In math class, I was told we need to take SAS as an axiom, otherwise we could not prove any congruences besides a triangle and itself. Is that really true? Is there a model of Hilbert's Euclidean geometry axioms (minus SAS and other congruence axioms), where the only way for triangles to be congruent is for them to be the same triangle?
Can one prove non-trivial congruences of triangles without SAS or other congruence axioms
axiomatic-geometryaxiomsgeometry
Related Solutions
Cut-the-Knot's SSS proof page has a number of solutions, including Euclid's. As the author indicates, however, only Hadamard's proof "goes through without a hitch", with the important aside: "assuming of course that isosceles triangles have been fairly treated previously".
I'll give a full development of Hadamard's argument, including the necessary bits needed about isosceles triangles. I include a couple of "obvious" sub-proofs just to make clear which axioms are in play.
Preliminaries:
- SAS triangle congruence is an axiom.
- (1) implies one direction of the Isosceles Triangle Theorem, namely: If two sides of a triangle are congruent, then the angles opposite those sides are congruent. $[\star]$
- (2) implies that A point equidistant from distinct points $P$ and $Q$ lies on the perpendicular bisector of the $\overline{PQ}$. $[\star\star]$
Now, we'll cut to the part of Hadamard's argument where we have $\triangle ABC$ and $\triangle A^\prime BC$ with corresponding edges congruent, constructed with $A$ and $A^\prime$ on the same side of $\overleftrightarrow{BC}$. (That last part is key.) We need only show that $A$ and $A^\prime$ coincide to prove SSS. Hadamard makes this argument by contradiction ...
- Assume $A \neq A^\prime$.
- $B$ is equidistant from $A$ and $A^\prime$, and therefore, $B$ lies on the perpendicular bisector of $\overline{AA^\prime}$ by (3) above. The same is true of $C$.
- Therefore, $\overleftrightarrow{BC}$ is the perpendicular bisector of $\overline{AA^\prime}$.
- Therefore, $\overleftrightarrow{BC}$ separates $A$ and $A^\prime$.
- This contradicts the fact that we constructed $A$ and $A^\prime$ to be on the same side of $\overleftrightarrow{BC^\prime}$. $[\star\star\star]$
- Therefore, the assumption that $A \neq A^\prime$ must be invalid, so that triangles $\triangle ABC$ and $\triangle A^\prime BC$ coincide; this gives us SSS. $\square$
$[\star]$ Proof: Isosceles $\triangle ABC$ with base $\overline{BC}$ is congruent to $\triangle ACB$ by SAS, and those opposite angles are corresponding angles of the congruent triangles.
$[\star\star]$ The midpoint, $M$, of $\overline{PQ}$ is certainly on the perpendicular bisector. A point $R \neq M$ equidistant from $P$ and $Q$ creates isosceles $\triangle RPQ$ with congruent angles at $P$ and $Q$ by (2). Thus, $\triangle RPM \cong \triangle RQM$ by SAS. Corresponding angles $\angle RMP$ and $\angle RMQ$ must be congruent; as supplements, they must be right. $\overleftrightarrow{MR}$, then, is the perpendicular bisector of $\overline{PQ}$. (I'll note that an easier proof could assert that, from the get-go, $\triangle RPM \cong RQM$ by SSS ... but we can't use that argument in a proof of SSS itself.)
$[\star\star\star]$ The contradiction here is one of Euclid's oversights later made explicit by Hilbert as the Plane Separation Axiom. The PSA has slightly-different formulations in different treatments of geometric foundations, but effectively it's the "Because I said so!" assertion that lines break planes into two disjoint "sides" and gives us our contradiction.
I think I have a correct proof now.
Proof by contradiction: Assume to the contrary that two lines parallel to the same line are not parallel to each other. Without loss of generality, assume line m and line n are parallel to a line l, but m and n are not parallel to each other. Then, m and n intersect at a point, P that is not on line l. However, this contradicts Axiom 5 because two lines would be containing P and be parallel to l. So the assumption that m and n are not parallel was incorrect. Thus, m and n are parallel to l and also parallel to each other.
Best Answer
Without the SAS axiom, there is very little that constrains the congruence relation on angles; it just has to be an equivalence relation that satisfies the "copying an angle" axiom (given any angle, there is a unique congruent angle on a given side of any ray). So, you could start with the usual model $\mathbb{R}^2$ of Hilbert's axioms (or $\mathbb{R}^3$ if you are doing the 3-dimensional version), and then redefine its angle congruence relation in some nasty way that still satisfies the copying axiom. For instance, suppose that for each $P\in\mathbb{R}^2$ you specify a bijection $A_P:(0,\pi)\to(0,\pi)$. Then you could define an angle $\alpha$ at a point $P$ to be congruent to an angle $\beta$ at a point $Q$ iff $A_P(a)=A_Q(b)$, where $a$ and $b$ are the usual radian angle measures of $\alpha$ and $\beta$, respectively.
In particular, by choosing all these bijections $A_P$ one element at a time by a transfinite recursion of length $\mathfrak{c}$, you can arrange that there are no non-equal triangles that are congruent. At each step where you need to define a new value of some $A_P$, there are fewer than $\mathfrak{c}$ different triangles whose angles you have already specified, and so you can pick a value that avoids repeating any of those angles. Similarly, at each step where you need to define a new value of some $A_P^{-1}$ (to make sure each $A_P$ is surjective), you can choose it to avoid being equal to the angle measure in any triangle with $P$ as a vertex such that you have already chosen the other two angles.