Well, there are general criteria for congruence of triangles:
- SAS (Side-Angle-Side)
- ASA (Angle-Side-Angle)
- SSS (Side-Side-Side)
- SAA (Side-Angle-Angle)
And there are some criteria for specific triangles:
- SSA (Side-Side-Angle) The side in front of the angle doesn't have to be minor of the adjacent one
- HC (Hypotenuse-Cathetus) In a right triangle
- HA (Hypotenuse-Angle) In a right triangle
And my question is about one case that I know is true in Euclidean geometry (which I would call AB—Apex-Base—):
If two isosceles triangles have in common the apex and the base, both are congruent
It's "obvious" but the proof is not that obvious. Here's my attempt.
Now, if AB = A'B' both triangles are congruent by SAS, but suppose that's not the case. Without loss of generality assume that AB<A'B', thus it's possible to mark a point D' in A'B' such that A'D' = AB, likewise it's possible to mark a point E' in A'C' such that A'E' = AC. Join the points D'E' to form the △A'D'E' ≅ △ABC by SAS.
Then we must conclude that D'E' = B'C', but this is a contradiction since D'E'<B'C', so assuming AB ≠ A'B' is absurd, hence AB = A'B' which derives △ABC ≅ △A'B'C' Q.E.D.
That's what I did, but I feel that the contradiction is not fully handled, especially because I cannot find another argument but visual that D'E' < B'C'. As I mentioned before, I know this is true in Euclid's (plane) geometry. I'm more interested in a proof (if exist) avoiding parallels (or equivalently the constant sum of the measures of the angles of any triangle), so that the result would be a neutral geometry like the SAA criterion (https://www.jstor.org/stable/27960835?seq=1).
Or I'm trying to prove a result that's not neutral geometry that in some cases D'E' =B'C'?
Thanks for your answers, they are very useful. But up this moment all of them depends on the parallel postulate which I try to avoid in order to find out if this is a neutral geometry result.