Saudi Arabia Mathematical Competitions

Let $ABCD$ be a parallelogram. If $ABCD$ is a rectangle, then it is clear that we can dissect it into $n$ rectangles by parallel lines to its sides.

Assume that $ABCD$ is not a rectangle. Without loss of generality we can assume that $AB\ge AD$. Let $E$ and $F$ be the midpoints of $BC$ and $AD$, respectively. There are two cases: $\angle A<90^{\circ }$ and $\angle A>90^{\circ }$. Because the second case is analogous to the first one, we shall study only the situation $\angle A<90^{\circ }$.



Construct the isosceles triangle $AFG$, with $AF=FG$ and vertex $G$ on line $AB$. Because $\angle A<90^{\circ }$ we have $$AG=2AF\cos A<2AF=AD\le AB,$$ hence $G$ belongs to the interior of the segment $AB$. Let $G^{\prime }$ be any point in the interior of segment $GB$. Construct the parallelogram $G{G}^{\prime }{F}^{\prime }F$. Then $AF{F}^{\prime }{G}^{\prime }$ and $G^{\prime }BE{F}^{\prime }$ are isosceles trapezoids, hence they are cyclic quadrilaterals. Similarly, we divide $FECD$ into two isosceles trapezoids, hence $ABCD$ is dissected into 4 isosceles trapezoids.

In order to pass from $n$ to $n+1$ cyclic quadrilaterals it is sufficient to notice that an isosceles trapezoid is divided into isosceles trapezoids by a parallel line to the basis.