3. Point $O$ is the center of the circumcircle of triangle $ABC$. On the circumcircle of triangle $BOC$, outside triangle $ABC$, point $X$ is chosen. On the rays $XB$ and $XC$ beyond points $B$ and $C$, points $Y$ and $Z$ are chosen respectively such that $XY = XZ$. The circumcircle of triangle $ABY$ intersects side $AC$ at point $T$. Find the angle $\angle Y T Z$.
Answer: $180^{\circ}$.

Solution. Note that $\angle B Y T=\angle B A T$ as inscribed angles subtending the same arc. Since quadrilateral $B O C X$ is cyclic, we get

$$
180^{\circ}-\angle B X C=\angle B O C=2 \angle B A C=2 \angle B Y T
$$

On the other hand, triangle $Y X Z$ is isosceles, so $180^{\circ}-\angle Y X Z=2 \angle X Y Z$. Therefore, $\angle X Y T=\angle X Y Z$, which means point $T$ lies on segment $Y Z$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-38.jpg?height=494&width=557&top_left_y=364&top_left_x=767)