We first show that the points $A, M, R, N$ are concyclic. Since $ABC$ is an acute-angled triangle, $M$ and $N$ are on the line segments $AB$ and $AC$ respectively. Let $R_1$ be the point such that the points $A, M, R_1, N$ are concyclic, where $R_1$ is on the ray $AR$. Since $AR_1$ bisects $\angle BAC$, we have $R_1M = R_1N$. Since $M$ and $N$ lie on the circle with centre $O$, we have $OM = ON$. It follows from $OM = ON$ and $R_1M = R_1N$ that $R_1$ is on the bisector of $\angle MON$. Since $AB \neq AC$, the bisectors of the angles $BAC$ and $MON$ intersect at the unique point $R$, and so $R_1 = R$, or $A, M, R, N$ are concyclic.

Let the bisector of $\angle BAC$ meet $BC$ at $K$. Since the points $B, C, N, M$ are concyclic, $\angle MBC = \angle ANM$. Moreover, since $A, M, R, N$ are concyclic, $\angle ANM = \angle MRA$. This implies $\angle MBK = \angle MRA$. Therefore, the points $B, M, R, K$ are concyclic. Using the same argument as above, we obtain that $C, N, R, K$ are concyclic. This completes the solution.