Question on isosceles triangle

Question

Let $ABC$ be a triangle such that there exists $N \in [AM]$ such that the angles $BAN$ and $CAN$ have the same measure, where $M$ is the midpoint of the segment $BC$. Then $ABC$ is isosceles. One hint?

Answer 1

The angles $BAN,CAN$ are the same as angles $BAM, CAM$ respectively. That is, they are the angles that the angle bisector make with the two other sides. There is a theorem saying that the bisector cuts the third side in the same ratio as that of the two sides going to the ends of the third side. So if $M$ is the midpoint that ratio is $1,$ and ...

Edit: The use of the above mentioned theorem can be avoided by using the law of sines. At one point one needs that $\sin(\pi-x)=\sin(x).$