In triangle $ABC$, we connect vertex $B$ to the point that divides side $AC$ into a ratio of $1:2$ closer to $A$. Let the intersection of this segment and the median from vertex $A$ be denoted by $P$, and the centroid of the triangle by $S$. In what ratio does $P$ divide segment $AS$?
We denote the trisection point by $H$, and the midpoint of $AB$ by $F_{C}$. Since $AH$ trisects $CA$, $S$ trisects the median - specifically, such that the $\frac{1}{3}$ part falls on the side of the midpoint - according to the converse of the intercept theorem,

$$
H S \| A B \quad \text{and} \quad \frac{A F_{C}}{H S}=\frac{C A}{C H}=\frac{3}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_9f6cee8b2f63db97e4a2g-1.jpg?height=393&width=637&top_left_y=443&top_left_x=733)

The triangles $ABP$ and $SHP$ are similar due to the equality of angles, and thus

$$
\frac{A P}{P S}=\frac{A B}{H S}=\frac{2 A F_{C}}{H S}=2 \cdot \frac{3}{2}=3
$$

Therefore, $P$ is the point that divides the segment $AS$ into a ratio of $3:1$, closer to $S$.