The vertices of the quadrilateral $A B C D$ lie on a circle. Each of its diagonals bisects one angle and trisects the opposite angle. Determine the measures of the angles of the quadrilateral.

![](https://cdn.mathpix.com/cropped/2024_05_01_96f63abf6bdef97495eag-33.jpg?height=547&width=568&top_left_y=1593&top_left_x=826)

Note: We say that a ray $O M$ trisects an angle $\angle A O B$ if the measure of one of the two angles it determines is one third of the measure of the angle $\angle A O B$.

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Solution

Suppose the diagonal $AC$ bisects the angle $\angle DCB$. The other diagonal bisects $\angle ABC$ or $\angle ADC$. Without loss of generality, we can assume it bisects the angle $\angle ADC$. Since $\angle ADB = \angle BDC$ and $\angle ACB = \angle ACD$, the arcs $AD, AB$, and $BC$ correspond to the same central angle value $x$. Analyzing the trisections of the angles, either $\angle DBC = 2 \cdot \angle ABD = x$ or $\angle DBC = \frac{1}{2} \cdot \angle ABD = \frac{x}{4}$. In the first case, we have

\[
\begin{aligned}
\angle DAB + \angle ABC + \angle BCD + \angle CDA & = 360^{\circ} \\
\frac{3x}{2} + \frac{3x}{2} + x + x & = 360^{\circ} \\
x & = 72^{\circ}
\end{aligned}
\]

Thus, the angles of the quadrilateral are $72^{\circ}, 72^{\circ}, 108^{\circ}$, and $108^{\circ}$. Proceeding similarly in the second case, we have $x = \frac{720^{\circ}}{7}$ and the angles of the quadrilateral are $\frac{720^{\circ}}{7}, \frac{720^{\circ}}{7}$, $\frac{540^{\circ}}{7}$, and $\frac{540^{\circ}}{7}$.