# If we know the congruence of the apex and the base of two isosceles, are both congruent triangles? (help with the proof)

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.

## Proof

Let △ABC and △A'B'C' be isosceles triangles like the next figure 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'?

## Edit

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.

– user822140
Jan 3, 2021 at 5:01

If you know that the sum of angles in a triangle are $$180^\circ$$, then, if the apex has value $$A$$, the other two angles are $$90^\circ-A/2$$. Then the problem is reduced to the angle-side-angle case. The side is the base.

In your case you use the same angles to show $$D'E'||B'C'$$. Then the triangles are similar, and the ratio of the sides is equal to the ratio of the bases. You know that the bases are the same, so then the sides must be the same. The problem is then reduced to side-side-side case.

Your proof is correct - good job there.

An easier way to see this is the following:

1. You have an isoceles triangle, so $$\angle ABC=\angle ACB$$ (for both).
2. You know $$\angle BAC=\angle B'A'C'$$ (by hypothesis).
3. Angles add up to $$180^{\circ}$$, so $$\angle ABC=\angle A'B'C'=90^{\circ}-\angle BAC$$.
4. Therefore, $$\triangle ABC$$ is similar to $$\triangle A'B'C'$$.
5. You additionally know that length $$BC=B'C'$$.

Well, they're now congruent (by SAA or equivalent).