Proof for SSS Congruence? I'm hoping that someone can provide a method for deducing the commonly known SSS congruence postulate? The postulate states

If the three sides of one triangle are pair-wise congruent to the three sides of another triangle, then the two triangles must be congruent.

I am a middle school math teacher (teaching a HS Geometry course) and would like to be able to explain/justify the triangle congruence theorems that I expect students to apply with more clarity. I understand that the Law of Cosines could be used to justify the SSS triangle congruence theorem but I wonder if a proof can use more basic properties. My students are currently learning the congruence theorems and using constructions to justify their validity. I know two ways to demonstrate the validity of SSS congruence: constructions and the Law of Cosines. Is there another way?
I've researched this and have not seen another answer so far. The Law of Cosines is the accepted answer to Proof of ASA , SAS , RHS , SSS congruency theorem. I've also seen a few other questions hinting at this without a clear answer, see: Why is SSS criterion for congruence of triangles referred to as "SSS postulate" in textbooks?. Another question on math.SE asks what I'm asking (how can we derive the triangle congruence postulates from basic axioms?) Need for triangle congruency axioms and the accepted answer suggests only a quick visit to http://en.wikipedia.org/wiki/Taxicab_geometry but does not discuss whether or not the triangle congruence axioms follow from the basic axioms or are independent of them.
 A: Assume the triangles are $ABC$ and $A'B'C'$ with sides $a,b,c$, and $a',b',c'$. First move vertex $A$ to vertex $A'$ (always possible). Then rotate to make coincide the sides $b$ and $b'$ (possible by assumption). The sides $a$ and $a'$ now part from the same point and are equal. So, they are radii of a circle with center at $C=C'$. A similar thing happens with the sides $c$ and $c'$, but with center at $A=A'$. These circles intersect at $B$ and $B'$. 
Applying symmetry with respect to the side $b$ we can assume the points $B$ and $B'$ are on the same side of the line $b$. The circles above are symmetric with respect to the line $b$. If $B$ doesn't coincide with $B'$. Then the circles intersect also at another two points on the other side of the line $b$. This is a contradiction because circles can't intersect at four points.
You were asking about the taxicab geometry. Suppose we have a segment that is diagonal with respect to "the grid" in the Taxicab plane (recall that in the taxicab geometry you first fix two axes to measure the distances so, diagonal means diagonal with respect to these axes). Now consider half the length of this segment. There are infinitely many isosceles triangles, with base in the segment and with sides equal to half this segment. The reason is that the circles with centers in the end-point of the diagonal segment and radius half of its length, contain both an entire segment of the perpendicular bisector of the segment. The picture is two squares like this (http://upload.wikimedia.org/wikipedia/commons/d/de/TaxicabGeometryCircle.svg) sharing a side and the segment if the one joining their centers. These squares are the circles in the Taxicab geometry. The Wikipedia article on the taxicab geometry mentions the taxicab geometry satisfies all Hilbert's axiom except the SAS congruence. 
A: 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. 
A: Proof of SSS is flawed. Once construct is drawn, the implicit assumption is that A and A' are at the same position, i.e, (A = A'). This is essentially assuming what you want to prove. (A and A' could be on the same side of line BC but their positions could be different).
Using the law of cosines may be possible but it is a bit ridiculous since first one needs to prove the Pythagorean theorem.
