Let's consider the given information and what we want to prove.

Given : Prove : \overline{A B} ≅ \overline{X Y}, \overline{B C} ≅ \overline{Y Z}, m ∠ B > m ∠ Y A C > X Z

The following diagram illustrates the given information.

To develop the proof of the Hinge Theorem, we will follow the given plan.

Locate Point $D$

We first measure $∠ Y$ by using a protractor. Then, we will place the protractor on $\overline{B C}$ so that $B$ is in the center of the protractor. We measure an angle equal to $m ∠ Y,$ and draw a ray starting from $B .$

Over this ray, by using a ruler we locate a point $D$ such that $B D = Y X .$ Also, we will draw $\overline{C D} .$

Next, let's list the congruent parts between

△ D B C

and

△ X Y Z .

\overline{B C} ≅ \overline{Y Z} ∠ C B D ≅ ∠ Y \overline{B D} ≅ \overline{Y X} S ide Included A ngle S ide

By the Side-Angle-Side (SAS) Congruence Postulate we conclude that

△ D B C ≅ △ X Y Z .

This implies that

\overline{X Z} ≅ \overline{D C} .

Locate Point $F$

By using a protractor we draw the angle bisector of $∠ A B D$ . Let $F$ the intersection point between the angle bisector and $\overline{A C} .$ Also, we draw $\overline{F D} .$

Proving that $△ A B F ≅ △ D B F$

From the latter diagram, we can list the following congruent parts between

△ A B F

and

△ D B F .

\overline{A B} ≅ \overline{D B} ∠ A B F ≅ ∠ D B F \overline{B F} ≅ \overline{B F} S ide Included A ngle S ide

By the SAS Congruence Postulate we get that

△ A B F ≅ △ D B F,

and so

\overline{A F} ≅ \overline{D F} .

Proving that $A C = F C + D F$

By the Segment Addition Postulate we have that

A C = A F + F C .

But, since

\overline{A F} ≅ \overline{D F},

we have that

A F = D F .

Let's substitute it into the previous equation.

A C = F C + D F ✓

Using the Triangle Inequality Theorem

Let's take a look at $△ F C D .$

By the Triangle Inequality Theorem we can write the following three relations.

F C + D F F C + C D C D + D F > C D > F D > F C

Remember that we proved before that

C D = X Z,

and also

A C = F C + D F .

Substituting them into the first inequality will give us the desired result.

A C > X Z ✓

Locate Point D

Locate Point F

Proving that △ABF≅△DBF

Proving that AC=FC+DF

Using the Triangle Inequality Theorem

Locate Point $D$

Locate Point $F$

Proving that $△ A B F ≅ △ D B F$

Proving that $A C = F C + D F$

{{ 'ml-heading-exercise' | message }} {{ exercise.name }} {{ 'ml-label-page' | message }} {{ exercise.page }}

{{ 'ml-solutions-missing-heading' | message }}