# Side opposite to greater angle is longer

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.