The median divides the triangle into two equal-area triangles.

## Solution

Since

$$
S_{\Delta \mathrm{A}_{1} \mathrm{BB}_{1}}=S_{\Delta \mathrm{A}_{1} \mathrm{AB}}=S_{\Delta \mathrm{ABC}}
$$

then $S_{\Delta \mathrm{AA}_{1} \mathrm{~B}_{1}}=2 S$. Similarly,

$$
S_{\Delta \mathrm{BB}_{1} \mathrm{C}_{1}}=S_{\Delta \mathrm{CC}_{1} \mathrm{~A}_{1}}=2 S .
$$

Therefore,

$$
S_{\Delta \mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}}=7 S
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-02.jpg?height=408&width=608&top_left_y=1814&top_left_x=725)

Answer

$7 S$.
8,9 On the extensions of the sides of triangle $A B C$, points $A_{1}, B_{1}$, and $C_{1}$ are taken such that $\overrightarrow{A B_{1}}=2 \overrightarrow{A B}, \overrightarrow{B C_{1}}=2 \overrightarrow{B C}$, and $\overrightarrow{C A_{1}}=2 \overrightarrow{A C}$. Find the area of triangle $A_{1} B_{1} C_{1}$, given that the area of triangle $A B C$ is $S$.