13. Points $M$ and $N$ are on sides $A B$ and $B C$ of $\triangle A B C$, respectively, and $\frac{A M}{B M}=m, \frac{C N}{B N}=n$. Line $M N$ intersects the median $B D$ of the triangle at point $O$. Find $\frac{\mathrm{DO}}{\mathrm{BO}}$.
13. For definiteness, let $m>n$. When $m<n$, the positions of vertices $\mathrm{A}$ and $\mathrm{C}$ need to be exchanged in the following discussion.

The lines $\mathrm{CE}$ and $\mathrm{DF}$ through points $\mathrm{C}$ and $\mathrm{D}$ are parallel to $\mathrm{MN}$ (Figure 15). At this time, $\frac{E M}{B M}=\frac{C N}{B N}$, and the desired ratio $\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{FM}}{\mathrm{BM}}$.

Now, take the length of $B M$ as the unit, then $A M=m, E M=n$. Since $D F$ is the median of $\triangle A C E$, we have $\mathrm{AF}=\mathrm{FE}$. Therefore,
$$
E F=\frac{1}{2} A E=\frac{1}{2}(m-n),
$$

Thus, $M F=E M+E F=\frac{1}{2}(m+n)$.
In this way, the desired ratio equals $\frac{1}{2}(m+n)$.