Exercise 2. 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 $\widehat{A B C}$?
## Solution to Exercise 2

![](https://cdn.mathpix.com/cropped/2024_06_04_0d4f01d8013de0f79932g-04.jpg?height=465&width=1265&top_left_y=567&top_left_x=407)

Since \(\widehat{C D B}=40^{\circ}\), by supplementary angles, we get \(\widehat{C D A}=180-\widehat{C D B}=140^{\circ}\). 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 \(n^{\circ} 1\): 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 a right triangle at \(C\). The angles \(\widehat{C A B}\) and \(\widehat{A B C}\) are thus complementary: \(\widehat{A B C}=90-\widehat{B A C}=70^{\circ}\).

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