Rectangle $ABCD$ has $AB=12$ and $BC=5$. Segment $EF$ is constructed through $B$ so that $EF$ is perpendicular to $DB$, and $A$ and $C$ lie on $DE$ and $DF$, respectively. What is $EF$?

1. $BD=13$. As in the problem, $\triangle BFC\sim \triangle BDA$ and $\triangle EBA\sim\triangle BDA$ (right triangles with $\angle FBC=\angle ABD$ and $\angle EBA=\angle BDA$).
2. $BF:BD=BC:BA=5:12$, so $BF=\frac{65}{12}$.
3. $EB:BD=AB:AD=12:5$, so $EB=\frac{12\cdot 13}{5}=\frac{156}5$.
4. $EF=EB+BF=\frac{156}5+\frac{65}{12}=\frac{2197}{60}$.