Love for geometry-4

Let the radius of the circle be $r$ . Then $AO=EO=r$ . Since $\red{BD} = \dfrac {DC}3$ and $BC = 2r$ , $\implies BD = \dfrac r2$ , $DC = \dfrac 32 r$ , and $DO=\dfrac r2$ . This means that $\triangle ABD$ and $\triangle ADO$ are similar. Therefore $AO = AB \implies r = 4$ . We note that $\triangle ADC$ and $\triangle CDE$ share a common base of $DC$ and have heights $AD$ and $EF$ respectively. Since $\triangle ADO$ and $\triangle OEF$ are similar, $AD=EF$ . Therefore $\triangle ADC$ and $\triangle CDE$ have the same area. Then

$\begin{aligned} [ADEC] & = [ADC] + [CDE] = 2[ADC] = 2 \times \frac 12 \times AD \times DC = 2 \times \frac 12 \times 2\sqrt 3 \times 6 = 12\sqrt 3 \\ \implies \frac {[ADEC]}{\sqrt 3} & = \boxed{12} \end{aligned}$