SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $BC$, $DE$, $AF$ intersect and bound triangle $MNP$. Because the sum of angles in a hexagon is $720^{\circ }$, we have $\angle A+\angle B=\angle C+\angle D=\angle E+\angle F=240^{\circ }$.



Therefore, we easily see triangle $MNP$ is equilateral. To build the equilateral triangle $DEO$ with $O$ lying inside the hexagon, we see $EOAF$ and $DOBC$ are parallelograms so $AB=EF=OA=CD=OB$, thus $OAB$ is an equilateral triangle. Hence, $$\angle AFE+\angle BCD=\angle AOE+\angle BOD=360^{\circ }-60^{\circ }-60^{\circ }=240^{\circ }.$$ But $\angle EDC+\angle BCD=240^{\circ }$. These imply $\angle AFE=\angle EDC$.

Similarly, $\angle EDC=\angle CBA$. Similarly, $\angle BAF=\angle FED=\angle EDC$.

Now from the condition $\angle A+\angle B=\angle C+\angle D=\angle E+\angle F$ we deduce that all angles of this hexagon are equal to $120^{\circ }$.