67th Romanian Mathematical Olympiad

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 $MAC$.