# Riemann proof involving non-Euclidean geometry

I've been searching the internet for a proof of Riemann's theorem involving Riemann (elliptic) geometry, all the perpendiculars to a straight line meet in a point, and I can't seem to find it.

I've been reading up on non-Euclidean geometry and while I think I understand the idea being put forth (an example being the equator as the base, and longitudinal lines extending from the equator to the north pole), I don't have the mathematical know-how to deduce the proof myself, and I think that reading over it would be helpful.

If anyone has a link to a site, or a chapter in a book, where I could find that proof I would really appreciate it!

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I don't understand the statement of the theorem. What kind of geometry is this line in? The statement is clearly false in the Euclidean plane. – Qiaochu Yuan Apr 4 '11 at 4:35
In Riemann (elliptic) geometry. I'll edit my post for clarification. – Ryan Apr 4 '11 at 4:37
What is your definition of elliptic geometry? There are axiomatic and non-axiomatic approaches which are quite distinct. – Qiaochu Yuan Apr 4 '11 at 4:41
"Parallel" lines eventually converge at a point. The angles of a triangle add up to more than 180 degrees. I've only read a couple of chapters on the topic, so I'm not really sure what the different approaches you speak of entail. Axiomatic seems like it would be more straight forward. Is that true? Either one would fine. – Ryan Apr 4 '11 at 5:04
Sometimes people just go ahead and use e.g. the 2-sphere as a model, in which case the statement is totally obvious. But the real thing to do would be to deduce the statement from definitions. I haven't looked at it in a while, but I'd imagine that what you're looking for is contained in Greenberg's "Euclidean and Non-Euclidean Geometries". – Aaron Mazel-Gee Apr 4 '11 at 7:40