Let $B'$ be reflection of $B$ across $AM$. Let $N'$ be where angle bisector of $\angle B'AC$ meets $BC$. Note that $AB' = AB = AC$ and $AN' = AN'$ and $\angle B'AN' = \angle CAN'$ so $B'AN'$ and $CAN'$ are congruent. $\angle MB'N' = \angle AB'M + \angle AAB'N' = \angle ABC + \angle ACB = \angle 90$ so $BM^2 + CN'^2 = MN'^2$ so $N$ is $N'$. Now we have $\angle MAN = \angle B'AM + \angle B'AN = \angle 45$.