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.
So that we will be looking at the same picture, let the vertices of the trapezoid be $A,B,C,D$, in counterclockwise order, with $AB$ and $CD$ the parallel sides, and $AB$ horizontal at the bottom. Let diagonals $AC$ and $BD$ meet at $X$.
Let the area of $\triangle ABX$ be $s^2$ and let the area of $\triangle CXD$ be $t^2$. Then corresponding sides in these two similar triangles are in the ratio $s$ to $t$. This is because if we scale linear dimensions by the factor $\rho$, areas get scaled by the factor $\rho^2$.
By similarity, we have $\frac{AX}{XC}=\frac{AB}{CD}=\frac{s}{t}$. So the area of $\triangle AXD$ is $\frac{s}{t}$ times the area of $\triangle XCD$. That's because these two triangles have the same height with respect to their two bases $AX$ and $XC$.
Thus $\triangle AXD$ has area $st$. Similarly, $\triangle BXC$ has area $st$, so the trapezoid has area $s^2+t^2+2st$.
Best Answer
Without loss of generality, |AD|<|BC|. Draw a line passing through A and meeting BC at G and EF at H. Since AGCD is a parallelogram, H is the midpoint of AG. Note that triangles AEH and ABG are similar, hence, EH is parallel to BG, and the result follows.