Geometry. Parallelogram. Let $A$, $B$, $C$, $D$ be $4$ fixed points on line $l$. Through $A$ and $B$ pass two arbitrary parallel lines, and two arbitrary parallel lines pass through $C$ and $D$. These $4$ lines form a parallelogram. Prove, that the diagonals of this parallelogram cross the line $l$ in two fixed points. Tried Menelaus' Theorem and  the Intercept theorem but couldn't work it out. How to solve this?
 A: Take the $y$-axis as the line $l$. We can write the four lines $l_1, l_2,l_3,l_4$ as $y=rx+a$, $y=rx+b$, $y=sx+c$, $y=sx+d$, respectively. The intersection point of $l_1$ with $l_3$, that of $l_1$ with $l_4$, that of $l_2$ with $l_3$, and that of $l_2$ with $l_4$ are $\displaystyle (\frac{c-a}{r-s}, \frac{rc-sa}{r-s})$, $\displaystyle (\frac{d-a}{r-s}, \frac{rd-sa}{r-s})$, $\displaystyle (\frac{c-b}{r-s}, \frac{rc-sb}{r-s})$, and $\displaystyle (\frac{d-b}{r-s}, \frac{rd-sb}{r-s})$, respectively.
So the diagonal passing through the intersection point of $l_1$ with $l_4$ and that of $l_2$ with $l_3$ has equation
$$y= \frac{(rd-sa-rc+sb)}{(d-a-c+b)}x+ \frac{(rd-sa)}{(r-s)}-  \frac{(rd-sa-rc+sb)}{(d-a-c+b)}\frac{(d-a)}{(r-s)}  $$
which for $x=0$ has solution
$$y= \frac{(rd-sa)}{(r-s)}-  \frac{(rd-sa-rc+sb)}{(d-a-c+b)}\frac{(d-a)}{(r-s)}$$
Expanding the products and simplifying, the last equation reduces to
$$ y=   \frac{  (rbd+sac-sbd-rac)  }{(r-s)(d-a-c+b)}=-\frac{  (r-s) (ac-bd) }{(r-s)(d-a-c+b)} =\frac{ (bd-ac) }{(d-a-c+b) }$$
Since the equation is of the form $y=k$, with $k$ constant and independent of the slopes $r$ and $s$, the diagonal of the parallelogram crosses the $y$-axis, i.e. the line $l$, in a fixed point whose $y$-coordinate is given by the last equation, no matter which slopes are considered.
The same considerations allow to show that the other diagonal passing through the intersections of $l_1$ with $l_3$ and of $l_2$ with $l_4$ also crosses the $y$-axis or line $l$ in a fixed point, whose $y$-value is given by
$$y=\frac{(bc-ad)}{(c-a-d+b)}$$
Finally, note that the cases where the denominators $d-a-c+b$ or $c-a-d+b$ are equal to zero represent those in which the corresponding diagonals are parallel to the $y$-axis.
A: Use coordinate geometry

Calculate the coordinates of the vertices V1 and V2 and then fit the equation of the diagonal. The intercept of the diagonal with the y axis is Y = [b(a-d) + a(c-b)] / [a - d + c - b], provided m not equal to n and provided (a - d + c -b) not equal 0. The first is the condition that the two sets of lines are not parallel and the second that the diagonal is not parallel to the y axis.
So, the y intercept is independant of the slope of the two sets of parallel lines.
