A geometry problem by ritwik jain

Let the circumradius be 1, the extension of $DE$ meets $BC$ at $F$ and $\angle ABC = \theta$ .

Then $\angle BAO = \dfrac 12 \angle BAC = \dfrac {180^\circ-2\theta}2 = 90^\circ-\theta$ and $\angle OCB = \angle OBC = \theta - (90^\circ - \theta) = 2\theta - 90^\circ$ .

Since $AC||DF$ , $\implies \triangle EBF \equiv \triangle ABC$ , so that $\angle EFB = \angle EBF = \theta$ ; $\angle DFC = 180^\circ - \theta$ and $\angle CDF = \angle EFB - \angle OCB = \theta - (2\theta - 90^\circ) = 90^\circ - \theta$ .

Since $AC || DF$ , $\implies CF:FB = AE:EB = 2:1$ , $\implies CF = \dfrac 23 BC = \dfrac {2\cdot 2\cos (2\theta - 90^\circ)}3 = \dfrac {4\cos (90^\circ - 2\theta)}3 = \dfrac {4\sin 2\theta}3$

Using sine rule , we have:

$\begin{aligned} \frac {\sin \angle DFC}{CD} & = \frac {\sin \angle CDF}{CF} \\ \frac {\sin (180^\circ - \theta)}2 & = \frac {\sin (90^\circ - \theta) }{\frac 43 \sin 2\theta} \\ \sin \theta & = \frac {3\cos \theta}{4\sin \theta \cos \theta} \\ \sin^2 \theta & = \frac 34 \\ \sin \theta & = \frac {\sqrt 3}2 \\ \implies \theta & = \boxed{60}^\circ \end{aligned}$