I know that OECD is a cyclic quadrilateral, but I'm not sure how to work from there.
Thanks for helping, BG
$\displaystyle \angle AEO= \angle OAE$ as $\displaystyle \triangle AOE$ is equilateral
This implies that $\displaystyle \angle DEB = \angle EBD$ since $\displaystyle \angle CAB+\angle ABC= 90^{\circ}$ and $\displaystyle \angle AEO+ \angle DEB=90^{\circ}$.