Congruence of angles axiom

Question

In my geometry book, the following statement is provided as axiom of congruence.

Let $ABC$ and $A'B'C'$ be two triangles. If $AB \equiv A'B'$, $AC \equiv A'C'$ and $ \angle BAC \equiv B'A'C'$ then $ \angle ABC \equiv A'B'C'$

I honestly have no idea where this comes from. Note that we use this axiom to prove $SSS$, $ASA$ and $SAS$.

Could someone provide intuition for something like this? It has to be obvious if it's used as an axiom.. right? I'm having trouble memorising something like that and using it to prove things. Not to mention the order here is extremely important.

EDIT: As an extra note in the book it says, if we change it to $AC \equiv A'C'$ , $ AB \equiv A'B'$ and $\angle CAB = \angle C'A'B'$ then $\angle ACB \equiv \angle A'C'B'$

Answer 1

Welcome to axiomatic mathematics!

Axioms are accepted not just on the basis of obviousness.

Yes, the truth of this statement should be obvious, as the comment of @MichaelHoppe remarks.

And yes, axioms and other mathematical statements can have a persnickety precision to them which requires you to memorize symbols in order. All I'll say to this is, you memorized your counting numbers once upon a time when you were much much younger, I'm sure you can memorize this axiom.

But the real point about formulating axioms is that they should be as narrowly formulated as possible, subject to the all important restriction that the entire rest of the theory can be derived from the axioms. Mathematicians HATE to accept things without proof. They just HATE it. If they have to accept an axiom, they want it to be formulated as narrowly as possible. In the case of this axiom, the hypotheses are the same as the stronger SAS theorem, but the conclusion is quite a bit narrower: all the conclusion asserts is equality of one specific angle pair.

The point is that from this narrow conclusion, many broader conclusions may be derived. For example, from the narrow conclusion of equality of one angle pair, one may prove equality of the other angle pair, as you have correctly noted in the edit at the bottom of your question. And then one can go on to prove other theorems about triangles such as SSS and so on. All from just one, narrow, very specifically formulated, special version of SAS! A mathematician loves that.

Answer 2

I am not sure this is a satisfactory answer.

That assertion is essentially Euclid's Proposition 4 (SAS) in Book I:

If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides.

The notes on Euclid's proof at
https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI4.html comment clearly on his use of an essentially undefined notion of congruence based on superimposing one triangle on another.

I suspect the author of your text was sufficiently unhappy with Euclid's argument that they created a new axiom instead, in order to get started on the interesting theorems.

You will probably never need to use it directly as an axiom once you've used it to prove SSS, ASA and SAS.

Answer 3

The assumptions are identical to $SAS$ so why is it an Axiom ? They should be able to prove it from previous theorems.