The area of $\triangle EBD$ is one fifth of the area of $5$-$12$-$13$ $\triangle ABC$ (with $BC=12$). Segment $DE$ is perpendicular to segment $AB$. What is $BD$?

Since $\angle BDE=\angle BCA=90^\circ$ and $\angle B$ is common, $\triangle BDE\sim \triangle BCA$, with $BD$ corresponding to $BC=12$. The ratio of areas is the square of the ratio of similarity, so $(BD:BC)^2=[EBD]:[ABC]=\frac15$ and $BD:BC=1:\sqrt5$.
So $BD=\frac{12}{\sqrt5}=\boxed{\frac{12\sqrt5}5}$.