How to prove SSS!

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Hello there, all you eager young Geometers!

As many of you know, there are lots of ways to prove that two triangles are congruent:

"Side-Angle-Side", or SAS: given two triangles, prove that two sides of one are congruent to two distinct sides of the other, and prove that the angle between the one pair of sides in the first triangle is congruent to the corresponding angle in the second.

"Side-Side-Side", or SSS: prove all three sides of one triangle are congruent to the sides of the other.

"Angle-Side-Angle", or ASA: prove that two angles in one triangle are congruent to two angles in the other, and prove that the common side in the first triangle is congruent to the corresponding side in the second triangle.

Now, in Geometry, ideally, we would like to prove everything. (Since we can't do this in real life, we like to do it in Geometry at least, to provide some feeling of security in our uncertain lives.) In particular, we would like to prove that all of the above criteria: SAS, SSS, and ASA, can be proved without assuming anything else. Unfortunately, as far back as 300 BC, roughly, Euclid realized that some things just can't be proved. Some projects like that are doomed to failure, including proving all three of SAS, SSS and ASA. Today, we use Axioms, which are springboard statements, from which we deduce as much as we can; the Axioms themselves, of course, have to be simply assumed.

Of the three criteria, SAS, SSS and ASA, it turns out that the easiest one to assume as an Axiom is: SAS. The other two can be proved from SAS! (and some simple theorems, such as the Isosceles Triangle theorem.)

Rest assured that we're not going to prove SSS here today, but the strategy is interesting! Here's the strategy.

We're given two triangles, ABC and DEF, and AB is congruent to DE, BC is congruent to EF, and the remaining pair of sides is also congruent.

Step 1. Make a third triangle! Copy triangle DEF onto the underside of line AC, using only SAS. In other words, copy the angle D over to the underside of A, mark off a line segment AB* exactly the same size as DE, and complete the triangle. Now DEF and this new triangle -- call it AB*C -- are congruent, by construction. Here's a picture of the situation now:Anyway, the rest of the work consists on proving that ABC and AB*C are congruent to each other. This is not hard, but involve using Isosceles triangles twice (the triangle with two red sides, and another triangle with two blue sides), and more serious work. I don't want to give the impression that things are as easy as just going ahead with the proof as described; for instance, we need to know where the line BB* falls; inside AC or outside? Still, the proof proceeds in essentially predictable ways.

Once we have shown that the two Siamese Twin triangles are congruent to each other, it follows that the two original triangles are congruent, using the principle of "transitivity", which says that if two objects are congruent to a common object, then they're congruent to each other.

So there you have it! Oh what a complex web we weave, when first we start proving theorems...

Archimedes, not the real one