The **Hinge Theorem** can be understood by exploring real hinges. If the two hinges are of the same size and the angle of the first hinge is opened wider than the second, then the distance between the edges of the first hinge, is farther than that of the second.

If a string is placed connecting the hinges, then a triangle is formed. As we shall see, hinges are connected to theorems about triangles.

**Theorem**

The Hinge Theorem states if two sides of one triangle is congruent, respectively, to two sides of another triangle, and the included angle of the first angle is larger than the included angle of the second, then the third side o f the first triangle is longer than the third side of the second.

The Hinge Theorem is illustrated in the first figure. Given triangle *ABC* and triangle *DEF*, with *AB* = *DE*, and *AC* = *DF*. If angle *A* > angle *D*, then *BC* > *EF*.

**Proof **

First we construct *AGC*, with *G* in the interior of angle *BAC* such that triangle *AGC* is congruent to triangle DEF. This can be done using compass and straightedge construction. First copy angle *EDC* to angle *BAC*, then locate *AG* = *DE*

Now, bisect angle *BAG* and let *M* be the intersection of the bisector and *BC*. By SAS Congruence, triangle *AMB* is congruent to triangle *AMG*. Therefore, *MB* = *MG*.

Now, by the **Triangle Inequality Theorem**,

*CG* < *CM* + *MG*

Therefore,

*CG* < *CM* + *MB*

because *MB* = *MG*.

Since *CG* = *EF*, and *CM *+ *MB* = *BC*,

We have *EF* < *BC* which is what we want to show.

This seems like such an obvious concept, yet the proof is surprisingly more complex! I think there is an error though. In the second paragraph of your proof, you say MD = MK, but I think MK should be MG?

You’re right. It should be MG. I have corrected it already. Thanks Shaun.

The proofs are more interesting than all the problems I had to solve in high school geometry ca.1965. I don’t recall any proofs, so it is wonderful to encounter them again. Much appreciate their beauty.