imc

WLOG, Let us assume that the diameter is of length $1$. The length of $AC$ is $\frac{1}{3}$ and $CO$ is $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$. $OD$ is the radius of the circle, which is $\frac{1}{2}$, so using the Pythagorean Theorem the height $CD$ of $△DCO$ is $\sqrt{{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{6}\right)}^2}=\frac{\sqrt{2}}{3}$. This is also the height of the $△ABD$. The area of $△DCO$ is $\frac{1}{2}\cdot \frac{1}{6}\cdot \frac{\sqrt{2}}{3}$ = $\frac{\sqrt{2}}{36}$. The height of $△DCE$ can be found using the area of $△DCO$ and $DO$ as base. Hence, the height of $△DCE$ is $\frac{\frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot \frac{1}{2}}$ = $\frac{\sqrt{2}}{9}$. The diameter is the base for both the triangles $△DCE$ and $△ABD$, Hence, the ratio of the area of $△DCE$ to the area of $△ABD$ is $\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}$ = $\boxed{\frac{1}{3}}$