Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National Olympiad Prove it North Macedonia

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 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 $CE$ .

Solution

As $AD = AC$ , $\triangle CDA$ is isosceles. If $\angle ADC = \angle ACD = \alpha$ and $\angle BCE = \beta$ then $\beta = 120^\circ - \alpha$ . The quadrilateral $ABED$ is cyclic, so $\angle ABE = 180^\circ - \alpha$ . Then $\angle CBE = 120^\circ - \alpha$ so $\angle CBE = \beta$ . Thus $\triangle CBE$ is isosceles, so $AE$ is the perpendicular bisector of $BC$ , so it bisects $\angle BAC$ . Now the arc $BE$ is intercepted by a $30^\circ$ inscribed angle, so it measures $60^\circ$ . Then $BE$ equals the radius of $k$ , namely $1$ . Hence $CE = BE = 1$ .

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