An algebraic proof that a convex quadrilateral has intersecting diagonals

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SUMMARY

The discussion centers on finding an algebraic proof that the diagonals of a convex quadrilateral intersect. Key points include the necessity to demonstrate that the diagonals are not parallel by showing they have different slopes. The use of parametric equations for the diagonals is suggested, along with the application of Caratheodory's Theorem to establish the intersection. The user ultimately resolves their query by recognizing the significance of choosing points within the convex hull, leading to a satisfactory proof of Radon's Theorem in two dimensions.

PREREQUISITES
  • Understanding of convex quadrilaterals and their properties
  • Familiarity with parametric equations in geometry
  • Knowledge of Caratheodory's Theorem in two dimensions
  • Basic concepts of linear algebra
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  • Study the proof of Radon's Theorem in detail
  • Learn about the applications of Caratheodory's Theorem in higher dimensions
  • Explore the geometric interpretations of convex hulls
  • Investigate parametric equations and their role in geometric proofs
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Mathematicians, geometry enthusiasts, and students studying algebraic proofs related to convex shapes and their properties.

LukeD
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Looking for an algebraic proof that a convex quadrilateral has intersecting diagonals

So I'm trying to find an algebraic proof that the diagonals of a convex quadrilateral intersect because I'm working on a proof of a generalization of this idea into higher dimensions and I really have no idea where to start.

Although this specifically isn't for a class, the ultimate proof that I'm working on is, so I'd rather not have anyone address the generalization.

I don't really know where to start. When discussing it in class, some people gave geometric arguments, but that's not quite what I'm looking for.

Could someone give me some pointers or at least an idea of where to start?

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Edited the title so that it wouldn't be misleading
 
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Well, first, you would want to prove, algebraically, that the diagonals are not parallel- that they do NOT have the same slopes. That means that they must intersect somewhere in the plane. Then use the definition of "convex" to show that, since the straight line segments between two points lies within the quadrilateral, so must the intersection of the two points. I think that writing parametric equations for the diagonals, so that endpoint of a diagonal corresponds to t=0 and the other to t= 1 should suffice.
 
I haven't thought about the problem in a while, but embarrassingly, I still have yet to prove it. Sorry, what I should have said is that I am trying to prove this using linear algebra.

I already know that given any point in the quadrilateral (convex hull of 4 points, no 3 collinear) that I can at least one way (in fact, I think I can always find at least 2 ways, though I haven't tried yet to prove this, and I'm not convinced it'd be all that useful) to choose 3 points so that the point is in the convex hull of those. (Caratheodory's Theorem in 2 dimensions)

It feels to me like I should be able to prove that the diagonals of the quadrilateral intersect fairly easy with Caratheodory's Theorem, but I'm just not seeing it for some reason.

By the way, just to be clear, I am trying to prove Radon's Theorem. We were posed Helly's Theorem in 1 and 2 dimensions in class, which were easy enough to prove, but we assumed Radon's Theorem in 1 and 2 dimensions for this because it's easy enough to prove using geometric arguments. In general though, I do not know how to prove it in n dimensions (which is why I'm still trying to prove it in 2 dimensions).

Yes, I know that there is a proof of Radon's Theorem on Wikipedia. I'm trying to resist all temptation to look at it.
(Also, I did not know that these theorems were named until I happened to stumble across articles about them on MahWorld, which thankfully did not have proofs)

Anyway, still trying to figure this out...
 
Actually, I just saw how there being two ways to pick the n+1 points so that your point is in the convex hull of those n+1 points immediately falls out of the proof of Caratheodory's Theorem. Since when you eliminate one of k > n+1 points, you have a choice of two different ones you can eliminate. Still don't know how this would help though. Also, that there are 2 ways seems pretty arbitrary when dealing with n dimensions, but it seems that there are only ever guaranteed to be 2 ways in general.
 
AHA! I got it! :D

Ha, and I thought that the existence of two ways to choose your n+1 points was meaningless, but that was all that I needed for the proof since I just set those equal, move all of the coefficients around and then divide by the proper thing.

Ah... I'm quite satisfied now.
 

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