I'm going to write an answer using vectors.
Let $O$ be the intersection point of $AC, BD$.
Let $$\vec{OA}=\vec{a}, \vec{OB}=\vec{b}, \vec{OC}=k\vec{a}, \vec{OD}=l\vec{b}$$
where $k,l\lt 0.$
Letting $E,F,G,H$ be the midpoints of $AB, BC, CD, DA$ respectively, we have
$$\vec{OE}=\frac{1}{2}\vec a+\frac 12\vec b,\vec{OF}=\frac 12\vec b+\frac k2\vec a, \vec{OG}=\frac k2\vec a+\frac l2\vec b, \vec{OH}=\frac 12\vec a+\frac l2\vec b.$$
Letting $I$ be the intersection point of $EG, FH$, there exist $m,n$ such that
$$\vec{EI}=m\vec{EG}, \vec{FI}=n\vec{FH}.$$
The former gives us
$$\vec{OI}-\vec{OE}=m\left(\vec{OG}-\vec{OE}\right)\iff \vec{OI}=(1-m)\vec{OE}+m\vec{OG}=\frac{1-m+mk}{2}\vec a+\frac{1-m+ml}{2}\vec b.$$
The latter gives us
$$\vec{OI}-\vec{OF}=n\left(\vec{OH}-\vec{OF}\right)\iff \vec{OI}=(1-n)\vec{OF}+n\vec{OH}=\frac{k-kn+n}{2}\vec a+\frac{1-n+nl}{2}\vec b.$$
Now since $\vec a$ and $\vec b$ are linearly independent, the following has to be satisfied :
$$\frac{1-m+mk}{2}=\frac{k-kn+n}{2}\ \text{and} \frac{1-m+ml}{2}=\frac{1-n+nl}{2}.$$
These give us $m=n=1/2$ since $(k,l)\not=(-1,-1).$
Hence, we get
$$\vec{OI}=\frac{k+1}{4}\vec a+\frac{l+1}{4}\vec b.$$
On the other hand, letting $P,Q$ be the midpoints of $AC, BD$, we have
$$\vec{OP}=\frac{k+1}{2}\vec a, \vec{OQ}=\frac{l+1}{2}\vec b.$$
Finally, we obtain
$$\vec{PI}=\frac 12\vec{PQ}.$$
Since this represents that $I$ is on the line $PQ$, we now know that we get what we want. Q.E.D.
P.S. If $(k,l)=(-1,-1)$, then $ABCD$ is a parallelogram, which is an easy case.
$EF$ is the basis media of triangle $ABC$, and then $A(BEF)=1/4A(ABC)$. Idem $A(GHD)=1/4A(ACD)$.
Now, note that $A(ACD)+A(ABC)=A(ABCD)\to A(BEF)+A(GHD)=1/4A(ABCD)[*_1]$.
Idem, $ A(HEA)+A(GFC)=1/4A(ABCD)[*_2]$.
Adding $ [*_1] +[*_2]$ the problem is solved
Best Answer
The midpoint of $AB$ is $\frac12(A+B)$.
The midpoint of $BC$ is $\frac12(B+C)$.
The vector from the midpoint of $AB$ to the midpoint of $BC$ is $\frac12(B+C)-\frac12(A+B) = \frac12(C-A)$.
Repeat for $AD$ and $DC$ to see that you get the same vector there.