# Hinge Theorem

## Theorem

If two triangles have two pairs of sides which are the same length, the triangle with the larger included angle also has the larger third side.

In the words of Euclid:

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

(*The Elements*: Book $\text{I}$: Proposition $24$)

## Proof

Let $\triangle ABC$ and $DEF$ be two triangles in which $AB = DE$, $AC = DF$, and $\angle CAB > \angle FDE$.

Construct $\angle EDG$ on $DE$ at point $D$.

Join $EG$ and $FG$.

Since $AB = DE$, $\angle BAC = \angle EDG$, and $AC = DG$, by Triangle Side-Angle-Side Congruence:

- $BC = GE$

By Euclid's first common notion:

- $DG = AC = DF$

Thus by Isosceles Triangle has Two Equal Angles:

- $\angle DGF = \angle DFG$

So by Euclid's fifth common notion:

- $\angle EFG \, > \, \angle DFG = \angle DGF \, > \, \angle EGF$

Since $\angle EFG > \angle EGF$, by Greater Angle of Triangle Subtended by Greater Side:

- $EG > EF$

Therefore, because $EG = BC$, $BC > EF$.

$\blacksquare$

## Also known as

This theorem is also known as the **side-angle side inequality theorem**, or **SAS inequality theorem**.

## Historical Note

This proof is Proposition $24$ of Book $\text{I}$ of Euclid's *The Elements*.

It is the converse of Proposition $25$: Converse Hinge Theorem.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 1*(2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions - 1968: M.N. Aref and William Wernick:
*Problems & Solutions in Euclidean Geometry*... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.19$