3. Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$, and a circle with center at point $O$ encompasses circles $\omega_{1}$ and $\omega_{2}$, touching them at points $C$ and $D$ respectively. It turns out that points $A, C$, and $D$ lie on the same line. Find the angle $A B O$.
Answer: $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c43b1f7fb831d10d42ecg-13.jpg?height=437&width=717&top_left_y=1992&top_left_x=721)

Solution. Let $O_{1}$ and $O_{2}$ be the centers of circles $\omega_{1}$ and $\omega_{2}$. Triangles $C O D$, $C O_{1} A$, and $A O_{2} D$ are isosceles. Since points $C, A, D$ lie on the same line, we get

$$
\angle O_{1} A C = \angle O_{1} C A = \angle O C D = \angle O D C
$$

Therefore, $O_{1} A \| O O_{2}$ and, similarly, $O_{2} A \| O O_{1}$. Thus, $O O_{1} A O_{2}$ is a parallelogram, from which we have

$$
O_{1} A = O_{2} O \quad \text{and} \quad \angle A O_{1} O_{2} = \angle O_{1} O_{2} O
$$

Since triangles $A O_{1} B$ and $A O_{2} B$ are isosceles, segment $O_{1} O_{2}$ is the perpendicular bisector of $A B$. Then

$$
O_{1} B = O_{1} A = O_{2} O \quad \text{and} \quad \angle B O_{1} O_{2} = \angle A O_{1} O_{2} = \angle O_{1} O_{2} O
$$

Therefore, $O_{1} O_{2} O B$ is an isosceles trapezoid, from which $B O \| O_{1} O_{2}$ and, hence, $B O \perp A B$.