\[ \angle OAB = \angle OBA \]
\[ \angle OCB = \angle OBC \]
\[ \angle BAC + \angle ABC + \angle BCA = 180^{\circ} \]
\[ \angle BAC \]
на\[ \angle OBA \]
и\[ \angle BCA \]
на\[ \angle OBC \]
\[ \angle OBA + \angle ABC + \angle OBC = 180^{\circ} \]
Так как
\[ \angle ABC = \angle OBA + \angle OBC \]
\[ \angle ABC + \angle ABC = 180^{\circ} \]
$$2 \angle ABC = 180^{\circ}$$
\[ \angle ABC = 90^{\circ} \]