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In all of the treatments of elementary Euclidean geometry which I've seen so far, the section about triangle congruences introduces S.A.S. criterion as the basic postulate from which A.S.A. and S.S.S. criteria are deduced. I remember reading somewhere that one could choose any one of these three as "the congruence postulate" and deduce others from it.

I am able to produce proofs for S.A.S. by taking A.S.A. as an axiom and vice versa, but S.S.S. seems to be the "odd" one since I cannot reach either S.A.S. or A.S.A. by taking it as the axiom. I was unable to find anything online that shows such a proof so my question is whether the premise that any one of these three criteria can be picked as the axiom is true or not. If it is, how can we prove, for example, S.A.S. through S.S.S.?

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    $\begingroup$ I disagree with the proposed approach. First, the postulates are not isolated statements but have basic geometrical axioms as background. Second, two triangles are congruent if their respective sides and angles are congruent. From this definition and basic geometry axioms we deduce all three triangle congruence criteria. See, for instance, Highest geometry by Efimov N.V. (7th edn., M.FIZMATLIT, 2004, in Russian) Theorem 14 (for S.A.S), Theorem 15 (for A.S.A), and Theorem 18 (for S.S.S), if I understood the abbreviations right. $\endgroup$ Nov 25, 2023 at 9:43
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    $\begingroup$ @AlexRavsky You are right, in principle. But Axiom III,5 in Efimov's book is in practice equivalent to SAS criterion, and could be replaced by it. I think the question could be rephrased as follows: is it possible to substitute Axiom III,5 with an equivalent axiom which doesn't mention the angles of the triangle? $\endgroup$ Nov 25, 2023 at 18:30
  • $\begingroup$ If nothing you know mentions angles, how can you prove anything about angles? $\endgroup$
    – GEdgar
    Nov 30, 2023 at 16:19
  • $\begingroup$ @GEdgar We know something about angles, from other axioms. But you are right: I don't think it's possible to prove what the OP asks. $\endgroup$ Nov 30, 2023 at 16:47
  • $\begingroup$ @AlexRavsky: For those of us without Efimov's book (or knowledge of Russian), would you please give the "basic geometry axioms" used therein, and perhaps a sense of how they prove SAS, ASA, SSS? An online reference would be fine. Thanks! $\endgroup$
    – Blue
    Nov 30, 2023 at 23:23

2 Answers 2

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Taking David Hilbert's axioms of geometry, without SAS (III.6), and adding SSS, does not recover SAS (as mentioned in the comments).

We take as evidence:

  1. The bounty has been offered a long time without a solution. (Cryptographers will sympathize with this reasoning.)
  2. If one examines the remaining axioms, there are too few that grant segment congruence.

The SAS postulate confers, in models of the geometry, all lengths constructible from the Law of Cosines. But, the remaining axioms without SAS allow one to only

  • (III.1) translate/rotate an existing segment from a point along a ray, or to
  • (III.3) combine two adjacent co-linear segments into one.

The SSS postulate grants congruence of angles, but not of segments. (E.g., two angles of an isosceles triangle are congruent.)

Is there a model of Hilbert's axioms with SSS but without SAS?

Consider a geometry of points in $\mathbb{R}^2$, where off-axis segments exist but are congruent only to themselves (not their translates). There, SSS holds vacuously, since only pairs of degenerate triangles have three corresponding congruent sides. SAS does not hold, since any pair of distinct triangles with an off-axis side are not congruent. Check that axiom III.3 (combining co-linear segments) holds in this model.

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  • $\begingroup$ So do you think it is right to conclude the following? Since we can deduce S.A.S. and A.S.A. from each other, in order to build our Euclidean geometry, we need to take either the first or the latter as our congruence axiom and move on from there. However, S.S.S. cannot be taken as the primary axiom since it cannot be used to deduce the other two, therefore one must take as a congruence axiom a criterion that includes at least one angle in its definition (namely, S.A.S. or A.S.A.) $\endgroup$
    – jacob78
    Nov 30, 2023 at 22:45
  • $\begingroup$ I modified the counterexample to use R^2 instead of Z^2. @jacob78, yes, provided you can derive SAS from ASA using only the other axioms. $\endgroup$ Nov 30, 2023 at 22:57
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Yep they all can be derived by taking one as an axiom. But I doubt that it can be done without trigonometry. Here's how I proved SAS and ASA using SSS as an axiom.

Page 1 Page 2 Page 3 Page 4

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  • $\begingroup$ Please don't post handwritten formulas as answers. Here's a mathjax tutorial :) $\endgroup$
    – Ricky
    Nov 30, 2023 at 15:23
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    $\begingroup$ A proof using trigometry or Pythagoras' theorem is hardly valid, unless you show that you can prove those theorems you used without SAS and ASA. $\endgroup$ Nov 30, 2023 at 15:27
  • $\begingroup$ @Ricky I'm sorry mate, I usually do type my answers, but this was just so long I didn't think I could do it with mathjax. Hope you'll understand. $\endgroup$
    – IshA
    Nov 30, 2023 at 15:30
  • $\begingroup$ @Intelligentipauca Pythagoras and Trigonometry are completely independent of SAS and ASA. Trigonometry is basically just a fundamental definition of the functions (or ratios) and Pythagoras has a lot of proofs, some of which do not require similarity or congruency of triangles. Here's an example - en.wikipedia.org/wiki/Pythagorean_theorem (Refer to algebraic proofs) $\endgroup$
    – IshA
    Nov 30, 2023 at 15:36
  • $\begingroup$ @IshA Really? As far as you consider trigonometry and Pythagoras in a purely algebraic way, then you cannot apply them to geometry. Just to prove that two triangles with the same angles have proportional sides requires a lot of geometry. $\endgroup$ Nov 30, 2023 at 16:43

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