jmc

We have $\angle CBD=90^{\circ }-\angle A=60^{\circ }$, so $△BDC$ and $△CDA$ are similar 30-60-90 triangles. Side $\overline{CD}$ of $△BCD$ corresponds to $\overline{AD}$ of $△CAD$ (each is opposite the $60^{\circ }$ angle), so the ratio of corresponding sides in these triangles is $\frac{CD}{AD}$. From 30-60-90 triangle $ACD$, this ratio equals $\frac{1}{\sqrt{3}}$. The ratio of the areas of these triangles equals the square of the ratio of the corresponding sides, or $${\left(\frac{1}{\sqrt{3}}\right)}^2=\boxed{\frac{1}{3}}.$$