Let $\angle A$ in a triangle $ABC$ be $60^\circ$. Let point $N$ be the intersection of $AC$ and perpendicular bisector to the side $AB$ while point $M$ be the intersection of $AB$ and perpendicular bisector to the side $AC$. Prove that $CB = MN$. (3 points)
Problem
Source: Spring 2006 Tournament of Towns Junior O-Level #1
Tags: geometry, perpendicular bisector