Problem 9.5. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 34^{\circ}$ and $\angle X D C = 52^{\circ}$.

(a) (1 point) How many degrees does the angle $A X C$ measure?

(b) (3 points) How many degrees does the angle $A C B$ measure?

![](https://cdn.mathpix.com/cropped/2024_05_06_1f4981519479532effc1g-04.jpg?height=319&width=713&top_left_y=301&top_left_x=366)
Answer: (a) $107^{\circ}$. (b) $47^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1f4981519479532effc1g-04.jpg?height=317&width=711&top_left_y=736&top_left_x=369)

Fig. 5: to the solution of problem 9.5

Solution. Given $A B=B X$ and $\angle B=34^{\circ}$, therefore $\angle A X B=\angle X A B=\frac{1}{2}\left(180^{\circ}-34^{\circ}\right)=73^{\circ}$. Then $\angle A X C=180^{\circ}-73^{\circ}=107^{\circ}$.

Since $A D=D C$, we get that arcs $A D$ and $D C$ are equal. This means that $B D$ is the bisector of angle $A B C$. From the equality of triangles $A B D$ and $X B D$, it follows that $X D=A D=C D$ (Fig. 5).

Since quadrilateral $A B C D$ is cyclic, $\angle A C D=\angle A B D=\frac{1}{2} \cdot 34^{\circ}=17^{\circ}$. Since $X D=C D$ and $\angle X D C=52^{\circ}$, we get that $\angle X C D=\angle C X D=\frac{1}{2}\left(180^{\circ}-52^{\circ}\right)=64^{\circ}$. We have

$$
\angle A C B=\angle X C D-\angle A C D=64^{\circ}-17^{\circ}=47^{\circ}
$$

Remark. The angle $A C X$ could also have been found by considering that point $D$ is the center of the circumcircle of triangle $A C X$.