5. Each of the diagonals of the inscribed quadrilateral $A B C D$ is the bisector of one of the angles from which it is drawn, and also divides the second angle in the ratio $1: 2$. What can the angle $\angle A$ of this quadrilateral be? Express the answer in degrees. List all possible options. If necessary, round the answer to the hundredths or write it as a common fraction.
Answer: $72, 108, \frac{720}{7}, \frac{540}{7}$.

Solution. Let in the inscribed quadrilateral $P Q R S$, the diagonals are angle bisectors of angles $P$ and $Q$, and also divide angles $R$ and $S$ in the ratio $1: 2$.

Recall that an inscribed angle is equal to half the arc it subtends. Since $P R$ and $Q S$ are bisectors (i.e., $\angle P Q S = \angle S Q R, \angle S P R = \angle R P Q$), the arcs $Q R, R S$, and $S P$ are equal. Since $P R$ divides angle $R$ in the ratio $1: 2$, the arc $P Q$ is either half the arc $Q R$ or twice the arc $Q R$.
![](https://cdn.mathpix.com/cropped/2024_05_06_05e9aaee929c8f4a3d1cg-2.jpg?height=550&width=1382&top_left_y=318&top_left_x=336)

- In the first case (left drawing), we get that the arc $P Q$ is $\frac{1}{7}$ of the entire circle, i.e., $\frac{1}{7} \cdot 360^{\circ}$, and the arcs $Q R, R S, S P$ are twice as large, i.e., $\frac{2}{7} \cdot 360^{\circ}$. Then the angles of the quadrilateral are $\angle P = \angle Q = \frac{4}{7} \cdot 180^{\circ}$ and $\angle R = \angle S = \frac{3}{7} \cdot 180^{\circ}$.
- In the second case (right drawing), we get that the arc $P Q$ is $\frac{2}{5}$ of the entire circle, i.e., $\frac{2}{5} \cdot 360^{\circ}$, and the arcs $Q R, R S, S P$ are half as large, i.e., $\frac{1}{5} \cdot 360^{\circ}$. Then the angles of the quadrilateral are $\angle P = \angle Q = \frac{2}{5} \cdot 180^{\circ}$ and $\angle R = \angle S = \frac{3}{5} \cdot 180^{\circ}$.

The angle $\angle A$ can be any of the found angles, so the answer is $\frac{3}{7} \cdot 180^{\circ} = \frac{540^{\circ}}{7}$, $\frac{4}{7} \cdot 180^{\circ} = \frac{720^{\circ}}{7}$, $\frac{2}{5} \cdot 180^{\circ} = 72^{\circ}$, or $\frac{3}{5} \cdot 180^{\circ} = 108^{\circ}$.