4. Through a random point inside a given triangle, three lines parallel to the sides of the triangle pass. These lines divide the area of the triangle into six parts, three of which are triangles with areas $s_{1}, s_{2}, s_{3}$. Determine the area of the given triangle.
Solution. Let $A, B, C$ be the vertices of the given triangle, $S$ the internal point, and the intersection points of the mentioned lines with the sides of the triangle be marked as in the diagram on the right. Clearly, triangles $P Q S, N S M$, and $S K L$ are similar to triangle $A B C$. Let $k_{1}, k_{2}, k_{3}$ be the respective similarity coefficients. We have:

$$
\frac{B K}{B C}=\frac{Q S}{B C}=k_{1}, \frac{L C}{B C}=\frac{S M}{B C}=k_{2}, \frac{K L}{B C}=k_{3} .
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_c53e313ed51c88c0c5c5g-5.jpg?height=279&width=409&top_left_y=267&top_left_x=1055)

However, $B K+L C+K L=B C$, so $k_{1}+k_{2}+k_{3}=1$. On the other hand, if $s$ denotes the area of triangle $A B C$, we have

$$
\frac{s_{1}}{s}=k_{1}^{2}, \frac{s_{2}}{s}=k_{2}^{2}, \frac{s_{3}}{s}=k_{3}^{2} \text {, }
$$

which means

$$
\sqrt{\frac{s_{1}}{s}}=k_{1}, \sqrt{\frac{s_{2}}{s}}=k_{2}, \sqrt{\frac{s_{3}}{s}}=k_{3}
$$

Therefore,

$$
\sqrt{s_{1}}+\sqrt{s_{2}}+\sqrt{s_{3}}=\sqrt{s}
$$

i.e.,

$$
s=\left(\sqrt{s_{1}}+\sqrt{s_{2}}+\sqrt{s_{3}}\right)^{2}
$$