SSS Inequality Theorem: Converse of Hinge Theorem

First, let's remind ourselves of the Hinge Theorem.

Hinge Theorem: Consider two triangles, $\triangle ABC$ and $\triangle DEF,$ with $\overline{AB} = \overline{DE},$ $\overline{AC} = \overline{DF},$ and $m\angle A > m\angle D.$ Then $\overline{BC} > \overline{EF}.$

The converse of the Hinge Theorem, referred to as the SSS Inequality Theorem, is also true.

SSS Inequality Theorem: Consider two triangles, $\triangle ABC$ and $\triangle DEF,$ with $\overline{AB} = \overline{DE},$ $\overline{AC} = \overline{DF},$ and $\overline{BC} > \overline{EF}.$ Then $m\angle A > m\angle D.$

Let's look at an example of the SSS Inequality Theorem. Consider these two triangles:

As you can see, $\overline{AB} = \overline{DE},$ $\overline{AC} = \overline{DF},$ and $\overline{BC} = 10 > \overline{EF} = 8.$ Thus, by the SSS Inequality Theorem, we know that $m\angle A$ is greater than $m\angle D.$

Here's another picture to demonstrate the SSS Inequality Theorem.

Notice that each of the triangles has two sides of length 5. Look at the triangles from left to right. As the side opposite $\angle A$ gets smaller and smaller, $\angle A$ gets smaller and smaller. Similarly, if you look at the triangles from right to left, as the side opposite $\angle A$ gets bigger and bigger, $\angle A$ gets bigger and bigger.

The converse of the Hinge Theorem also holds; this theorem is more formally named the SSS Inequality Theorem.

Given two triangles $\triangle ABC$ and $\triangle DEF$ such that $AB=DE$, $AC=DF$, and $BC\gt EF$, it can be shown that $m\angle A\gt m\angle D$.

The proof of this theorem is essentially the reverse of the proof of the Hinge Theorem.

First, we use the Law of Cosines on both triangles: $$BC^2=AB^2+AC^2-2AB\cdot AC\cos A$$ $$EF^2=DE^2+DF^2-2DE\cdot DF\cos D$$ Subtract the first equation from the second and note that $AB=DE$ and $AC=DF$: $$BC^2-EF^2=2AB\cdot AC(\cos D-\cos A)$$ Since $BC\gt EF$, we know that $\cos D-\cos A\gt 0$.

Since the cosine function is decreasing on the interval $[0^\circ,180^\circ]$, and since both $m\angle A$ and $m\angle D$ obviously lie in this interval, we have that $m\angle A\gt m\angle D$.

Thus, the proof is complete, and we have shown that the converse of the Hinge Theorem also holds.

Combining the Hinge Theorem, its converse, and the SSS congruence postulate, we have the following:

Given two triangles $\triangle ABC$ and $\triangle DEF$ such that $AB=DE$ and $AC=DF$, $$BC\gt EF\iff m\angle A\gt m\angle D$$ $$BC\lt EF\iff m\angle A\lt m\angle D$$ $$BC=EF\iff m\angle A=m\angle D$$