KöMaL 2009 K229

Show that a given line segment $AB$ can be divided into three equal parts by the following method: A $30^o$ angle is drawn to each end of the line segment so that their other arms intersect at point $C$. Then the perpendicular bisectors of the line segments $AC$ and $BC$are constructed. These will intersect $AB$ at two points that cut it into three equal parts.