Let $P$ be the point where the perpendicular bisector of the segment $[BC]$ meets $AB$. Then $m(\angle PCB) = 30^\circ$, $m(\angle PCA) = 15^\circ$ and $m(\angle MPC) = 60^\circ$.

Since $PC = PB$ and $NC = AB$, it follows that $\frac{AP}{NC} = \frac{BP}{BC}$, that is $\frac{PA}{PB} = \frac{CN}{CB}$, whence $AN \parallel PC$.
Since $PC = 2PM$ (the right triangle $MPC$ has an angle of $30^\circ$), $\frac{PA}{AB} = \frac{PC}{BC} = \frac{2PM}{2BM} = \frac{PM}{BM}$ so, using the converse of the Bisector Theorem, $[MA$ is the bisector of the angle $BMP$.

Then $45^\circ = m(\angle AMB) = m(\angle ANB) + m(\angle MAN)$, and, since $AN \parallel PC$ yields $m(\angle ANB) = 30^\circ$, $m(\angle MAN) = 15^\circ$. From $m(\angle BAN) = m(\angle BPC) = 120^\circ$ and $m(\angle BAC) = 135^\circ$ follows that $m(\angle CAN) = 15^\circ = m(\angle MAN)$, that is $[AN$ is the bisector of the angle $MAC$.