2013 IMO SL #G5

 Let $ABCDEF$ be a convex hexagon with $AB=DE$, $BC=EF$, $CD=FA$, and $\angle A-\angle D = \angle C -\angle F = \angle E -\angle B$. Prove that the diagonals $AD$, $BE$, and $CF$ are concurrent.

Let $$AF \cap BC=P, BC \cap DE=Q, DE \cap AF=R$$ and let

$$CD \cap AB=W, AB \cap FE = V, FE \cap CD = U.$$

This forms $\triangle PQR$ and $\triangle WVU$. Next, consider points $X, Y, Z$ such that $XFAB, YBCD, ZDEF$ are parallelograms. 

Now, note that $$\angle RPQ = \angle APB = 180-(\angle PAB+\angle PBA) = \angle FAB + \angle CBA - 180$$ $$=\angle CDE+\angle FED-180=\angle DUE = \angle WUV.$$ We can similarly do this for the other pairs. So, we know that $\triangle{PQR} \sim \triangle{WUV}$.

Next, we can easily see that $BX=AF=CD=BY$ and similarly $FZ=FX$ and $DZ=DY$. Also, we have $$\angle ABY = \angle B - (180^{\circ} - \angle C) = \angle B + \angle C - 180^{\circ}$$ $$ = \angle E + \angle F - 180^{\circ} - \angle F - (180^{\circ} - \angle E) = \angle ZFA.$$ So, by SAS, we have $\triangle EXF \cong \triangle YED$ which implies $XE=YE$.

Hence, we know that line $\overline{BE}$ is the perpendicular bisector of the line $\overline{XY}$, from which it follows that $AD, BE, CF$ intersect, namely at the circumcenter of $\triangle XYZ$. 

We are done. $\square$