Exercise 2. Let $A B C D$ be a square with side length 1. Let $E$ be the point on the segment $[A C]$ such that $A E=1$. Calculate the value, in degrees, of the angle $\widehat{B E C}$.
## Solution to Exercise 2

![](https://cdn.mathpix.com/cropped/2024_06_04_17def46ea4e6119381e4g-04.jpg?height=711&width=717&top_left_y=370&top_left_x=675)

First, since segment $[A C]$ is a diagonal of the square, it is an axis of symmetry for which points $B$ and $D$ are symmetric. In particular, it bisects angle $\widehat{B A D}$ into two equal angles. We can deduce that $\widehat{B A C}=45^{\circ}$.

Since $A E=A B=1$, triangle $A B E$ is isosceles at $A$. We can then deduce that angles $\widehat{B E A}$ and $\widehat{E B A}$ are equal. Since the sum of the angles in triangle $A E B$ is $180^{\circ}$, it follows that

$$
180^{\circ}=\widehat{E A B}+\widehat{A E B}+\widehat{A B E}=45^{\circ}+2 \widehat{A E B}
$$

so that $\widehat{A E B}=\frac{180^{\circ}-45^{\circ}}{2}=67.5^{\circ}$.

We can then deduce that $\widehat{B E C}=180^{\circ}-\widehat{B E A}=112.5^{\circ}$.

Alternative Solution $n^{\circ} 1$ We propose a solution involving the intermediate calculation of $\widehat{E B C}$.

In the same way as the previous solution, we show that $\widehat{A B E}=\widehat{A E B}=67.5^{\circ}$.

Since angle $\widehat{A B C}$ is a right angle, we have $\widehat{E B C}=90^{\circ}-\widehat{A B E}=22.5^{\circ}$. Since we also have $\widehat{E C B}=45^{\circ}$ and the sum of the angles in triangle $B C$ is $180^{\circ}$, we find

$$
\widehat{B E C}=180^{\circ}-\widehat{E B C}-\widehat{E C B}=112.5^{\circ}
$$

Graders' Comment The exercise is very well done. However, it is regrettable that a significant number of students do not provide a figure to support their reasoning, or provide figures that are illegible. There are also a number of calculation errors accompanying otherwise correct reasoning, which is unfortunate. It is good to proofread carefully to avoid such mistakes.