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.

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 $D$.

Place $G$ so that $DG = AC$.

Join $EG$ and $FG$.

Since $AB = DE$, $\angle BAC = \angle EDG$, and $AC = DG$, $BC = GE$.

By Euclid's first common notion, $DG = AC = DF$.

Thus, $\angle DGF = \angle DFG$.

So by Euclid's fifth common notion, $\angle EFG \, > \, \angle DFG = \angle DGF \, > \, \angle EGF$.

Since $\angle EFG > \angle EGF$, $EG > EF$.

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

$\blacksquare$

Historical Note

This is Proposition 24 of Book I of Euclid's The Elements.

This theorem is the converse of Proposition 25: Converse Hinge Theorem.

This theorem is also known as the SAS Inequality Theorem.