6. (5 points). Point $\boldsymbol{D}$ lies on the extension of side $\boldsymbol{A} \boldsymbol{C}$ of triangle $\boldsymbol{A B C}$, the area of which is equal to $\boldsymbol{S}$; point $\boldsymbol{A}$ is located between $\boldsymbol{D}$ and $\boldsymbol{C}$. Let $\boldsymbol{O}$ be the point of intersection of the medians of triangle $\boldsymbol{A B C}$. It is known that the area of triangle $\boldsymbol{D O C}$ is $\boldsymbol{S}_{l}$. Express the area of triangle $\boldsymbol{D O B}$ in terms of $\boldsymbol{S}$ and $\boldsymbol{S}_{\boldsymbol{l}}$.
Solution: We will show that

![](https://cdn.mathpix.com/cropped/2024_05_06_f8af2824a37d1a011ad7g-5.jpg?height=411&width=780&top_left_y=660&top_left_x=247)

$$
S_{\triangle D O C} + S_{\triangle D O A} = S_{\triangle D O B}
$$

To do this (since these triangles share the side $D O$), it is sufficient to show that the lengths of the perpendiculars dropped from the vertices of the triangles to $D O$ satisfy the equation $h_{1} + h_{3} = h_{2}$.

Indeed, $h_{1}^{\prime} = (h_{1} + h_{3}) / 2$ (property of the midline of a trapezoid), and from the similarity of the gray-marked triangles, it follows that $h_{2} = 2 h_{1}^{\prime}$ (medians are divided in the ratio 2:1, counting from the vertex). This confirms the validity of $(* *)$.

$$
\text{Since } S_{\triangle D O C} = S_{\triangle D O A} + S_{\triangle A O C} \text{, and } S_{\triangle A O C} = \frac{S}{3} \text{, we get}
$$

Answer: $S_{\triangle D O B} = 2 S_{1} - \frac{S}{3}$.