[Segment visible from two points at the same angle]  [Perpendicular bisector of the segment (LMT).]

Inside a non-isosceles triangle $A B C$, a point $O$ is taken such that $\angle O B C=\angle O C B=20^{\circ}$. Additionally, $\angle$ BAO + $\angle O C A=70^{\circ}$. Find the angle $A$.
Through point $C$, draw a line perpendicular to $AC$. Let this line intersect the extension of segment $AO$ at point $E$. Then

\[
\begin{gathered}
\angle B C E=90^{\circ}-\angle B C O-\angle O C A= \\
=90^{\circ}-20^{\circ}-\angle O C A=70^{\circ}-\angle O C A=\angle B A O=\angle B A E
\end{gathered}
\]

This means that from points $A$ and $C$, lying on the same side of line $B E$, segment $B E$ is seen at the same angle, so quadrilateral $ABEC$ is inscribed in a circle. Since $\angle A C E=90^{\circ}$, $AE$ is the diameter of this circle. From the problem statement, it follows that point $O$, lying on the diameter of the circle, is equidistant from the ends of chord $BC$, so it lies on the perpendicular bisector of $BC$. Since triangle $ABC$ is not isosceles, $O$ is the center of the circle. Therefore, $OA=OC$. Consequently,

\[
\angle B A C=\angle B A O+\angle O A C=\angle B A O+\angle O C A=70^{\circ}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-03.jpg?height=646&width=640&top_left_y=49&top_left_x=732)

## Answer

$70^{\circ}$.