ELMO 2013 Problem 4

Triangle $ABC$ is inscribed in circle $\omega$ . A circle with chord $BC$ intersects segments $AB$ and $AC$ again at $S$ and $R$ , respectively. Segments $BR$ and $CS$ meet at $L$ , and rays $LR$ and $LS$ intersect $\omega$ at $D$ and $E$ , respectively. The internal angle bisector of $\angle BDE$ meets line $ER$ at $K$ . Prove that if $BE = BR$ , then $\angle ELK = \tfrac{1}{2} \angle BCD$ .

Proposed by Evan Chen