SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $M,N,P,Q,R,S$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$, respectively, and $X,Y,Z$ be the midpoints of $AD$, $BE$, $CF$.



Since $AE=BD$ and the midsegments in some triangles, we get $$XQ=YM=\frac{1}{2}\cdot AE=\frac{1}{2}\cdot BD=XM=YQ,$$ so $XMYQ$ is a rhombus, then $MQ$ is the perpendicular bisector of the segment $XY$. Similarly, $NR$, $PS$ are the perpendicular bisectors of $XZ$, $YZ$, so $MQ$, $NR$, $PS$ are concurrent at the circumcenter of the $△XYZ$. $\square$