jmc

Since $AE$ and $AF$ are tangents from the same point to the same circle, $AE=AF$. Let $x=AE=AF$. Similarly, let $y=BD=BF$ and $z=CD=CE$.



Then $x+y=AB=13$, $x+z=AC=15$, and $y+z=BC=14$. Adding all these equations, we get $2x+2y+2z=42$, so $x+y+z=21$. Subtracting the equation $x+z=15$, we get $y=6$.

By Heron's formula, the area of triangle $ABC$ is $$K=\sqrt{21(21-14)(21-15)(21-13)}=84,$$so the inradius is $r=K/s=84/21=4$.

Hence, by Pythagoras on right triangle $BDI$, $$BI=\sqrt{B{D}^2+D{I}^2}=\sqrt{y^2+r^2}=\sqrt{6^2+4^2}=\sqrt{52}=\boxed{2\sqrt{13}}.$$