Let $E$ be the point of intersection of the diagonals of convex quadrilateral $ABCD$, and let $P,Q,R$, and $S$ be the centers of the circles 
circumscribing triangles $ABE, BCE, CDE$, and $ADE$, respectively. Then
$\textbf{(A) }PQRS\text{ is a parallelogram}\\ \textbf{(B) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rhombus}\\ \textbf{(C) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rectangle}\\ \textbf{(D) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a parallelogram}\\ \textbf{(E) }\text{none of the above are true}$
Let $L$, $M$, $N$, and $O$ be the intersections of $AC$ and $PS$, $BD$ and $PQ$, $CA$ and $QR$, and $DB$ and $RS$ respectively. Since the [circumcenter](https://artofproblemsolving.com/wiki/index.php/Circumcenter) of a triangle is determined by the intersection of its perpendicular bisectors, each of the pairs mentioned earlier are perpendicular to each other. Let $\angle SPQ=\alpha$. Then $\angle LEM=180^\circ-\alpha$ since $\angle PLC=PMD=90^\circ$. This means $\angle DEC=180^\circ-\alpha$, which similarly implies $\angle QRS=\alpha$. Through a similar method, we find that $\angle PSR=\angle PQR=180^\circ-\alpha$.
Since the opposite angles of $PQRS$ are congruent, $PQRS$ is indeed a parallelogram - and we showed this without any assumption about the type of quadrilateral $ABCD$ was. Therefore, $PQRS$ is always a parallelogram, and so our answer is $\fbox{A}$.