Let $A B C$ be a triangle such that $\widehat{C A B}=20^{\circ}$. Let $D$ be the midpoint of the segment $[A B]$. Suppose that $\widehat{C D B}=40^{\circ}$. What is the value of the angle $\widehat{A B C}$?
Given that $\widehat{C D B}=40^{\circ}$, we obtain by supplementary angles that $\widehat{C D A}=180-\widehat{C D B}=140^{\circ}$. Since the sum of the angles in triangle $A D C$ is $180^{\circ}$, we get $\widehat{D C A}=180-\widehat{C A D}-\widehat{A D C}=180-\widehat{C A B}-\widehat{A D C}=180-20-140=20^{\circ}$. Thus, triangle $C D A$ is isosceles at $D$, so $D B=D A=D C$: triangle $C D B$ is also isosceles at $D$. Since the sum of the angles in triangle $C D B$ is $180^{\circ}$, we have $180^{\circ}=\widehat{D C B}+\widehat{D B C}+\widehat{C B D}=2 \widehat{D B C}+40^{\circ}$, so $\widehat{D B C}=\frac{180-40}{2}=70^{\circ}$. Thus $\widehat{A B C}=70^{\circ}$.

Alternative solution: We similarly prove that $D B=D A=D C$. Thus, the midpoint of segment $[A B]$ is the center of the circumcircle of $A B C$. Therefore, triangle $A B C$ is right-angled at $C$. The angles $\widehat{C A B}$ and $\widehat{A B C}$ are thus complementary: $\widehat{A B C}=90-\widehat{B A C}=70^{\circ}$.

Examiner's comment: The exercise is very well done. However, a number of students confuse the median and the angle bisector: just because in a triangle, a line passing through a vertex cuts the opposite side in half does not mean that the angle at the vertex is also bisected. We also regret that a significant number of students do not provide a figure to support their reasoning, which is a pity. Some students also try to invoke the converse of the inscribed angle theorem, which unfortunately is not true in the general case.