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 larger 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 $AB = DF$ and $AC = DE$ and $BC > EF$.

Assume that $\angle BAC \not> \angle EDF$. Then either $\angle BAC = \angle EDF$ or $\angle BAC < \angle EDF$.

If $\angle BAC = \angle EDF$, then it must be the case that $BC = EF$, but we know this is not the case, so $\angle BAC \neq \angle EDF$.

If $\angle BAC < \angle EDF$, then it must be the case that $EF > BC$, but we know this is not the case, so $\angle BDC \not< \angle EDF$.

Thus, $\angle BAC > \angle EDF$.