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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?

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  • $\begingroup$ Prove that the diagonals are not parallel to the line. $\endgroup$ – Dale M Sep 27 '14 at 7:35
  • $\begingroup$ @DaleM actually, they can be parallel. but we need to prove that for all other cases if the points are fixed, no matter what pairs of parallel lines we draw, they will respectively pass through the same point on the line l $\endgroup$ – Jackie Poehler Sep 27 '14 at 8:09
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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.

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Use coordinate geometry

enter image description here

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.

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