Can one prove non-trivial congruences of triangles without SAS or other congruence axioms? 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?
 A: 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.
