Consider an irregular quadrilateral $ABCD$. Let $E,F,G,H$ be the midpoints of its edges. It seems that there is a point $K$ such that
$$
S_{AHKE} = S_{EKFB} = S_{KHDG} = S_{KGCF} \left(= \frac{1}{4} S_{ABCD}\right)
$$
I'm curious whether the point $K$ has any other interesting properties.
Here's the proof that this point does exist:
Assuming that $A,B,C,D,I$ have coordinates $\mathbf p_1, \mathbf p_2, \mathbf p_3, \mathbf p_4, \mathbf p$, respectively. Then
$$
\mathbf S_{AHKE} = \frac{1}{2} (\mathbf p – \mathbf p_1) \times \frac{\mathbf p_2 – \mathbf p_4}{2} = \frac{1}{4} (\mathbf p – \mathbf p_1) \times (\mathbf p_2 – \mathbf p_4)\\
\mathbf S_{EKFB} = \frac{1}{4} (\mathbf p_3 – \mathbf p_1) \times (\mathbf p_2 – \mathbf p) = \frac{1}{4} (\mathbf p – \mathbf p_2) \times (\mathbf p_3 – \mathbf p_1)\\
\mathbf S_{KHDG} = \frac{1}{4} (\mathbf p_3 – \mathbf p_1) \times (\mathbf p – \mathbf p_4) = \frac{1}{4} (\mathbf p_4 – \mathbf p) \times (\mathbf p_3 – \mathbf p_1)\\
\mathbf S_{KGCF} = \frac{1}{4} (\mathbf p_3 – \mathbf p) \times (\mathbf p_2 – \mathbf p_4)
$$
It is easy to see that
$$
\mathbf S_{AHKE} + \mathbf S_{KGCF} = \frac{1}{2} \mathbf S_{ABCD}\\
\mathbf S_{EKFB} + \mathbf S_{KHDG} = \frac{1}{2} \mathbf S_{ABCD}
$$
thus there is exactly two linear equations
$$
\mathbf S_{AHKE} – \mathbf S_{KGCF} = 0\\
\mathbf S_{EKFB} – \mathbf S_{KHDG} = 0
$$
to determine two components of $\mathbf p$. And they are
$$
(2\mathbf p – \mathbf p_1 – \mathbf p_3) \times (\mathbf p_2 – \mathbf p_4) = 0\\
(2\mathbf p – \mathbf p_2 – \mathbf p_4) \times (\mathbf p_3 – \mathbf p_1) = 0
$$
which is equivalent to
$$
\mathbf p = \frac{\mathbf p_1 + \mathbf p_3}{2} + \lambda(\mathbf p_2 – \mathbf p_4) = \frac{\mathbf p_2 + \mathbf p_4}{2} + \mu(\mathbf p_3 – \mathbf p_1), \quad \lambda,\mu \in \mathbb R
$$
The geometrical definition of $K$ should be obvious now: the point $K$ is reflection of diagonal intersection point $M = AC \cap BD$ about the vertices' centroid $P$
Best Answer
From your definition of $K$ it follows that triangles $EBK$ and $GCK$ have the same area, which is the same as saying that triangles $ABK$ and $DCK$ have the same area. In other words: $K$ divides $ABCD$ into two couples of triangles of equal area.
Given two segments $AB$ and $CD$ whose extensions meet at $Q$ (see picture below), the locus of points $X$ such that $XAB$ and $XCD$ have the same area, with $X$ belonging to $\angle AQD$, is line $QR$, where $R$ is the vertex opposite to $Q$ of the parallelogram with sides parallel and congruent to $AB$ and $CD$. A different but obvious construction is needed if $AB$ and $CD$ are parallel.
In the same way one can construct line $ST$, which is the locus of the points forming with $AD$ and $BC$ triangles of equal area. Point $K$ is then the intersection of $QR$ and $ST$.