Exercise 3. Let $A B C D$ be a square and $E$ a point on the segment $[B D]$ such that $E B=A B$. We define the point $F$ as the intersection of the lines $(C E)$ and $(A D)$. Find the value of the angle $\widehat{F E A}$.
## Solution to Exercise 3

![](https://cdn.mathpix.com/cropped/2024_04_29_bad7460d4ce66061b968g-05.jpg?height=608&width=620&top_left_y=410&top_left_x=775)

We will use the following central property multiple times: in a triangle, the sum of the angles is $180^{\circ}$.

Since the quadrilateral $A B C D$ is a square, we get that $E B=A B=C B$. Therefore, the triangles $A B E$ and $B E C$ are both isosceles at $B$. Since $\widehat{A B E}=45^{\circ}$, we obtain by the sum of the angles in triangle $A B E$ that $180^{\circ}=\widehat{A B E}+\widehat{B A E}+\widehat{A E B}=45^{\circ}+2 \cdot \widehat{A E B}$, so $\widehat{A E B}=\frac{135^{\circ}}{2}=67.5^{\circ}$. Since $\widehat{C B E}=45^{\circ}$, we obtain by the sum of the angles in triangle $C B E$ that $180^{\circ}=\widehat{C B E}+\widehat{B C E}+\widehat{C E B}=45^{\circ}+2 \cdot \widehat{C E B}$, so $\widehat{C E B}=\frac{135^{\circ}}{2}=67.5^{\circ}$.

In particular, we have $\widehat{A E B}=\widehat{C E B}=67.5^{\circ}$.

Finally, since $F, E, C$ are collinear, we get

$$
\widehat{F E A}=180^{\circ}-\widehat{A E C}=180^{\circ}-(\widehat{A E B}+\widehat{B E C})=180^{\circ}-135^{\circ}=45^{\circ}
$$