In any given triangle, **the side opposite to greater angle is always greater.**

In this chapter, we will prove the above theorem using two methods.

This is an important theorem on triangle, so make sure you understand and remember the concept for examination purpose.

**Method 01**

**Given**:

Given above is the triangle ABC in which ∠ABC > ∠ACB

**Construction:**

Construct the line BN on side AC such that ∠NBC = ∠ACB

**To prove:**

Prove that side opposite to greater angle is larger.

Here ∠ABC is larger angle and AC is the side opposite to it.

So prove** AC > AB**

**Proof****Consider triangle BNC.**

Its given that ∠NBC = ∠ACB

We know that side opposite to equal angles are equal.

Hence, **NB = NC.**

Adding side AN in above equation.

AN + NB = AN + NC

AN + NB = AC – – -eq(1)**Now take triangle ABN.**

We know that in triangle sum of two side is always greater than third side.

AN + NB > AB

Using equation (1), the above expression can be written as;

AC > AB

Hence, we proved that side opposite to greater angle is larger.

**Method 02**

Given below is the simple method to prove the same theorem

**Given;**

ABC is a triangle in which ∠ABC > ∠ACB

**To prove:**

AC > AB

**Solution**

Let us **assume that both sides equal to each other (i.e. AB = AC)**

If both sides are equal then the angle opposite to the sides will also be equal.

i.e. ∠ABC = ∠ACB

But such is not the case as it is given that ∠ABC > ∠ACB, hence our assumption is wrong.

Hence, \mathtt{AB\neq \ AC}

Now let’s **assume that AC < AB.**

W know that angle opposite to equal side is greater.

So, ∠ACB > ∠ABC.

But such is not the case as it’s given that ∠ABC > ∠ACB. Hence, our assumption is wrong.

**So there is only now possibility left.**

i.e. AC > AB

Hence Proved.

**Next chapter :** **Prove that in triangle sum of two sides if greater than third side**