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 triangle $ABC$.
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 point $D$, 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 point $A$, 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 relied on the fact that the base angles of an isosceles triangle are equal. A vast majority of students solved the exercise completely. Nevertheless, a significant number made errors such as confusing bisector with altitude, bisector with median, or getting the vertex of the bisector wrong. It is worth noting that a large number of students did not submit a figure with the exercise. We would like to emphasize the importance of the figure in a geometry problem; it is the main tool for working! Some students might have avoided certain errors by drawing a neat figure.