5. The segment $AC = a$ with its internal point $B$ is divided in the ratio $3:2$. Squares $ABDE$ and $CBFG$ are constructed on segments $AB$ and $BC$, respectively, on different sides relative to segment $AC$. Let $O$ and $O'$ be the intersections of the diagonals of these squares. Determine the ratio of the area of quadrilateral $OO'C D$ to the area of the square with side length $AC$.
Solution. The quadrilateral $O O^{\prime} C D$ is composed of three right-angled triangles, where $O B D$ and $O^{\prime} B C$ are quarters of corresponding squares (diagram on the right). The required area is

$$
\begin{aligned}
P & =\frac{1}{4}\left(\frac{3}{5} a\right)^{2}+\frac{1}{4}\left(\frac{2}{5} a\right)^{2}+\frac{1}{2} \frac{3 a}{5} \cdot \frac{2 a}{5} \\
& =\frac{9 a^{2}}{100}+\frac{4 a^{2}}{100}+\frac{6 a^{2}}{50}=\frac{a^{2}}{4}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_e0d95f5510204c7c0f0fg-3.jpg?height=397&width=428&top_left_y=354&top_left_x=1036)

which is a quarter of the area of a square with side length $a$. Therefore, the required ratio is $1: 4$.

## Eighth Division