Angle-Side-Angle (ASA) Congruence:

Case 1:
We need to show that if A, B, C are distinct points, and A', B', C' are distinct points, with AB = A'B', a(BAC) = a(B'A'C') and a(CBA) = a (C'B'A'), then the triangles ABC and A'B'C' are positively congruent.

Proof:
By definition, we can construct triangle ABC and triangle A'B'C' from the distinct points A, B, C and A', B', C'. Now, a construction is needed Create the rays A'C' and B'C'. Create c1, the circle with center B' and radius length BC. Create D, the intersection of ray B'C' and circle c1.

So we have AB= A'B', a(ABC) = a(A'B'D), and (from the construction) BC=B'D. And by SAS Congruence, we can say that triangle ABC is positively congruent to triangle A'B'D. Now, by this theorem we know that a(BAC) = a(B'A'D). Now, from the construction, we can use the Addition Angle Axiom to say that a(B'A'D) = a(B'A'C') + a(C'A'D). Recall that we were given a(BAC) = a(B'A'C'). So, we could say that a(B'A'C') = a(B'A'D). The only way that a(B'A'D) = a(B'A'C') + a(C'A'D) and that a(B'A'C') = a(B'A'D) is if a(C'A'D) = 0. Now from the Ray Angle Axiom this happens if and only if D lies on the ray A'C'. So now we have D laying on the ray A'C' and on the ray B'C'. Well C lies on the ray A'C' and on the ray B'C', so, D and C' must be the same point. So now we have triangle A'B'C =triangle A'B'D and by transitivity we can say that triangle A'B'C is also positively congruent to triangle ABC.

Case 2:
We need to show that if A, B, C are distinct points, and A', B', C' are distinct points, with AB = A'B', -a(BAC) = a(B'A'C') and -a(CBA) = a(C'B'A'), then the triangles ABC and A'B'C' are negatively congruent.

Proof:
By definition, we can construct triangle ABC and Triangle A'B'C' from the distinct points A, B, C and A', B', C'. Now, the same construction as above is made for this case Create the rays A'C' and B'C'. Create c1, the circle with center B' and radius length BC. Create D, the intersection of ray B'C' and circle c1.

So we have AB= A'B', -a(ABC) = a(A'B'D), and (from the construction) BC=B'D. And by SAS Congruence, we can say that DABC is negatively congruent to DA'B'D. Now, by this theorem we know that -a(BAC) = a(B'A'D). Now, from the construction, we can use the Addition Angle Axiom to say that a(B'A'D) = a(B'A'C') + a(C'A'D). Recall that we were given -a(BAC) = a(B'A'C'). So, we could say that a(B'A'C') = a(B'A'D). The only way that a(B'A'D) = a(B'A'C') + a(C'A'D) and that a(B'A'C') = a(B'A'D) is if a(C'A'D) = 0. Now from the Ray Angle Axiom this happens if and only if D lies on the ray A'C'. So now we have D laying on the ray A'C' and on the ray B'C'. Well C lies on the ray A'C' and on the ray B'C', so, D and C' must be the same point. So now we have triangle A'B'C =triangle A'B'D and by transitivity we can say that triangle A'B'C is also negatively congruent to triangle ABC.

Thus, by utilizing the SAS Congruence Axiom, we have shown that the Angle-Side-Angle (ASA) Congruence Theorem is, in fact, a valid congruence theorem.

Side-Side-Side (SSS) Congruence:
We need to show that if A, B, C are distinct points, and A', B', C' are distinct points with AB=A'B, AC = A'C' and BC = B'C', then the triangles ABC and A'B'C' are positively or negatively congruent.

Proof:
By definition, we can construct the triangles ABC and A'B'C'. Now, a construction is needed create the angle BCA'' so that a(BCA'') = -a(B'C'A'). Also, create the angle CBA'' so that a(CBA'') = -a(C'B'A'). Essentially, the Completeness Angle Axiom let's us do this. Now, create A'', the intersection of the rays CA'' and BA''. Now, construct the segment A''A.

So, we have BC=B'C' and from our construction a(BCA'') = -a(B'C'A') and a(CBA'') = -a(C'B'A'). So, by ASA Congruence we can say that triangle A''BC is negatively congruent to triangle A'B'C'. Now, we can say that A''C = A'C' and that A''B = A'B'. Recall that AB = A'B' and that AC = A'C'. So, by transitivity we can say that A"C = AC and that A"B = AB. So, by definition, triangles ACA'' and ABA'' are isosceles triangles. Now, the isosceles triangle theorem states that a(BAA'') = -a(A''AB), and a(CAA'') = -a(A"AC). So, consider the Addition Angle Axiom: we have a(BAC) = a(BAA'') + a(CAA''). This can be re-written as a(BAC) = -a(A"AB) + -a(A"AC) and a(BA"A) + a(CA"A) = a(BA''C). So, a(BAC) = -a(A"AB) + -a(A"AC) = -(a(A"AB) + a(A"AC)) = -a(BA"C), i.e. a(BAC) = -a(BA"C). So, by SAS congruence, we have triangle A"BC is negatively congruent to triangle ABC. So, by transitivity, we have triangle A'B'C' positively or negatively congruent to triangle ABC.

The axioms used during this investigation were:
1) Congruence (SAS)
2) Addition Angle
3) Ray Angle
4) Completeness Angle

The definitions/properties used during this investigation were:
1) Segment
2) Triangle
3) Positively Congruent
4) Ray
5) Circle
6) Transitivity
7) Negatively Congruent
8) Isosceles Triangle

The theorems used during this investigation were:
1) Isosceles Triangle
2) ASA Congruence (once proven)