Small correction: Another math-fun member has kindly pointed out that the intersection of AB with BC in the first paragraph below is not necessarily in the interior of both of them. If the intersection of AB and CD occurs at one of the 4 endpoints (but necessarily not at a common endpoint), the difference with what I wrote below is that one of the triangles ACD or BCD (but not both) will be degenerate. Fortunately, the conclusion still follows. —Dan

On Oct 13, 2015, at 3:55 PM, rcs@xmission.com wrote:

Quoting Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>>:

...

Assume A, B, C, D are distinct points in the plane such that segments AB and CD intersect (necessarily in their interiors).

Then ACD and BCD are triangles sharing the edge CD but otherwise disjoint, since A and B are on opposite sides of the line extending CD.

Thus BC and AD are disjoint.

?Dan

...

On Oct 13, 2015, at 8:02 AM, James Propp <jamespropp@gmail.com <mailto:jamespropp@gmail.com>> wrote:

3) Although this is more relevant to my mistake from last night than to the questions I'm raising here, I'd still like to know the right way to see that, given points A, B, C, and D in the plane, it isn't possible for line segment AB to intersect line segment CD *and* for line segment BC to intersect line segment AD unless two or more of the points coincide.