ASA and AAS
Up until now, sides have been hogging the spotlight (they're really hammy). If we give angles the floor for a bit, maybe they'll make some interesting discoveries for us.
If we have two angles whose sum is strictly less than 180° and we draw them on either side of a segment, we'd have something that looks like this.
Imagine pointing a laser off of each end at the desired angle. There's only one point where these laser lines will intersect. In other words, specifying two angles and the length of the line segment between them can give us only one triangle.
That means we can determine whether two triangles are congruent by knowing two angles and the included side. That's called the Angle Side Angle Postulate, abbreviated ASA (or ASAP, which is when we should use it).
Is it true that ∆ABC ≅ ∆DEF? How do you know?
Let's take a look at what we have. We have two sets of congruent angles: ∠A ≅ ∠D and ∠C ≅ ∠F. The segments in between them, AC and DF, are also congruent. If all of that is true (which it is), then we can say that ∆ABC ≅ ∆DEF by ASA (and no, we don't mean the American Society of Anesthesiologists).
One important fact to consider: Knowing two angles in a triangle can automatically give us the third angle, thanks to the triangle Angle Sum Theorem.
This means that knowing any two angles and one side is essentially the same as the ASA postulate. Since the only other arrangement of angles and sides available is two angles and a non-included side, we call that the Angle Angle Side Theorem, or AAS.
A quick thing to note is that AAS is a theorem, not a postulate. Since we use the Angle Sum Theorem to prove it, it's no longer a postulate because it isn't assumed anymore. Basically, the Angle Sum Theorem for triangles elevates its rank from postulate to theorem.
Don't believe us?
In the image, we can see that AC ≅ DF. We also know that ∠CAB ≅ ∠FDE and ∠ABC ≅ ∠DEF. So this is a generic AAS situation.
The triangle Angle Sum Theorem tells us that all the interior angles in a triangle add up to 180°. We can rewrite that into equations.
m∠BCA + m∠ABC + m∠CAB = 180
m∠EFD + m∠DEF + m∠FDE = 180
Solving for ∠BCA and ∠EFD shouldn't be a problem.
m∠BCA = 180 – m∠ABC – m∠CAB
m∠EFD = 180 – m∠DEF – m∠FDE
It follows by substitution that ∠BCA ≅ ∠EFD.
Since knowing any two angles is equivalent to knowing all 3 angles, the ASA postulate can be generalized to this: If we have two triangles ∆ABC and ∆DEF and AB is congruent to DE, the two triangles are congruent as long as any of two corresponding angles are congruent.