$O_1, O_2, O_3, O_4$ circumcenters for triangles $AQM , MBN, NCP, PDQ$. Let ABCD a parallelogram and $M, N, P, Q$ on $AB, BC, CD, DA$ and $O_1, O_2, O_3, O_4$  circumcenters for triangles $AQM , MBN, NCP, PDQ$. Show that $O_1O_2,O_3,O_4$ is a parallelogram. In a rectangle it's easy to prove.
I tried to costruct line bisectors but I didn't find anything.
 A: The signs in this solution will depend on location of points relatively to parallelogram sides. Using vector projection one can eliminate this disadvantage.

$$GA=AI+IG=AI+EJ=AE \cos A+O_1E\sin A \Rightarrow O_1E=\frac{GA-AE\cos A}{\sin A}$$
$$HB=HL=KF-LF=O_2F \sin A - BF \cos A \Rightarrow O_2F=\frac{HB+BF \cos A}{\sin A}$$
$$EF=EM+MF=AE+BF=\frac{AE+EM}{2}+\frac{BF+MF}{2}=\frac{AB}{2}$$
$$O_2F-O_1E=\frac{HB-GA+(AE+BF) \cos A}{\sin A}=\frac{NB-QA+AB\cos A}{2\sin A}$$
Slope of line $O_1O2$ about line $AB$ is $$k_{12}=\frac{O_2F-O_1E}{EF}=\frac{NB-QA+AB\cos A}{AB \sin A}$$
Using analogy: slope of line $O_3O_4$ about line $CD$ is $$k_{34}=\frac{O_2F-O_1E}{EF}=\frac{QD-NC+CD\cos C}{CD \sin C}$$
$$QA+QD=AD=BC=NB+NC \Rightarrow QD-NC=NB-QA \Rightarrow k_{34}=k_{12}$$
$CD$ is parallel to $AB$, therefore $O_1O_2$ is parallel to $O_3O_4$.
A: Trivially, $O_1O_2O_3O_4$ is a parallelogram when $M,N,P,Q$ are the midpoints of the sides. Let's call this the middle parallelogram.

(Sorry for not labeling the points, I'll use the names for the points in this drawing.)
From the middle parallelogram, let's move $E$ to where we want it to be.
$I$ is the intersect of the perpendicular bisectors of $CE$ and $CF$, and $K$ is the intersect of the perpendicular bisectors of $DE$ and $DH$. When we move $E$, only the line segments $CE$ and $DE$ change.
Therefore, when we move $E$, the gradient of $IK$ will stay constant, so it will always be parallel to $JL$ just like it was in the middle parallelogram.
Repeat this process for $F,G,H$ and all the sides of $IJLK$ will be parallel to the sides of the middle parallelogram.
