Good problem. Let $T$ be the circumcenter of triangle $BAO$ and $K$ be the midpoint of $AB$. By the sine rule radius of $(BAO)$ equal to radius of $(BOC)$, so $BTON$ is rhombus. Now $BK = MN$, $BT = ON$ and $KT = MO$ (since $TKMO$ is parallelogram) so $\triangle BTK = \triangle NOM$ and $\angle BKT = \angle NMO = 90^{o}$ (since $TK$ is perpendicular bisector of $AB$). So we are done.

Here is another solution. Let $K $ be the point on the extension on $CO$ such that $KO=CO$. Then, $NO||KB, MO||AK.$ Also, $\triangle BOC \equiv \triangle BOK$. This implies $BKO=BCO=BAO$, so $BKAO$ is cyclic, hence $KAB=90^\circ $. Because of the parallel lines, we are done.

See my solution on my Youtube channel here: https://www.youtube.com/watch?v=7Uo0ItjfFDo&feature=youtu.be

Let S be reflection of C across O. we have SAB and OMN are similar so we need to prove SAB = 90. ∠SBC is isosceles so ∠BAO = ∠BCO = ∠BSO ---> BSAO is cyclic. ∠SAB = ∠SEB = 90. we're Done.

Mahdi_Mashayekhi wrote: Let S be reflection of C across O. we have SAB and OMN are similar so we need to prove SAB = 90. ∠SBC is isosceles so ∠BAO = ∠BCO = ∠BSO ---> BSAO is cyclic. ∠SAB = ∠SEB = 90. we're Done. what a nice way to solve it!

We use one more method. Take a point \( O^* \) such that \( A O C O^* \) is a parallelogram. By the Parallelogram Isogonality Lemma, \( B O \) and \( B O^* \) are isogonals. Let the angle \( \angle A B O = \angle O^* B C = \beta \), and assume that \( \angle B A O = \angle B C O = \alpha \). Note that by simple angle chasing, \( \angle O B O^* = 90^\circ - \alpha - \beta \). Also, \( \angle A O C = 90^\circ + \alpha + \beta \), which implies \( \angle O^* C O = 90^\circ - \alpha - \beta \). Thus, we have \( \angle O B O^* = \angle O C O^* \), making quadrilateral \( O B C O^* \) cyclic with center \( N \), since \( \angle B O C = 90^\circ \). As \( O M = O^* M \), this implies that \( N M \perp O O^* \). Hence, \( \angle N M O = 90^\circ \). And, ladies and gentlemen, we are done.