Solution:

Without loss of generality, $AB < AC$. It follows that $\angle BAH = 90^{\circ} - \angle B$, since the extension of $AH$ is perpendicular to $BC$. Moreover, we also have $\angle AOC = 2\angle B$; but since $OA = OC$, this implies $\angle OAC = \frac{1}{2}\left(180^{\circ} - \angle AOC\right) = 90^{\circ} - \angle B$. So we conclude that $\angle BAH = \angle CAO$. Since $\angle BAI = \angle CAI$ as well, it follows that $\angle HAI = \angle OAI$, which is what we wanted to prove.