Is there a link between the sideangleside 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?

2$\begingroup$ You would probably do better to use either Hilbert's or Tarski's axioms, since Euclid's axioms aren't rigorous. In fact, nonPasch geometries arguably satisfy Euclid's axioms. math.stackexchange.com/questions/352566/… $\endgroup$– Chill2MachtJul 4, 2017 at 12:24
8 Answers
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:
http://en.wikipedia.org/wiki/Absolute_geometry
The first 15 propositions of Euclid hold in Elliptic geometry(no parallels) see:
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI16.html
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:
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.

1$\begingroup$ "Sufficient unto the day is the rigor thereof." [E. H. Moore] $\endgroup$ Dec 3, 2009 at 19:38
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.
Sideangleside 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.

7$\begingroup$ Before anyone comes and closes this question, I'd like to say that I am a "fulltime mathematician" and I was interested to read Michael's answer (I don't mean by this preemption to complain about the moderators who are, of course, extremely awesome). $\endgroup$ Nov 12, 2009 at 21:59
On reflection, SAS tells me that Euclidean geometry has a strong semilocal 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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1$\begingroup$ So when you say that SAS gives local isotropy, does that sort of mean/imply a Riemannian metric structure? (An "inner product" locally around every point?) But then one needs the parallel postulate to get the "flatness" required for the inner product to be defined on the entire space? $\endgroup$ Jul 3, 2017 at 11:05
You would probably do better to use either Hilbert's or Tarski's axioms, since Euclid's axioms aren't rigorous. In fact, nonPasch geometries (see here) arguably satisfy Euclid's axioms.
In other words, since Euclid had a muddled understanding of Euclidean geometry, attempting to use his axiomatic system without modifications will lead to you having a muddled understanding of Euclidean geometry as well.
They are separate axioms in two rigorous axiomatizations of Euclidean geometry.
In both of the (rigorous) axiomatic systems mentioned above, the SAS postulate (fivesegment axiom in Tarski's system and the last congruence axiom in Hilbert's system) and the parallel postulate (axiom 10 of Tarski, and here in Hilbert) are different axioms.
I don't know how much has been proven for Hilbert's axioms, but since Tarski's axiom system is simpler, most of the axioms have been proven to be independent of one another, including the SAS and the parallel postulate. (See e.g. here.)
There exist valid geometries satisfying SAS but not the parallel postulate.
Both Hilbert's and Tarski's axioms, which include SAS as one of the axioms, can also be used to create axiom systems for neutral geometry (by omitting the parallel postulate) and for hyperbolic geometry (by negating the parallel postulate). Since these axiom systems for hyperbolic geometry, which contain both SAS and the negation of the parallel postulate, are also consistent, this also shows that the parallel postulate is independent of SAS. (This is essentially the same argument made by Kristal Cantwell in another answer.)
There exist valid geometries satisfying the parallel postulate but not SAS.
Finally, in taxicab geometry, the parallel postulate does hold, but SAS doesn't. See here:
...the Taxicab Plane satisfies all of the usual axioms (rules) of the Euclidean plane except one of the congruence axioms... We have two triangles obeying SAS which are not congruent!
Since there exist consistent systems of geometry in which (1) the parallel postulate doesn't hold but SAS does, and (2) the parallel postulate does hold but SAS doesn't, we can conclude with certainty that the two are independent of one another.
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.

2$\begingroup$ 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. $\endgroup$ Dec 3, 2009 at 19:42

1$\begingroup$ 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. $\endgroup$ Dec 3, 2009 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!

3$\begingroup$ Interestingly, I read somewhere that you shouldn't believe everything you read. $\endgroup$ Nov 13, 2009 at 22:48