$\begin{aligned} & {\left[\begin{array}{l}\text { Angles between bisectors } \\ \text { [Inscribed angle is half of the central angle] }\end{array}\right]}\end{aligned}$

In a circle with center at point $O$, a triangle $E G F$ is inscribed, where angle $\angle E F G$ is obtuse. Outside the circle, there is a point $L$ such that $\angle L E F = \angle F E G, \angle L G F = \angle F G E$. Find the radius of the circumcircle of triangle $E L G$, if the area of triangle $E G O$ is $81 \sqrt{3}$ and $\angle O E G = 60^{\circ}$.
Prove that $\angle E F G=90^{\circ}+\frac{1}{2} \angle E L G$.

## Answer

$6 \sqrt{3}$.