\left.\begin{array}{l}{[\text { Auxiliary area. The area helps to solve the task] }} \\ {[\quad \text { The ratio of the areas of triangles with a common angle }}\end{array}\right]

On the lateral sides $A B$ and $B C$ of the isosceles triangle $A B C$, points $M$ and $N$ are located such that $\frac{A M}{B M}=m, \frac{C N}{B N}=n$. The line $M N$ intersects the height $B D$ of the triangle at point $O$. Find the ratio $\frac{D O}{B O}$.
Let the area of triangle $ABC$ be $S$. Then the areas of triangles $ABD$ and $CBD$ are each $\frac{1}{2} S$. Let $\frac{BO}{BD}=x$. Then

$$
S_{\triangle BMN}=\frac{BM}{AB} \cdot \frac{BN}{BC} S=\frac{1}{m+1} \cdot \frac{1}{n+1} \cdot S
$$

On the other hand,

$$
S_{\triangle BMN}=S_{\triangle BMO}+S_{\triangle BNO}=\frac{1}{m+1} \cdot x \cdot \frac{1}{2} S + \frac{1}{n+1} \cdot x \cdot \frac{1}{2} S
$$

From the equation

![](https://cdn.mathpix.com/cropped/2024_05_06_d69082500b56170eb8ddg-29.jpg?height=91&width=637&top_left_y=1049&top_left_x=716)

we find that $x=\frac{2}{m+n+2}$. Therefore,

$$
\frac{DO}{BO}=\frac{BD-BO}{BO}=\frac{BD}{BO}-1=\frac{1}{x}-1=\frac{m+n+2}{2}-1=\frac{1}{2}(m+n)
$$

## Answer

$\frac{1}{2}(m+n)$