Converse Hinge Theorem

Consider $△ABC$ and $△A^{\prime }{B}^{\prime }C$ such that $\overline{AB}=\overline{A^{\prime }{B}^{\prime }}$ and $\overline{AC}=\overline{A^{\prime }{C}^{\prime }},$ where $\overline{BC}>\overline{B^{\prime }{C}^{\prime }}.$

This theorem can be proven by contradiction. Since the goal is to prove that $m\angle BAC>m\angle B^{\prime }{A}^{\prime }{C}^{\prime }$ the opposite statement will be assumed, that is, $m\angle BAC\le m\angle B^{\prime }{A}^{\prime }{C}^{\prime }.$ Because this is a non-strict inequality, both $m\angle BAC=m\angle B^{\prime }{A}^{\prime }{C}^{\prime }$ and $m\angle BAC<m\angle B^{\prime }{A}^{\prime }{C}^{\prime }$ have to be considered.

$m\angle BAC=m\angle B^{\prime }{A}^{\prime }{C}^{\prime }$

If $m\angle BAC=m\angle B^{\prime }{A}^{\prime }{C}^{\prime },$ then, $△ABC$ and $△A^{\prime }{B}^{\prime }{C}^{\prime }$ share two congruent sides and their included angles are congruent. Because of the Side-Angle-Side Congruence Theorem, they are congruent. Since the triangles are congruent, their sides are congruent as well. This means that $\overline{BC}=\overline{B^{\prime }{C}^{\prime }},$ but this contradicts the fact that the given triangle is such that $\overline{BC}>\overline{B^{\prime }{C}^{\prime }}.$

$m\angle BAC<m\angle B^{\prime }{A}^{\prime }{C}^{\prime }$

If $m\angle BAC<m\angle B^{\prime }{A}^{\prime }{C}^{\prime }$ then the Hinge Theorem states that $\overline{BC}<\overline{B^{\prime }{C}^{\prime }},$ but this also contradicts the fact that $\overline{BC}>\overline{B^{\prime }{C}^{\prime }}.$

Conclusion

The assumption that $m\angle BAC$ is less than or equal to $m\angle B^{\prime }{A}^{\prime }{C}^{\prime }$ contradicts the hypothesis. Therefore, this assumption must be false. Consequently, the initial conclusion of the theorem is true.