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

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 $\angle BAC \not< \angle EDF$.

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

$\blacksquare$

Historical Note

This is Proposition 25 of Book I of Euclid's The Elements.

This theorem is the converse of Proposition 24: Hinge Theorem.

This theorem is also known as the SSS Inequality Theorem.