Let $A B C$ be a triangle such that the median, the altitude, and the angle bisector from $A$ divide the angle $\widehat{A}$ into four equal angles $\alpha$. Express all the angles of the figure in terms of $\alpha$ and calculate $\alpha$.

Express all the angles of the figure in terms of $\alpha$ and calculate $\alpha$.
Let's draw the diagram:

![](https://cdn.mathpix.com/cropped/2024_05_10_60ac669be8d11ef61a82g-27.jpg?height=420&width=1043&top_left_y=180&top_left_x=560)

It is easy to calculate all the angles in terms of $\alpha$: $\widehat{B C A} = \widehat{H D A} = 90^{\circ} - \alpha$, $\widehat{D M A} = 90^{\circ} - 2 \alpha$, and $\widehat{C B A} = 90^{\circ} - 3 \alpha$. Now let's focus on calculating $\alpha$. Let $O$ be the center of the circumcircle of $A B C$. Thanks to the previous exercise, we know that $\widehat{B A O} = \widehat{C A H} = \alpha$. This means that $O$ is on the median $(A M)$. Since $O$ is also on the perpendicular bisector of $[B C]$, this implies that $O = M$. The radius of the circumcircle lies on the segment $[B C]$, so the triangle is right-angled at $A$ and $\alpha = \frac{90^{\circ}}{4} = 22.5^{\circ}$.