Prove that $ABCD$ is a parallelogram. Let $E , G , F , H$ be the midpoints of sides $AB , BC , CD , DA ,$ respectively of quadrilateral $ABCD.$ The common points of segments $AF , BH,CE,DG $ divide
each of into three parts, as shown in the figure.

Given that $$\frac{LK}{GD}=\frac{LI}{AF}=\frac{IJ}{BH}=\frac{JK}{CE}=k$$
Prove that $ABCD$ is a parallelogram.
Note that $EFGH$ is a parallelogram. And $Ar[ABCD]=2Ar[EFGH].$ And define $M=AC\cap BD.$ Note that $$2k^2\times Ar[MIJ]=2 Ar[MHB]=2Ar[MEB]=Ar[MAB].$$
Hence, we have $2k^2\times Ar[IJKL]=Ar[ABCD]\implies 2k^2\times Ar[IJKL]=Ar[EFGH].$
To prove that $ABCD$ is a paralleogram. Areas can help. It is enough to show that $Ar[MAB]=Ar[MDC]$ or show $Ar[MLK]=Ar[MIJ].$
Can we also show $IJKL$ parallelogram? It would help us.
 A: Let $M$ be the intersection point of $AC$ and $BD$ with $\frac{MA}{CA}=s$ and $\frac{MB}{DB}=t$ with $s,t\in (0,1) $.

In the coordinate system centered at $M$ with base vectors $\vec a=\vec{CA}$ and $\vec b=\vec{DB}$ the points have the following coordinates (those of the last four points were calculated as the intersection points of the corresponding lines: $I=(AF)\cap(BG)$ and so on):
$\def\ds{\displaystyle}$
$$\begin{array}{c|c|c}
&\vec a&\vec b\\
\hline
A& s& 0\\
B& 0& t\\
C& s-1& 0\\
D& 0& t-1\\
E&\frac12s&\frac12t\\
F&\frac12(s-1)&\frac12t\\
G&\frac12(s-1)&\frac12(t-1)\\
H&\frac12s&\frac12(t-1)\\
I&\ds\frac{(s-1)t}{s+2t+1}&\ds\frac{(s+t)t}{s+2t+1}\\
J&\ds\frac{(s-1)(s-t-1)}{2s-t-3}&\ds\frac{(s-1)(t-1)}{2s-t-3}\\
K&\ds\frac{s(t-1)}{s+2t-4}&\ds\frac{(s+t-2)(t-1)}{s+2t-4}\\
L&\ds\frac{s(s-t+1)}{2s-t+2}&\ds\frac{st}{2s-t+2}\\
\end{array}$$
We should find such $s,t$ that:
$$
\frac{IJ}{BG}=\frac{JK}{CH}=\frac{KL}{DE}=\frac{LI}{AF}:=k.\tag1
$$
Computing the above ratios as $\frac{IJ}{BG}=\frac{[IMJ]}{[BMG]}=\frac{|I\times J|}{|B\times G|}$ the relations $(1)$ can be transformed into:
$$\begin{align}
&\frac{s^2-st-t^2-1}{(s+2t+1)(2s-t-3)}\\
=&\frac{s^2-st-t^2-2s+t+2}{(2s-t-3)(s+2t-4)}\\
=&\frac{s^2-st-t^2-s+3t-2}{(s+2t-4)(2s-t+2)}\\
=&\frac{s^2-st-t^2+s+2t}{(2s-t+2)(s+2t+1)}.\\
\end{align}$$
The solutions of the system are all pairs $(s,t)$ satisfying one of the two following linear equations:
$$
t+3s=2;\quad\text{or}\quad 3t-s=1,\tag2
$$
which means: the quadrilateral $ABCD$ need not be a parallelogram ($s=t=\frac12$).
Both solutions $(2) $ imply $k=\frac25$. This value can be obtained also by elementary geometric means without using the coordinate algebra.

EDIT: Some additional geometrical insight gives the following Lemma:
The relations $(1)$ hold
if and only if the quadrilateral $IJKL$ is a trapezoid. Particularly:
$$
t+3s=2\iff IJ\parallel KL;\quad 3t-s=1 \iff LI\parallel JK.
$$
To prove the Lemma it is sufficient to compute the cross products $IJ\times KL$ and $LI\times JK$ using the coordinates given in the above table.
