5. The two sides enclosing a vertex of a triangle are $\mathrm{a}$ and $\mathrm{b}(\mathrm{a}<\mathrm{b})$, and the angle at this vertex is trisected by straight lines. The ratio of the segments of these lines inside the triangle is $\mathrm{m}: \mathrm{n}(\mathrm{m}<\mathrm{n})$. Find the lengths of these segments.
5. $\angle \mathrm{ACD}=$
$$
\begin{array}{l}
\angle \mathrm{DCE}=\angle \mathrm{ECB}=\alpha, \text { given } \\
\mathrm{HCE}=\mathrm{x}, \mathrm{CD}=\mathrm{v} \text { (Figure 5). }
\end{array}
$$
The area of $\triangle \mathrm{ABC}$ can be expressed in three different ways:
$$
\begin{array}{l}
S_{A C D}+S_{D C D}=\frac{1}{2} b y \sin \alpha+\frac{1}{2} \text { aysin } 2 \alpha, \\
S_{A C B}+S_{B C D}=\frac{1}{2} b x \sin 2 \alpha+\frac{1}{2} a x \sin \alpha, \\
S_{\triangle C D}+S_{D C Z}+S_{\triangle C D} \\
=\frac{1}{2} \text { bysin } \alpha+\frac{1}{2} x y \sin \alpha+\frac{1}{2} a x \sin \alpha .
\end{array}
$$

By comparing the first and second expressions, we get: $2 a \cos \alpha=x+a \frac{x}{y}$, and by comparing the second and third expressions, we get: $2 b \cos \alpha=y+b_{x}$
Given that: $\frac{x}{y}=\frac{m}{n}$.
Solving it, we obtain:
$$
x=\frac{\left(n^{2}-m^{2}\right) a b}{n(b m-a n)}, y=\frac{\left(n^{2}-m^{2}\right) a b}{m(b m-a n)} .
$$