Solution

Let $l$ be the length of the sides of the squares. By the Pythagorean Theorem, $A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{2} l$. Let $H$ be the intersection of $A D$ and $E F$. Since $C E=l$, it follows that $A E=\sqrt{2} l-l=(\sqrt{2}-1) l$. From $\angle D A C=45^{\circ}$ and $\angle A E H=90^{\circ}$, we can conclude that $A E H$ is an isosceles right triangle, and thus $H E=A E=(\sqrt{2}-1) l$ and $A H=\sqrt{2} \cdot A E=(2-\sqrt{2}) l$. Therefore, $D H=l-(2-\sqrt{2}) l=(\sqrt{2}-1) l$ and the areas of triangles $H E C$ and $H D C$ are:

$$
\begin{aligned}
{[C E H] } & =\frac{E H \cdot C E}{2} \\
& =\frac{(\sqrt{2}-1) l^{2}}{2} \\
{[C D H] } & =\frac{D H \cdot D C}{2} \\
& =\frac{(\sqrt{2}-1) l^{2}}{2}
\end{aligned}
$$

Thus, the shaded area measures $[A B C D]-[C E H]-[C D H]=l^{2}-(\sqrt{2}-1) l^{2}=(2-\sqrt{2}) l^{2}$. Finally, the desired quotient is

$$
\frac{[A B C E H]}{[A B C D]}=\frac{(2-\sqrt{2}) l^{2}}{l^{2}}=2-\sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-05.jpg?height=582&width=497&top_left_y=248&top_left_x=843)
In the following figure, the squares $A B C D$ and $C E F G$ have the same side length. Determine the ratio between the shaded area and the area of square $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-04.jpg?height=577&width=497&top_left_y=502&top_left_x=632)

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