In the triangle $A B C$, we have the angles $\angle A C B=65^{\circ} \text{ and } \angle A B C=70^{\circ}$. On the extensions of the side $B C$, mark the point $P$ such that $B P=A B$ and that $B$ is between $P$ and $C$; and the point $Q$ such that $C Q=A C$ and that $C$ is between $B$ and $Q$. If $O$ is the center of the circle passing through $A, P$, and $Q$, find the values of the angles $\angle O A Q$ and $\angle O A P$.
Solution

Let $\angle B A O=x$ and $\angle O A C=y$. From $B P=A B$ and $A C=C Q$, we can conclude that $\angle A P B=$ $\angle P A B=\beta$ and $\angle C A Q=\angle A Q C=\gamma$. Since $O$ is the center of the circle passing through $A$, $P$ and $Q$, we have $O P=O A=O Q$ and thus the triangles $A O P$ and $A O Q$ are isosceles. Consequently,

\[
\begin{aligned}
\angle A P O & =\angle P A O \\
\alpha+\beta & =x+\beta \\
x & =\alpha
\end{aligned}
\]

Similarly, we can conclude that $y=\alpha$. By the Exterior Angle Theorem, $\angle A B C=2 \beta$ and $\angle A C B=2 \gamma$. Thus, $\beta=70 / 2=35^{\circ}$ and $\gamma=65 / 2^{\circ}=32^{\circ} 30'$. Moreover, $2 \alpha=$ $\angle B A C=45^{\circ}$ and thus $\alpha=22^{\circ} 30'$. Finally,

\[
\begin{aligned}
\angle O A P & =\beta+\alpha \\
& =57^{\circ} 30'
\end{aligned}
\]

and

\[
\begin{aligned}
\angle O A Q & =\gamma+\alpha \\
& =55^{\circ}
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-06.jpg?height=563&width=764&top_left_y=1905&top_left_x=527)