The number of rhombi that can be inscribed in a parallelogram such that each side of the parallelogram contains precisely one vertex of the rhombus is [ ]

Let the parallelogram be $$ABCD$$ with centre $$O$$. Take any line through $$O$$ meeting sides $$AB$$ and $$CD$$ at $$P$$ and $$R$$, and the perpendicular line through $$O$$ meeting sides $$BC$$ and $$DA$$ at $$Q$$ and $$S$$. By the central symmetry of the parallelogram $$OP = OR$$ and $$OQ = OS$$, so $$PQRS$$ has perpendicular diagonals bisecting each other and is a rhombus with one vertex on each side. The first line must point inside $$\angle AOB$$ and the second inside $$\angle BOC = 180^\circ - \angle AOB$$, and a whole open interval of directions satisfies both conditions. Each direction gives a different rhombus, so there are infinitely many.