10. In $\triangle A B C$, $A B=\sqrt{30}, A C=\sqrt{6}, B C$ $=\sqrt{15}$, there is a point $D$ such that $A D$ bisects $B C$ and $\angle A D B$ is a right angle, the ratio $\frac{S_{\triangle A D B}}{S_{\triangle A B C}}$ can be written as $\frac{m}{n}$. ($m, n$ are coprime positive integers). Then $m+n=$
10.65.

Let the midpoint of $BC$ be $E$, and $AD=\frac{x}{2}$.
By the median formula, we get $AE=\frac{\sqrt{57}}{2}$.
By the Pythagorean theorem, we get $120-15+57=2 \sqrt{57} x$, solving for $x$ gives $x=\frac{81}{\sqrt{57}}$.
Thus, $\frac{m}{n}=\frac{AD}{2AE}=\frac{x}{2\sqrt{57}}=\frac{27}{38}$.
Therefore, $m+n=27+38=65$.