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.
1) Let the line $MO$ intersects the sides of the trapezoid at the points $E$ and $F$. Then $$\triangle BME \sim \triangle AMF \Rightarrow\frac {BE}{AF}=\frac {ME}{MF}$$
2) $$\triangle EMC \sim \triangle FMD \Rightarrow\frac {ME}{MF}=\frac {EC}{FD}\Rightarrow\frac {BE}{AF}=\frac {EC}{FD}$$
3) $$\triangle AOF \sim \triangle COE \Rightarrow\frac {EC}{AF}=\frac {EO}{OF}$$
$$\triangle BEO \sim \triangle DFO \Rightarrow\frac {BE}{FD}=\frac {EO}{OF} \Rightarrow\frac {BE}{FD}=\frac {EC}{AF}$$
So
$$\frac {BE}{AF}=\frac {EC}{FD}$$ and $$\frac{BE}{FD}=\frac {EC}{AF}$$
Then the four points $M,E,O,F$ lie on a straight line
Best Answer
Consider quadrilateral ABCD. As can be seen in figure, quadrilateral MPNQ is a parallelogram because points M, N, P and Q are midpoints of sides of ABCD. So segments PQ and MN bisect each other. In other words, segment MN crosses the midpoint of segment PQ. Also quadrilateral PFQE is parallelogram because $PE||FQ$ and $PE=FQ$, so it's diagonals also bisect each other. That is, EF cross the mid point of PQ; in this way lines MN, PQ and EF are concurrent.