9-4-1. In the figure, $O$ is the center of the circle, $A B \| C D$. Find the degree measure of the angle marked with a «?».

![](https://cdn.mathpix.com/cropped/2024_05_06_ec2625783a617a5ecc6bg-3.jpg?height=540&width=640&top_left_y=1649&top_left_x=708)
Answer: $54^{\circ}$.

Solution variant 1. Quadrilateral $A D C B$ is an inscribed trapezoid in a circle. As is known, such a trapezoid is isosceles, and in an isosceles trapezoid, the angles at the base are equal: $\angle B A D=\angle C B A=63^{\circ}$. Triangle $D O A$ is isosceles ($O A$ and $O D$ are equal as radii), so its base angles $D A O$ and $A D O$ are equal. Therefore,

$$
\angle D O A=180^{\circ}-\angle D A O-\angle A D O=180^{\circ}-63^{\circ}-63^{\circ}=54^{\circ}
$$