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Desargues' Theorem

Date: 07/03/98 at 13:27:44
From: Jos Verhulst
Subject: Theorem of Desargues

Is there any (fundamental) reason why Desargues' two-triangle theorem 
is so easy to prove in three dimensions, and seems impossible to prove 
when we restrict ourselves to two dimensions?

Date: 07/04/98 at 13:57:22
From: Doctor Floor
Subject: Re: Theorem of Desargues

Hi Jos,

Thank you for sending your question to Dr. Math!

I will try to give you an impression of what is known on Desargues' 
theorem, but a lot of it is quite difficult.

Proving Desargues' two-triangle theorem in two-dimensional geometry is 
usually done using Menelaus' theorem.

Menelaus' theorem says:

   Given a triangle ABC, and points P on BC, Q on AC and R on AB, then  
   P, Q and R are collinear if and only if:

     AR   BP   CQ
     -- * -- * -- = -1
     RB   PC   QA

This theorem can be used in Euclidean and affine geometry, so 
Desargues' theorem can be proved there without jumping to three 
dimensions. But Menelaus' theorem does not make any sense in 
projective geometry, because ratios of distances of 3 collinear 
points, like AR/RB, are not invariant under central projection.

So there has been a big question whether Desargues' theorem could be 
proved without ratios and without third dimension, but just using the 
axioms on connection of points and intersection of lines. In Hilbert's 
investigations in 1899, this is proved to be impossible. I don't know 
the proof though.

The same (it cannot be proved by the mentioned axioms alone) goes for 
Pappus' theorem:

   Given a hexagon A1A2A3A4A5A6, if A1, A3 and A5 are on a line l1, 
   and A2, A4, A6 are one a second line l2, then the points of 
   intersection of opposite sides of the hexagon are collinear.

I tell you this about Pappus' theorem, because it was proved by 
Hessenberg in 1905 that Pappus' theorem implies Desargues' theorem. 
I don't know this proof either, but it is said to be very ingenious. 
So usually Pappus' theorem is added to the axioms in projective 

I hope to have given you a little idea of the answer to your question.

- Doctor Floor, The Math Forum
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Associated Topics:
College Euclidean Geometry
College Higher-Dimensional Geometry

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