Triangle Angle-Side-Angle Equality
If two triangles have:
In the words of Euclid:
- If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle equal to the remaining angle.
Let $\angle ABC = \angle DEF$, $\angle BCA = \angle EFD$, and $BC = EF$.
Aiming for a contradiction, suppose that $AB \ne DE$.
If this is the case, one of the two must be greater.
Without loss of generality, we let $AB > DE$.
We construct a point $G$ on $AB$ such that $BG = ED$.
Using Euclid's first postulate, we construct the segment $CG$.
Now, since we have:
- $BG = ED$
- $\angle GBC = \angle DEF$
- $BC = EF$
it follows from Triangle Side-Angle-Side Equality that:
- $\angle GCB = \angle DFE$
But from Euclid's fifth common notion:
- $\angle DFE = \angle ACB > \angle GCB$
which is a contradiction.
Therefore, $AB = DE$.
So from Triangle Side-Angle-Side Equality:
- $\triangle ABC = \triangle DEF$