waitt isn't this mean AMB collinear im quite confused

Let $O'$ be the center of $(ANC)$. Notice that $A$ is the center of spiral similarity taking $\overline{BC}$ to $\overline{MN}$, and hence $A$ is also the center of the spiral similarity which takes $\overline{BM}$ to $\overline{CN}$. This spiral sim also takes $(AMB)$ to $(ANC)$, and thus $O$ to $O'$, so $\triangle AOO' \sim \triangle ABC \sim \triangle AMN$ and $AO = AO'$. Therefore $O$ lies on $(ANC)$ iff $OO' = AO' = AO \iff \triangle AOO'$ is equilateral, which by spiral sim happens only when $ABC$ is equilateral.

We assume that $A,B,M$ are not collinear, so the circumcentre is defined (else, even if we allow the circumcentre to be the point at infinity, the problem is false). Now use complex numbers with $A=0$, $B=1$, $C=\alpha$ and $M=z$, then the condition $AB=AC$ implies $|\alpha |=1$, the direct similarity implies $N=\alpha z$ and the fact that $A,M,B$ are not collinear implies $z-\overline z\neq 0$. We have $O=\dfrac{z(1-\overline z)}{z-\overline z}$, so $\dfrac{O-A}{C-A}\dfrac{C-N}{O-N}=\dfrac{(1-z)(1-\overline z)}{(1-\overline z)-\alpha (z-\overline z)}$, which gives $\overline{\left (\dfrac{O-A}{C-A}\dfrac{C-N}{O-N}\right )}=\dfrac{(1-z)(1-\overline z)}{(1-z)+\overline \alpha (z-\overline z)}$. The points $O$, $C$, $N$ and $A$ lie on a circle of and only if $\dfrac{O-A}{C-A}\dfrac{C-N}{O-N}\in \mathbb{R}$, if and only if $(1-\overline z)-\alpha (z-\overline z)=(1-z)+\overline \alpha (z-\overline z)$, if and only if $(z-\overline z)(1-\alpha -\overline \alpha )=0$, if and only if $1-\alpha -\overline \alpha =0$, if and only if $\alpha +\overline \alpha =1$, if and only if $(x-\alpha )(x-\overline \alpha )=x^2-x+1$, if and only if $ABC$ is equilateral.

The similarity implies $\angle BAC = \angle MAN$, hence $\angle BAM = \angle CAN$ and so $\triangle ABM \cong \triangle ACN$, implying $AO = AO_1$, where $O_1$ is the circumcenter of $ANC$, as well as $\angle BAC = \angle OAO_1$. But now $AOCN$ being cyclic (with center $O_1$) implies $O_1O = O_1A$ - thus $AOO_1$ is equilateral and $\angle BAC = \angle OAO_1 = 60^{\circ}$, as desired.