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 $\triangle XYZ$. $\square$