# 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.

In the words of Euclid:

If two triangles have two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.

## Proof

Let $\triangle ABC$ and $\triangle DEF$ be two triangles in which:

$AB = DF$
$AC = DE$
$BC > EF$

Aiming for a contradiction, suppose that $\angle BAC \not > \angle EDF$.

Then either:

$\angle BAC = \angle EDF$

or:

$\angle BAC < \angle EDF$

Let $\angle BAC = \angle EDF$.

$BC = EF$

But we know this is not the case, so by Proof by Contradiction:

$\angle BAC \ne \angle EDF$

Suppose $\angle BAC < \angle EDF$.

$EF > BC$

But we know this is not the case, so by Proof by Contradiction:

$\angle BAC \not < \angle EDF$

Thus:

$\angle BAC > \angle EDF$

$\blacksquare$

## Also known as

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

## Historical Note

This proof is Proposition $25$ of Book $\text{I}$ of Euclid's The Elements.
It is the converse of Proposition $24$: Hinge Theorem..