Saudi Arabian 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 $AB \geq AD$. 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
$$AG = 2AF \cos A < 2AF = AD \leq AB,$$
hence $G$ belongs to the interior of the segment $AB$. Let $G'$ be any point in the interior of segment $GB$. Construct the parallelogram $GG'F'F$. Then $AFF'G'$ and $G'BEF'$ are isosceles trapezoids, hence they are cyclic quadrilaterals. Similarly, we divide $F E C D$ 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.