Let $AB$ be a diameter of a circle and $C$ a point on $AB$ with $3\cdot AC=BC$. Let $D,E$ be points on the circle with $DC\perp AB$ and $DE$ a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?

1. As in the problem, let $O$ be the center. Since $O$ is the midpoint of $DE$, $[DCE]=2[DCO]$, and $\triangle DCO$ and $\triangle DAB$ have the same height $DC$, so $\frac{[DCE]}{[ABD]}=\frac{2\,CO}{AB}$.
2. With $AB=1$ and $BC=k\cdot AC$, $AC=\frac1{k+1}$ and $CO=\left|\frac12-\frac1{k+1}\right|$, so the ratio is $2\left|\frac1{k+1}-\frac12\right|$.
3. With $k=3$: $2\left|\frac14-\frac12\right|=2\cdot\frac14=\frac12$.
4. The answer is $\boxed{\dfrac12}$.