Proof of the ASA triangle congruence criterium

Question

So we studied the triangle congruence criteria and we proved the ASA criterium: If two triangles have respectively congruent two angles and the side included in them, then the two triangles are congruent.

The proof is a proof of contradiction, and starts with: Assume the triangles ABC and A'B'C' aren't congruent. Then one side, for example AC, must not be congruent to its equivalent, in this case A'C'.

My question is: How do we know that one side must obligatory not be congruent if the two triangles aren't congruent? I mean, there's the third criterium, but its proof relies on the second. Can someone please illuminate me?

Answer 1

It flows from the definition of congruence: two triangles are congruent when they have same three sides and exactly the same three angles. When we look at ASA case, we can see that both triangles have the same three angles (if two angles are the same, the third one will be the same) so for triangles not be congruent, at least one pair of sides must be different.

Answer 2

Think about it this way: if two triangles are congruent, then in terms of their three sides, it's enough to make one of the sides a little bit shorter or a little bit longer to make these two triangles incongruent. The converse of this is that if two triangles are not congruent, then there must be at least one pair of corresponding sides in both triangles whose lengths are not congruent.

Does that answer your question?