Geometry is the study of lines, angles and shapes. When we go ahead in geometry, we learn about triangles and their congruency. Two triangles are said to be congruent if they have exactly similar in shape and size. Two congruent triangles, when placed on top of another, they are able to cover each other completely. There are various theorems based on triangle congruency.

There is a theorem called **Hinge theorem **which plays quite an important role. Hinge theorem deals with inequalities in two triangles. When a side in one triangle is bigger or smaller than the corresponding side in other triangle and even an angle in a triangle is bigger or smaller than the corresponding angle in some other triangle, the hinge theorem is proved to be very useful in order to prove congruency of such triangles. The inequalities in two triangles means the triangles are not equal. Here, we are going to compare two triangles by their sides and their angles by using some special rule. The inequality of two triangles is found by **SAS inequality** and **SSS inequality**,** **which is known as **Hinge Theorem**. These inequalities are very similar to congruency rules of SAS equality and SSS equality. Let us go ahead and learn more in detail about this theorem and its applications.

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If two sides of the triangle are
congruent with the two sides of another triangle, then the triangle with
the larger included angle also has the larger third side.

Given that,

In $\triangle$ ABC and $\triangle$DEF,

AB = DE, AC = DF and $\angle$BAC > $\angle$EDF.**To Prove:** BC > EF

Construct DG such that DG = DF = AC and join EG and FG by dotted lines.

Since AB = DE, $\angle$BAC = $\angle$EDG and AC = DG, BC = EG.

=> DG = AC = DF

Thus, $\angle$DGF = $\angle$DFG.

So, $\angle$EFG > $\angle$DFG = $\angle$DGF > $\angle$EGF (By Euclid's lemma)

Since $\angle$EFG > $\angle$EGF, EG > EF.

Therefore, EG = BC, BC > EF (Hence Proved).

**Given: **In $\triangle$ABC and $\triangle$DEF

AB = DE, AC = DF and BC > EF.**To Prove:** $\angle$BAC > $\angle$EDF

**Proof:**

Let us assume that $\angle$BAC $\not{>}$ $\angle$EDF.

Then, either $\angle$BAC = $\angle$EDF or $\angle$BAC < $\angle$EDF

- If $\angle$BAC = $\angle$EDF then BC = EF (not possible)

- If $\angle$BAC < $\angle$EDF then BC < EF (which is not possible)

Since our assumption is wrong.

=> $\angle$BAC > $\angle$EDF (Hence Proved).

Given below are some of the problems based on Hinge Theorems.

A car leaves a mall, first north direction up to 5 mile. Then move west upto 3 mile. Car B left the same mall moves towards south for 5 mile the turn east by 100 degrees for 3 mile. Which one is far from the mall?

Two triangle of the given figure have one sides congruent, one triangle have sides 5 mile and another have 3 miles.

Since 100 degrees > 90 degrees (By Hing's Theorem)

Car B is far from the mall than car A.

** **Use the Hinge theorem and find which angle is greater?

From figure, we know that AD = BC, AB = 36 and DC = 34

=> AB > DC

And, BD is common side

So, two sides of $\triangle$ ADB are congruent with the other $\triangle$ CBD.

The rest side or third side of the $\triangle$ ADB is larger.

By Hinge theorem, $\angle$ 1 is greater than $\angle$ 2.

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