Ceva, Desargues and Pascal's theorems for conics

Question

I was told in class today that these three theorems are valid in projective geometry and with conic sections (I'm taking a modern geometry class) but I can't seem to find proofs anywhere online, and I've googled them. Would they be very different from their standard forms? Can you please help me with the proofs?
Thanks

Answer 1

The theorem of Desargues is true in the real projective plane, but not in every projective plane: there exist non-Desarguesian planes. I'm not perfectly sure, but I assume that Pascal's theorem should be not only a generalization of but also a consequence of Pappos' theorem, which in turn means you have a projective plane over some field. For Ceva's theorem you obviously need lengths, unless you reformulate it in terms of cross ratios. In any case, when you write “valid in projective geometry”, I guess what you really mean is “valid in the real projective plane”.

In a certain sense, all of these theorems are not only valid in that real projective plane, but essentially live in projective geometry. If you reformulate them for Euclidean geometry, you'll have to handle a bunch of special cases which are not special at all in projective geometry.

Take Desargues' theorem, for example. You can formulate that like this:

If $Aa$, $Bb$ and $Cc$ meet in a single point, then the intersections of $AB$ with $ab$, $AC$ with $ac$ and $BC$ with $bc$ lie on a single line.

But that's the projective formulation. The corresponding Euclidean formulation would be something like this:

If $Aa$, $Bb$ and $Cc$ either meet in a single point or are parallel to one another, then one of three things can happen. Either the intersection points of $AB$ with $ab$, $AC$ with $ac$ and $BC$ with $bc$ all exist and lie on a single line. Or none of these intersection points exists, since all three pairs of lines are parallel. Or exactly one pair of lines is parallel, in which case the line joining the other two points of intersection is parallel to them as well.

Obviously this looks a lot more complicated and bulky. That's one good reason to interpret the theorem in projective geometry, where parallelism is no special case. The proof in projective geometry is easier as well, since one doesn't have to proove the special cases either.

Answer 2

Personally, I'm not aware of any connection of these theorems with conic sections other than the possibility of a conic section appearing in the diagram describing the configuration. The theorem itself would be in whatever geometry (real projective/Euclidean) that we are talking about.

And even then, the only one I'm aware of at present is for Pascal's theorem. The following scan is taken from Kaplansky's Linear algebra and geometry: a second course page 119, an awesome and inexpensive book which I would recommend to anyone:

Here are the basic things to keep in mind when thinking about these theorems:

1. affine and projective geometry (in the sense of Artin or Kaplanksy) in 3 or more dimensions always satisfy Desargues' theorem.
2. Some planes do not satisfy Desargues theorem or Pascal's theorem
3. Planes which can be built as $D\times D$ for a division ring $D$ are Desarguesian, in fact, this characterizes Desarguesian planes.
4. If a plane $D\times D$ for a division ring $D$ satisfies Pascal's theorem, $D$ is actually a field.
5. A chief tool for converting affine or Euclidean theorems to projective theorems is to replace orthogonality, midpoints and distances with the cross ratio. It's rather involved for me to explain here, but Kaplansky's book does a great job, as I imagine Coxeter's books do too.