Two similar equilateral triangles in a circle

An elegant approach, Vijay. My approach was a slight variation on Niranjan's ...

Let $O$ be the center of the circle, and focus on $\Delta ODE.$ Now $OD$ is a radius of the circle, which is $\frac{2}{3}$ the altitude of $\Delta ABC,$ i.e., $|OD| = \frac{2}{3}(8\sqrt{3}) = \frac{16}{3}\sqrt{3}.$

We also have that $|OE| = \frac{1}{2}|OD| = \frac{8}{3}\sqrt{3}$ and $\angle OED = 150^{\circ}.$

Then with $|DE| = x,$ by the Cosine rule we see that

$(\frac{16}{3}\sqrt{3})^{2} = x^{2} + (\frac{8}{3}\sqrt{3})^{2} - 2x(\frac{8}{3}\sqrt{3})\cos(150^{\circ}) \Longrightarrow x^{2} + 8x - 64 = 0$

$\Longrightarrow x = \dfrac{-8 \pm \sqrt{64 + 4*64}}{2} = -4 \pm 4\sqrt{5} \Longrightarrow x = 4(\sqrt{5} - 1)$ since $x \gt 0.$

Nice approach. Up voted.The common approach would be as under.
Let O be the center of the circumcircle. R its radius. S=ED. M midpoint of DF. $R^2=(OE+EM)^2+DM^2,~~\\\therefore ~\left (\dfrac{16}{\sqrt3} \right )^2=\left \{ \dfrac{8}{\sqrt3}+~\dfrac{\sqrt3}{2}*S \right \}^2+\left (\dfrac S 2 \right )^2\\ \therefore ~\dfrac{256}{3} = \dfrac{64}{3}+~\dfrac{3}{4}*S^2+2* \dfrac{8}{\sqrt3}*\dfrac{\sqrt3}{2}*S + \dfrac {S^2}{4} \\\implies~S^2+8*S-64=0 \\Solving ~ for~S>0, ~~S= \color{#D61F06}{ 4.944} \\\text{Missed it due to calculation error. }$