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Sorry for a silly question,

I am trying to prove the fact of intersection of two segments on the plane.

For example, $(d_1,d_2)$ is the first segment, where $d_1$ and $d_2$ are endpoint of the segment, and $(d_3,d_4)$ is the second segment.

One of the ways to show the existence of intersection is to show that $d_1$ and $d_2$ are placed on the different sides of the line $l(d_3,d_4)$ and $d_3$, $d_4$ on the different sides of $l(d_1,d_2)$.

It's very intuitive statement and obviously this should be true, but how to prove this.

How to prove that if endpoints of the one segment are located on the different sides of the another segment and vice versa show there is a intersection of segments.

Thanks!

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I assume you're working in the plane (ie. $\mathbb{R}^2$)? –  Najib Idrissi Mar 17 '12 at 8:52
    
@zulon: Yes, thanks for pointing this out –  fog Mar 17 '12 at 8:54
    
Hint: try the intermediate value theorem. –  dtldarek Mar 17 '12 at 11:16

2 Answers 2

up vote 1 down vote accepted

WLOG (we can translate, rotate and scale), we can assume that $d_1 = (0,0)$ and $d_2 = (1,0)$. Define $\gamma(t) = d_3 + t (d_4 - d_3)$.

If $d_3$ and $d_4$ are on the same side of the line $l(d_1,d_2)$, eg. if they both have positive y-coordinate, then for $t\in[0,1]$, the y-coordinate of $\gamma(t)$ will be positive and there will be no intersection. By symmetry, if $d_1,d_2$ are on the same side of $l(d_3,d_4)$ there will be no intersection either.

Now if all the points are on different sides of the respective lines, then by the mean value theorem there is a $t_0$ st. the y-coordinate of $\gamma(t_0)$ is zero (it takes positive and negative values and it is obviously continuous). Since $d_1,d_2$ are on different sides of the line $l(d_3,d_4)$, $\gamma(t_0)$ must be between $d_1$ and $d_2$, therefore there is an intersection.

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A line divides the plane into two half planes which are both convex. Consider the two lines $g:=d_1\vee d_2$ and $h:=d_3\vee d_4$.

If $d_3$ and $d_4$ both lie in the same half plane determined by $g$ then the complete segment $[d_3,d_4]$ will lie in this half plane and so will not intersect $g$, let alone $[d_1,d_2]\subset g$, and similarly: If $d_1$ and $d_2$ both lie in the same half plane determined by $h$ then the complete segment $[d_1,d_2]$ will lie in this half plane and so will not intersect $h$, let alone $[d_3,d_4]\subset h$.

Assume now that $d_3$ and $d_4$ lie on different sides of $g$ and that $d_1$ and $d_2$ lie on different sides of $h$. Then there are two points $p:=[d_3,d_4]\cap g$ and $q:=[d_1,d_2]\cap h$. As both $p$, $q\in g\wedge h=:c$ it follows that $p=q=c$, whence $c=[d_1,d_2]\cap[d_3,d_4]$.

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