How to show that if the centres of the circumscribed circle and the inscribed circle in $\triangle ABC$ match (coincide), $\triangle ABC$ is equilateral?
I have an idea but I'm not sure. Let $\angle OCB = \angle ACO = \gamma$, thus $\angle OBC = \angle OBA = \gamma$ ($\triangle OBC$ is isosceles; $B, C \in k$). Similarly $\angle OAB = \angle OAC = \gamma$ and $\angle OAC = \angle OAB = \gamma$.