Exercise 10. 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 10

![](https://cdn.mathpix.com/cropped/2024_04_29_bad7460d4ce66061b968g-13.jpg?height=608&width=620&top_left_y=407&top_left_x=772)

Since quadrilateral $ABCD$ is a square, we obtain that $EB = AB = CB$. Therefore, triangles $ABE$ and $BEC$ are both isosceles at $B$. Since $\widehat{ABE} = 45^{\circ}$, we get from the sum of angles in triangle $ABE$ that $180^{\circ} = \widehat{ABE} + \widehat{BAE} + \widehat{AEB} = 45^{\circ} + 2 \cdot \widehat{AEB}$, so $\widehat{AEB} = \frac{135^{\circ}}{2} = 67.5^{\circ}$. Since $\widehat{CBE} = 45^{\circ}$, we get from the sum of angles in triangle $CBE$ that $180^{\circ} = \widehat{CBE} + \widehat{BCE} + \widehat{CEB} = 45^{\circ} + 2 \cdot \widehat{CEB}$, so $\widehat{CEB} = \frac{135^{\circ}}{2} = 67.5^{\circ}$.

In particular, we have $\widehat{AEB} = \widehat{CEB} = 67.5^{\circ}$.

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

$$
\widehat{FEA} = 180^{\circ} - \widehat{AEC} = 180^{\circ} - (\widehat{AEB} + \widehat{BEC}) = 180^{\circ} - 135^{\circ} = 45^{\circ}
$$