Geometry – Can S.S.S. Congruence Criterion Prove S.A.S. and A.S.A.?

axiomatic-geometrygeometry

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.?

Best Answer

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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