Let $ABC$ be an isosceles triangle at $A$. The bisector of angle $\widehat{ABC}$ intersects the side $[AC]$ at $D$. Suppose that $BD = DA$. Determine the angles of the triangle.
Let $\alpha=\widehat{A B D}$. Let's examine the different angles in the figure and try to express them in terms of $\alpha$. Since the line $(B D)$ is the bisector of the angle $\widehat{C B A}$, we know that $\widehat{A B D}=\widehat{D B C}$ and thus $\widehat{A B C}=2 \widehat{A B D}=2 \alpha$. Since $D B=D A$, triangle $A D B$ is isosceles at $D$, which means $\widehat{B A D}=\widehat{A B D}$, so $\widehat{B A C}=\widehat{D A B}=\alpha$. Finally, triangle $A B C$ is isosceles at $A$, which means $\widehat{A C B}=\widehat{A B C}=2 \widehat{A B D}=2 \alpha$. Since the sum of the angles in triangle $A B C$ is $180^{\circ}$, we find

$$
180^{\circ}=\widehat{A B C}+\widehat{A C B}+\widehat{B A C}=2 \alpha+2 \alpha+\alpha=5 \alpha
$$

Thus, $\alpha=\frac{180^{\circ}}{5}=36^{\circ}$. This means that $\widehat{A B C}=\widehat{A C B}=72^{\circ}$ and $\widehat{B A C}=36^{\circ}$.

Comment: The exercise was generally well solved, with particularly effective and meticulous proofs and writing. Be careful not to confuse the different notable lines in the triangle. It is important to remember that the bisector of an angle is the line that divides this angle into two adjacent and equal angles. It should not be confused with the altitude. The altitude from vertex $A$ is the line perpendicular to the line $(B C)$ passing through point $A$. The median is the line that connects the midpoint of a side to the opposite vertex. In an isosceles triangle at $A$, only the bisector of $\widehat{B A C}$ cuts the opposite side perpendicularly at its midpoint. These confusions were the source of some errors.