Triangle Side-Angle-Angle Equality

Theorem
If two triangles have:
 * two angles equal to two angles, respectively
 * the sides opposite one pair of equal 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 a pair of opposite sides are equal, then the triangles are congruent.

Proof

 * Euclid-I-26-2.png

Let:
 * $\angle ABC = \angle DEF$
 * $\angle BCA = \angle EFD$
 * $AB = DE$

that $BC \ne EF$.

If this is the case, one of the two must be greater.

, let $BC > EF$.

We construct a point $H$ on $BC$ such that $BH = EF$, and then we construct the segment $AH$.

Now, since we have:
 * $BH = EF$
 * $\angle ABH = \angle DEF$
 * $AB = DE$

from Triangle Side-Angle-Side Equality we have:
 * $\angle BHA = \angle EFD$

But from External Angle of Triangle is Greater than Internal Opposite, we have:
 * $\angle BHA > \angle HCA = \angle EFD$

which is a contradiction.

Therefore $BC = EF$.

So from Triangle Side-Angle-Side Equality:
 * $\triangle ABC = \triangle DEF$