[ Properties of medians. The centroid of a triangle.] [ The ratio of areas of similar triangles ]

In triangle $F G H$, angle $G$ is right, $F G=8, G H=2$. Point $D$ lies on side $F H$, $A$ and $B$ are the points of intersection of the medians of triangles $F G D$ and $D G H$. Find the area of triangle $G A B$.
Let $G C$ and $G E$ be the medians of triangles $G D F$ and $G D H$ respectively. Then triangle $G A B$ is similar to triangle $G C E$ with a similarity coefficient of $\frac{2}{3}$.

## Solution

Let $G C$ and $G E$ be the medians of triangles $G D F$ and $G D H$ respectively. Since the medians of a triangle are divided by the point of intersection in the ratio 2:1, counting from the vertex, we have $\frac{G A}{G C}=\frac{G B}{G E}=\frac{2}{3}$. Therefore, triangle $G A B$ is similar to triangle $G C E$ with a similarity coefficient $k=\frac{2}{3}$. Consequently,

$$
\begin{gathered}
S_{\Delta \mathrm{GAB}}=k^{2} \cdot S_{\Delta \mathrm{GCE}}=\frac{4}{9}\left(S_{\Delta \mathrm{CGD}}+S_{\Delta \mathrm{EGD}}\right)=\frac{4}{9}\left(\frac{1}{2} \cdot S_{\Delta \mathrm{FGD}}+\frac{1}{2} \cdot S_{\Delta \mathrm{DGH}}\right)= \\
=\frac{4}{9} \cdot \frac{1}{2}\left(S_{\Delta \mathrm{FGD}}+S_{\Delta \mathrm{DGH}}\right)=\frac{2}{9} \cdot S_{\Delta \mathrm{FGH}}=\frac{2}{9} \cdot \frac{1}{2} \cdot G H \cdot G F=\frac{1}{9} \cdot 2 \cdot 8=\frac{16}{9}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-38.jpg?height=232&width=715&top_left_y=1519&top_left_x=677)

## Answer

$\frac{16}{9}$