Let $P$ be the intersection of the tangents to the circuncricle of $\triangle ABC$ at $B$ and $C$ Let $K$ be the second point of intersection of the circumcircle of $\triangle BPC$ and $AP$. $\angle BKP = \angle BCP = \angle A, \angle CKP = \angle CBP = \angle A$ It is well-known that $AP$ is the symedian, then $\angle BAK = \angle MAC, \angle CAK = \angle MAB$ $\angle KBA = \angle BNP - \angle BAN = \angle A - \angle MAC = \angle MAB$ $\angle KCA = \angle CNP - \angle CAN = \angle A - \angle MAB = \angle MAC$ Then $K=N$ and then we are done

My solution: Denote $\angle BAN=a, \angle NAM=b, \angle CAM=c$. We want to prove that $a=c$. Firstly, notice that $\frac{NA}{\sin (a+b)}=\frac{AB}{\sin \angle ANB}$ and $\frac{NA}{\sin c}=\frac{AC}{\sin \angle ANC}$ Thus $\frac{\sin (a+b)}{\sin c}=\frac{AC}{AB}\cdot \frac{\sin \angle ANB}{\sin \angle ANC}$ On the other hand, $\frac{BM}{MC}=\frac{AB}{AC}\cdot \frac{\sin (a+b)}{\sin c}\Rightarrow \frac{AC}{AB}=\frac{\sin (a+b)}{\sin c}$ Using the previous result, we get that $\angle ANB=\angle ANC\Rightarrow 2a+b=2c+b\Rightarrow a=c$, and the conclusion follows.