Exercise 10. Let $A B C$ be an isosceles triangle at $A$. The bisector of angle $\widehat{A B C}$ intersects side $[A C]$ at $D$. Suppose that $B D = D A$. Determine the angles of triangle $A B C$.
## Solution to Exercise 10

![](https://cdn.mathpix.com/cropped/2024_06_04_8ce8798f2c48d3d66a92g-17.jpg?height=1028&width=717&top_left_y=377&top_left_x=675)

Let $\alpha = \widehat{ABD}$. Let's examine the different angles in the figure and try to express them in terms of $\alpha$. Since the line $(BD)$ is the bisector of the angle $\widehat{CBA}$, we know that $\widehat{ABD} = \widehat{DBC}$, and therefore $\widehat{ABC} = 2 \widehat{ABD} = 2\alpha$. Since $DB = DA$, triangle $ADB$ is isosceles at $D$, which means $\widehat{BAD} = \widehat{ABD}$, so $\widehat{BAC} = \widehat{DAB} = \alpha$. Finally, triangle $ABC$ is isosceles at $A$, which means $\widehat{ACB} = \widehat{ABC} = 2 \widehat{ABD} = 2\alpha$. Since the sum of the angles in triangle $ABC$ is $180^\circ$, we find

$$
180^\circ = \widehat{ABC} + \widehat{ACB} + \widehat{BAC} = 2\alpha + 2\alpha + \alpha = 5\alpha
$$

Thus, $\alpha = \frac{180^\circ}{5} = 36^\circ$. This means that $\widehat{ABC} = \widehat{ACB} = 72^\circ$ and $\widehat{BAC} = 36^\circ$.

Grader's 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 and altitude, bisector and 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 some errors by drawing a neat figure.