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

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

If $$\angle BDC = \angle EDF$$, then $BC = EF$, but we know this is not the case, so $$\angle BDC \neq \angle EDF$$.

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

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

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

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