Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?

![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-43.jpg?height=448&width=519&top_left_y=234&top_left_x=465)
Answer: 58.

Solution. As is known, in a circle, inscribed angles subtended by equal chords are either equal or supplementary to $180^{\circ}$. Since $B C=C D$ and $\angle B A D<180^{\circ}$, we get that $\angle B A C=\angle D A C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-43.jpg?height=449&width=507&top_left_y=912&top_left_x=465)

Fig. 9: to the solution of problem 11.5

Since $\angle B D A=\angle B C A=64^{\circ}$, we get that $\angle B D O=\angle A D O=32^{\circ}$. Therefore, point $O$ lies on the two angle bisectors of triangle $A B D$, i.e., it is the point of intersection of the angle bisectors (Fig. 9). Then

$$
\angle B O C=\angle B A O+\angle A B O=\frac{\angle B A D+\angle A B D}{2}=\frac{180^{\circ}-\angle B D A}{2}=\frac{180^{\circ}-64^{\circ}}{2}=58^{\circ}
$$

Remark. The solution could also have been completed differently: since $O$ is the point of intersection of the angle bisectors of triangle $A B D$, by the trident lemma $C D=C B=C O$, from which $\angle C O B$ is easily found.