It's well known that consecutively connecting midpoints of an arbitrary quadrilateral forms a parallelogram. Is it possible to inscribe other parallelograms inside a quadrilateral? I didn't find an example for that, so tried to come up with some argument which shows that parallelogram is unique but didn't get result. After that I used complex numbers to describe the problem. Let $ABCD$ be a quadrilateral with points $M$, $P$,$N$ and $Q$ on its sides. Also $MPNQ$ is a parallelogram.
So we have
$$\cases{\frac{A - M}{A^* - M^*}= \frac{M - B}{M^* - B^*} \\ \frac{A - P}{A^* - P^*} = \frac{P - D}{P^* - D^*} \\ \frac{D - N}{D^* - N^*} = \frac{N - C}{N^* - C^*} \\ \frac{C - Q}{C^* - Q^*} = \frac{Q - B}{Q^* - B^*} \\ M - Q + N - P = 0 }$$
I don't know how to proceed further.
Considering you are interested in the following question:
The above statement is not true. For instance, consider the following construction:
Draw two squares $ABCD$ and $A'B'C'D'$. Consider the quadrilateral formed by $$\{AA'\cap BB',BB'\cap CC',CC'\cap DD', DD'\cap AA'\}$$This quadrilateral will obviously have at least two distinct parallelogram (more precisely squares).