Triangle $ABC$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with right angle at $C$. The rays from $C$ that trisect $\angle ACB$ meet $\overline{AB}$ at $D$ and $E$. What is the ratio of the area of $\triangle CDE$ to the area of $\triangle ABC$?

1. Let $BC=1$ and $AC=\sqrt3$. The trisecting rays meet $AB$ at $D$ and $E$ with $CD=1$ and $CE=\frac{\sqrt3}{2}$.
2. Since $\angle DCE=30^\circ$, $[CDE]=\frac12\cdot1\cdot\frac{\sqrt3}{2}\cdot\sin30^\circ=\frac{\sqrt3}{8}$, while $[ABC]=\frac{\sqrt3}{2}$.
3. Therefore $\dfrac{[CDE]}{[ABC]}=\boxed{\dfrac14}$.