## Task B-2.5.

On the side $\overline{A B}$ of triangle $A B C$, a point $D$ is given. Let $T_{1}$ and $T_{2}$ be the centroids of triangles $C A D$ and $C D B$, respectively. Determine the length of the segment $\overline{T_{1} T_{2}}$ if $|A B|=6 \mathrm{~cm}$.
## Solution.

Let $P_{1}$ and $P_{2}$ be the midpoints of segments $\overline{A D}$ and $\overline{D B}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_30_964e2f56f6f4ff54223dg-09.jpg?height=397&width=436&top_left_y=2052&top_left_x=730)

The centroid of a triangle divides the median of the triangle in the ratio 2:1, counting from the vertex of the triangle, so we have

$\frac{\left|C T_{1}\right|}{\left|C P_{1}\right|}=\frac{\left|C T_{2}\right|}{\left|C P_{2}\right|}=\frac{2}{3}$.

1 point

According to the SAS similarity theorem, triangles $C T_{1} T_{2}$ and $C P_{1} P_{2}$ are similar with a similarity coefficient of $\frac{2}{3}$ (they share a common angle, and the sides enclosing this angle are proportional).

Thus, $\frac{\left|T_{1} T_{2}\right|}{\left|P_{1} P_{2}\right|}=\frac{2}{3} \Rightarrow\left|T_{1} T_{2}\right|=\frac{2}{3}\left|P_{1} P_{2}\right|$.

$\left|P_{1} P_{2}\right|=\left|P_{1} D\right|+\left|D P_{2}\right|=\frac{|A D|}{2}+\frac{|D B|}{2}=\frac{|A B|}{2}$. 1 point

$\left|T_{1} T_{2}\right|=\frac{2}{3}\left|P_{1} P_{2}\right|=\frac{1}{3}|A B|=2 \mathrm{~cm}$