCD}=180-(60+\angle{ACD})=120-a$. Note that $\angle{EBC}=180-a-\angle{ABC}=120-a$. Therefore $\triangle{ECB}$ is isosceles (1). By (1) and because $ABC$ is equilateral, $EA$ bisects $\angle{CAB}$ and therefore $\angle{EAB}=30$. Hence $\angle{EOB}=60$ and $\triangle{EOB}$ is equilateral (2). By (1) and (2), it follows that $CE=EB=OE=1$.

How does everyone know that ABED is cyclic here?

Ahiles wrote: The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD=AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$. By construction, both $D$ and $E$ are on $k$, the same as $A$ and $B$, duh.

Let <ABD=x. By angle chase, we have <CBE=90-x and <CEB=2x. Now by sine law, we have BC/sin(2x)=CE/cos(x) and AB/sin(x)=2. So BC/2sin(x)cos(x) =1/cos(x)=CE/cos(x) and so CE=1.

How did you get that BE = 1?

Very nice problem here! I used angle chasing for the solution. Konigsberg wrote: How does everyone know that ABED is cyclic here? We are given that they lie on circle k.