4. On the circumscribed circle of square $A B C D$, a point $M$ is marked such that $\angle M A B=18^{\circ}$. Segment $M D$ intersects diagonal $A C$ at point $Q$.

What values can the measure of angle $\angle A Q D$ take? Express your answer in degrees. List all possible answer options.
Answer: 72 or 108.

Solution. The first thing to remember in this problem is that for point $M$, there are two possible positions: in the same half-plane relative to line $AB$ as the square (point $M_{1}$ in the diagram), and in the other half-plane (point $M_{2}$ in the diagram).

![](https://cdn.mathpix.com/cropped/2024_05_06_b124656272d750e4349eg-2.jpg?height=731&width=768&top_left_y=1077&top_left_x=658)

By the exterior angle theorem for triangle $C Q_{1} D, \angle A Q_{1} D = \angle A C D + \angle M_{1} D C = 45^{\circ} + \angle M_{1} D C$. Note that $\angle M_{1} D C = \angle B D C - \angle B D M_{1} = 45^{\circ} - \angle B A M_{1} = 27^{\circ}$: angles $\angle B D M_{1}$ and $\angle B A M_{1}$ are equal, as they subtend the same arc $B M_{1}$. We get $\angle A Q_{1} D = 45^{\circ} + 27^{\circ} = 72^{\circ}$.

Similarly, $\angle A Q_{2} D = \angle A C D + \angle M_{2} D C = 45^{\circ} + (\angle M_{2} D B + \angle B D C) = 90^{\circ} + \angle M_{2} A B = 108^{\circ}$.