Solution. Let $O$ be the center of $\omega$, $M$ be the intersection point of lines $A B$ and $C D$. Note that

$$
\angle A O C=2\left(180^{\circ}-\angle A B C\right)=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$

Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secant theorem,

$$
D M^{2}=M C^{2}=B M \cdot A M, \quad \text { that is } \quad \frac{D M}{A M}=\frac{B M}{D M} .
$$

Therefore, triangles $D B M$ and $A D M$ are similar, from which

$$
\angle A B D=180^{\circ}-\angle M B D=180^{\circ}-\angle A D M=90^{\circ}
$$
3. Triangle $A B C$ with angle $\angle A B C=135^{\circ}$ is inscribed in circle $\omega$. The lines tangent to $\omega$ at points $A$ and $C$ intersect at point $D$. Find $\angle A B D$, given that $A B$ bisects segment $C D$. Answer: $90^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_8ca9121536825a6be283g-21.jpg?height=454&width=448&top_left_y=2052&top_left_x=861)