(solved by Marc Coiffier). Let $A, B, C, D, E, F$ and $G$ be points in the plane such that $A B=$ $B C=C D=D E=E F=F G=G A$ and $A, B, F, D$ on the one hand and $A, G, C, E$ on the other hand are collinear. Calculate the angle $\widehat{E A D}$.
. The figure is entirely determined by the statement and in particular, it is symmetric with respect to the perpendicular bisector of $[D E]$. Let $2 \alpha$ be the angle we are looking for. Then $\widehat{A E D}=$ $\widehat{A D E}=\frac{\pi}{2}-\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_10_8d7011708e480f09f0d6g-104.jpg?height=678&width=363&top_left_y=2039&top_left_x=848)

The triangles $D A E$ and $C D E$ are isosceles with vertices at $A$ and $D$ respectively. Since they share the angle $\widehat{A E D}$, they are similar. It follows that $\widehat{C D E}=2 \alpha$. Moreover, $A G F$ and $B C D$ are also isosceles triangles with vertices at $G$ and $C$ respectively. It follows that $\overline{A C B}=2 \alpha$ and:

$$
\widehat{B C D}=\pi-2 \widehat{A D C}=\pi-2\left(\frac{\pi}{2}-\alpha-2 \alpha\right)=6 \alpha
$$

Since the points $A, C$ and $E$ are collinear, we have the relation $\widehat{A C B}+\widehat{B C D}+\widehat{D C E}=\pi$, that is $2 \alpha+6 \alpha+$ $\frac{\pi}{2}-\alpha=\pi$, hence $\alpha=\frac{\pi}{14}$ and the angle we are looking for is $2 \alpha=\frac{\pi}{7}$.