Parallelograms $OAEB$, $OBFC$, $OCGD$, $ODHA$ are complated from any point $O$ in parallelogram $ABCD$. Show $EFGH$ is a parallelogram.

Question

$ABCD$ is a parallelogram and $O$ is any point. The parallelograms $OAEB$, $OBFC$, $OCGD$, $ODHA$ are completed, Prove that $EFGH$ is a parallelogram.

How can this be proved using the concept of congruence of triangles?

I have thought about this a lot and I still can't figure out the proof, someone please help me solve this.

Answer 1

EDIT 1:

A vector diagram as needed is drawn, labelling the vectors and adding them as per their color named parallels ( g for green, r for red etc.). The unconnected quadrilateral EFGH is a parallelogram.

( However note that internal point $O$ does not lie on straight lines AC or BD which are not diagonals, and these considerations are unnecessary).