Exercise 2. Let ABCD be a square. Suppose there exists a point E on the segment $[A D]$ and a point $F$ on the segment $[B C]$ such that $B E=E F=F D=1$. What is the area of the square?
## Solution to Exercise 2

![](https://cdn.mathpix.com/cropped/2024_06_04_3e4a8a24fa1d8fa6f702g-1.jpg?height=493&width=482&top_left_y=2022&top_left_x=819)

Let \( E' \) be the midpoint of the segment \([B F]\). We will first show that \( A E E' B \) is a rectangle. Since triangle \( B E F \) is isosceles at \( E \), \( E' \) is also the foot of the altitude from \( E \). Therefore, \((E E') \parallel (A B)\). Moreover, \((A E) \parallel (B E')\) because the opposite sides of a square are parallel. The quadrilateral \( A E E' B \) has its opposite sides parallel, so it is a parallelogram. Since \((A E) \perp (A B)\), we know that a parallelogram with a right angle is a rectangle. Therefore, \( A E E' B \) is a rectangle. In particular, since the opposite sides of a rectangle are equal, we have \( A E = B E' \).

If we denote \( F' \) as the midpoint of the segment \([E D]\), we can show in the same way that \((F F') \parallel (A B)\), and that \( E F' F E' \) and \( F' D C F \) are rectangles, and finally that \( E F' = E' F \) and \( F' D = F C \).

Now, since \( E' \) is the midpoint of \([B F]\), we have \( B E' = E' F \), and since \( F' \) is the midpoint of \([E D]\), we have \( E F' = F' D \). Therefore, \( B E = E' F = F C \). Let \( a \) be the length of the side of the square. We deduce that the widths of the rectangles \( A E E' B \), \( E F' F E \), and \( F' D C F \) are all of length \(\frac{a}{3}\).

Applying the Pythagorean theorem in the right triangle \( A E B \):

\[
1 = \mathrm{BE}^2 = A E^2 + \mathrm{AB}^2 = a^2 + \frac{a^2}{9} = \frac{10}{9} a^2
\]

Thus, \( a^2 = \frac{9}{10} \). We conclude that the area of the square is equal to \(\frac{9}{10}\).