Triangle Angle-Side-Angle Equality

Theorem
If two triangles have:
 * two angles equal to two angles, respectively
 * the sides between the two angles equal

then the remaining angles are equal, and the remaining sides equal the respective sides.

That is to say, if two pairs of angles and the included sides are equal, then the triangles are congruent.

Proof

 * Triangle ASA Equality.png

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.

WLOG, 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:
 * $\triangle ABC = \triangle DEF$