In triangle $ABC$, take point $P$ on the median $CC_{1}$ such that $\frac{CP}{PC_{1}}=\frac{m}{n}(m, n \geq 1$ arbitrary integers). In what ratio does $P$ divide the segments of lines $AP$ and $BP$ that lie within the triangle?
Reflect the triangle over the point $C_{1}$. From the properties of central reflection, it follows that the resulting $A C B C^{\prime}$ quadrilateral is a parallelogram. The condition that $\frac{C P}{P C_{1}}=\frac{m}{n}$ means that $C P=m \cdot x, P C_{1}=n \cdot x$ can be written in the form $(x>0)$. Then the length of $C_{1} C$ will be $(m+n) x$. The line $A P$ intersects the side $C B$ at $A_{1}$, and the line $B P$ intersects the side $A C$ at $B_{1}$. The triangles $C P A_{1}$ and $C^{\prime} P A$ are similar (their angles are equal). Therefore, the ratio of their corresponding sides is the same. That is,

$$
\frac{P A}{P A_{1}}=\frac{C^{\prime} P}{P C}=\frac{n x+(n+m) x}{m x}=\frac{m+2 n}{m}
$$

this is the desired ratio.

Similarly, from the similarity of the triangles $B_{1} P C$ and $B P C^{\prime}$, we can read that

$$
\frac{P B}{P B_{1}}=\frac{m+2 n}{m}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_f4da5e66dabd234d396dg-1.jpg?height=469&width=486&top_left_y=709&top_left_x=814)