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 $120^\circ$.