Problem 9.4. Given a cyclic quadrilateral $A B C D$. It is known that $\angle A D B=48^{\circ}, \angle B D C=$ $56^{\circ}$. Inside triangle $A B C$, a point $X$ is marked such that $\angle B C X=24^{\circ}$, and ray $A X$ is the angle bisector of $\angle B A C$. Find the angle $C B X$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ce887724b4f1896529bag-3.jpg?height=396&width=377&top_left_y=868&top_left_x=542)
Answer: $38^{\circ}$.

Solution. Angles $B D C$ and $B A C$ are equal since they subtend the same arc. Similarly, angles $A D B$ and $A C B$ are equal. Then

$$
\angle A C X=\angle A C B-\angle X C B=\angle A D B-\angle X C B=48^{\circ}-24^{\circ}=24^{\circ}=\angle X C B .
$$

We obtain that $C X$ is the bisector of angle $A C B$, so $X$ is the point of intersection of the angle bisectors of triangle $A B C$.

Now it is easy to find the required angle:

$$
\angle C B X=\frac{\angle A B C}{2}=\frac{180^{\circ}-\angle B A C-\angle A C B}{2}=\frac{180^{\circ}-56^{\circ}-48^{\circ}}{2}=38^{\circ}
$$