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.
First consider the triangles with two vertices on the first line. There are $10$ possible pairs of points on this line, so that they can form $60$ triangles ($50$ with the third vertex on the second line and $10$ with the third vertex on the non-collinear point).
Now consider the triangles with two vertices on the second line. By the symmetry of the problem, these are $60$ as well.
Finally, consider the triangles that do not have two vertices on the same line. These triangles clearly have one vertex on the first line, one vertex on the second line, and the third one on the non-collinear point. Because we can chose any of the $5$ points on each line, the number of these triangles is $5^2=25$.
Thus, the total number is $60+60+25=145$.
Best Answer
As said before, there are $x={n \choose 2}$ intersections of the lines as long as no two are parallel and not three meet at a point. Each line has $n-1$ points on it, so there are ${n-1 \choose 2}$ pairs that give that existing line. The number of new lines is then ${x\choose 2}-n{n-1 \choose 2}$ We can rework the last term as $n{n-1 \choose 2}=\frac {n(n-1)(n-2)}2=(n-2){n \choose 2}$ so the final number becomes $${x\choose 2}-(n-2){n \choose 2}=\frac 18(n^4-6n^3+11n^2-6n)$$