Vector Geometry - Proving that a polygon is parallelogram

Question

Question states: "The sides $AD, AB, CB, CD$ of the quadrilateral $ABCD$ are divided by the points $E, F, G, H$ so that $AE: ED = AF: FB = CG: GB = CH: HD$. Prove that $EFGH$ is a parallelogram."

The question, although seemingly easy, is quite difficult for me. I expressed the sides of $EFGH$ as vectors with

$$HG = kGB - kHD$$ $$EF = kFB - kED$$ $$EH = \frac{1}{k}(HC - EA)$$ $$FG = \frac{1}{k}(GC - FA)$$

But I do not know how to proceed from here.

Answer 1

Demonstrate that $EF$ is parallel to $DB$ and a fraction $k$ of it.

Then demonstrate that $HG$ is also parallel to $DB$ and is the same fraction of it.

So oposite sides of $EFGH$ are parallel - enough to prove that it is a parallelogram.