jmc

From 30-60-90 right triangle $ACD$ with hypotenuse $\overline{AC}$ and shorter leg $\overline{CD}$, we have $AC=2CD=2\sqrt{3}$.

From 30-60-90 triangle $ABC$ with shorter leg $\overline{BC}$ and longer leg $\overline{AC}$, we have $AC=BC\sqrt{3}$. Since $AC=2\sqrt{3}$, we have $BC=2$. Therefore, the area of $△ABC$ is $$\frac{(AC)(BC)}{2}=\frac{(2\sqrt{3})(2)}{2}=\boxed{2\sqrt{3}}.$$