jmc

Since $AC=AD+DC=13$ and $DC=2$, $AD=11$. We also have two pairs of similar triangles: $△BCD\sim △ACB$ and $△ABD\sim △ACB$. From the first similarity, we have the equation $$\frac{DC}{BC}=\frac{BD}{AB}$$ Rearranging this equation and plugging in the known value of $DC$ gives $$\frac{AB}{BC}=\frac{BD}{2}$$ From the second similarity, we have the equation $$\frac{AD}{AB}=\frac{BD}{BC}$$ Rearranging this equation and plugging in the known value of $AD$ gives $$\frac{AB}{BC}=\frac{11}{BD}$$ But we know that $\frac{AB}{BC}=\frac{BD}{2}$ from above, so we have the equation $\frac{11}{BD}=\frac{BD}{2}$, or $BD=\boxed{\sqrt{22}}$.