Answer: 65

Small Hint:

Let $E$ be the midpoint of $\overline{BC},$ so $A,$ $E,$ and $D$ are collinear

Big Hint:

Find $AE$ with the median formula, then compare two right triangles sharing $BD$

Solution:

Let $E$ be the midpoint of $\overline{BC}.$ Then $A,$ $E,$ and $D$ are collinear. The median formula gives $$\begin{aligned}AE^2&=\frac{2AB^2+2AC^2-BC^2}{4}\\&=\frac{57}{4}.\end{aligned}$$ Since $AB^2>AE^2+BE^2,$ angle $AEB$ is obtuse, so the perpendicular foot $D$ lies beyond $E.$ Both $\triangle ABD$ and $\triangle EBD$ are right at $D.$ Therefore $$\begin{gathered}AB^2-BE^2\\=(AE+DE)^2-DE^2\\=AE^2+2AE\cdot DE.\end{gathered}$$ Substitution gives $30-\frac{15}{4}=\frac{57}{4}+2AE\cdot DE,$ so $\frac{DE}{AE}=\frac{8}{19}.$

Triangles $ABE$ and $DBE$ have bases $AE$ and $DE$ on the same line and share the altitude from $B,$ while $[ABC]=2[ABE].$ Hence $$\begin{aligned}\frac{[ADB]}{[ABC]}&=\frac{[ABE]+[DBE]}{2[ABE]}\\&=\frac12\left(1+\frac8{19}\right)\\&=\frac{27}{38}.\end{aligned}$$ Thus $m+n=65.$