Converse Hinge Theorem

Theorem
If two triangles have two pairs of sides which are the same length, the triangle in which the third side is longer also has the larger angle contained by the first two sides.

Proof

 * Converse Hinge Theorem.png

Let $\triangle ABC$ and $\triangle DEF$ be two triangles in which:
 * $AB = DF$
 * $AC = DE$
 * $BC > EF$

that $\angle BAC \not > \angle EDF$.

Then either:
 * $\angle BAC = \angle EDF$

or:
 * $\angle BAC < \angle EDF$

Let $\angle BAC = \angle EDF$.

Then by Triangle Side-Angle-Side Equality:
 * $BC = EF$

But we know this is not the case, so by Proof by Contradiction:
 * $\angle BAC \ne \angle EDF$

Suppose $\angle BAC < \angle EDF$.

Then by Greater Angle of Triangle Subtended by Greater Side:
 * $EF > BC$

But we know this is not the case, so by Proof by Contradiction:
 * $\angle BAC \not < \angle EDF$

Thus:
 * $\angle BAC > \angle EDF$

Also known as
This theorem is also known as the side-side-side inequality theorem, or SSS inequality theorem.