4. Given a rhombus $A B C D$. The intersection point of the angle bisectors of $\measuredangle B A C$ and $\measuredangle B D C$ lies on one of its sides. Determine the angles of the rhombus.
Solution. The intersection of the angle bisectors of $\measuredangle B A C$ and $\measuredangle B D C$ is denoted by $M$, where $M$ lies on the side $B C$. According to the theorem on the angle bisectors of the interior angles of a triangle, the following ratios hold:

$$
\begin{aligned}
& \overline{B M}: \overline{M C}=\overline{A B}: \overline{A C} \\
& \overline{B M}: \overline{M C}=\overline{B D}: \overline{C D}
\end{aligned}
$$

From this, we obtain

![](https://cdn.mathpix.com/cropped/2024_06_05_4826f81c3e0b08e4acb7g-2.jpg?height=435&width=680&top_left_y=1349&top_left_x=804)

$$
\overline{A B}: \overline{A C}=\overline{B D}: \overline{C D} \text {, i.e., } \overline{A C} \cdot \overline{B D}=\overline{A B} \cdot \overline{C D}
$$

If we denote the side of the rhombus by $a$ and the height by $h$, then the last equality takes the form: $2 a h=a^{2}$, i.e., $a=2 h$, which means that the angles of the rhombus are $60^{\circ}$ and $120^{\circ}$.

## II year