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Is there a link between the side-angle-side congruence of triangles and the parallel postulate? Specifically, does it follow from Euclid's first four axioms alone? In fact, does it even follow from all five?

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The first 28 propositions of Euclid's geometry use the first four axioms and are theorems in both hyperbolic(many parallels through a point parallel to a given line) and euclidean geometry(one parallel through a point parallel to a given line). see the following:

The first 15 propositions of Euclid hold in Elliptic geometry(no parallels) see:

So since side angle side is proposition 4 it can hold in all three systems. But as mentioned elsewhere it is one of Hilbert's axioms that were added as part of the process of formalizing Euclid's geometry. Hilbert's book The foundations of geometry is available here:

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Side-angle-side congruence is the fourth proposition of the Elements. Euclid doesn't use the fifth postulate until I.29, which is the equality of alternate interior angles.

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Before anyone comes and closes this question, I'd like to say that I am a "full-time mathematician" and I was interested to read Michael's answer (I don't mean by this pre-emption to complain about the moderators who are, of course, extremely awesome). – David Jordan Nov 12 '09 at 21:59

My understanding was that SAS does not follow from the Euclid's postulates and should instead be considered an axiom itself. See Hilbert's axioms for more detail.

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You don't need the parallel postulate, but you are right to be a bit worried about triangle congruences following from Euclid's axioms. The link provided by Michael Lugo explains the issue: Euclid's proof uses "superposition" but his axioms do not allow him to draw conclusions on this basis.

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"Sufficient unto the day is the rigor thereof." [E. H. Moore] – Gerald Edgar Dec 3 '09 at 19:38

On reflection, SAS tells me that Euclidean geometry has a strong semi-local homogeneity, in that every neighborhood of every point is isotropically isomorphic with some neighborhood of any other point --- once you find a good way to say "neighborhood", that is.

The parallel postulate, on the other hand, can be used to construct *canonical* isomorphisms of point(ed)-neighborhoods --- by parallel translation of course; but since the constructed isomorphisms are all parallel in a good sense, we don't get the isotropy structure without SAS. (edit/add:ed): in the other direction, SAS doesn't give any canonical isomorphisms, which is just as well because hyperbolic and elliptical space both have SAS, but not the parallel postulate. (end edit)

The related postulate that Euclid states properly --- that all right angles are equal --- only gives a pointwise isotropy; it doesn't help much for segments subtended by respectively equal segments at equal angles.

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"On reflection...." Pun intended? – Gerry Myerson Oct 7 '14 at 22:50

Ironically, I had intentions of working the other way around. As Hugh Thomas said, Euclid used a superposition of one figure onto another to demonstrate SAS congruence. My idea was to use a superposition of one figure onto another to demonstrate the parallel postulate, as well, but it never seemed to work well enough. However, regarding what Kristal Cantwell says, using the superposition "axiom" is contrary to hyperbolic geometry, and thus should be able to restrict it to our geometry.

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Your description of Kristal Cantwell's comment seems wrong to me. Though maybe I misunderstand what you mean. Superposition (or, equivalently, SAS) works fine in hyperbolic geometry. So it's good that you couldn't prove the parallel postulate using it. – Hugh Thomas Dec 3 '09 at 19:42
This case, it was a "regarding" rather than "about". She talked about how far into book 1 each of those geometries still holds, and I would have thought that the superposition method doesn't hold for hyperbolic (but that doesn't mean that the proof doesn't hold). Perhaps it's me who's thinking about it properly, because straight lines in hyperbolic geometry don't look the same. I guess it's that I am thinking about hyperbolic in Euclidean space, then, because on the cone, when viewed from above, lines do appear straight. – Gabriel Benamy Dec 3 '09 at 20:08

The proof given by Euclid is incorrect. Interestingly, I read somewhere that by using techniques used by Euclid in his Elements, it is possible to prove that every triangle is equilateral!

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Interestingly, I read somewhere that you shouldn't believe everything you read. – Kevin H. Lin Nov 13 '09 at 22:48

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