In this chapter we will try to prove that angle opposite to greater side is greater than the other angle.

**Given :**

Given above is triangle ABC in which side **AC > AB.****Construction:**

Draw line BM intersecting at side AC such that AB = AM

**To prove:**

Since AC > AB, prove that angle opposite to greater side is larger.

i.e. **∠ABC > ∠ACB**

**Proof:**

Consider triangle BMC.**Using exterior angle theorem**, we can say that;

∠AMB = ∠MCB + ∠MBC

This means that, **∠AMB > ∠MCB** – -> eq(1)

It’s given that AB = AM.

We know that **angle opposite to equal sides are equal.**

∠ABM = ∠AMB

Using this expression in equation (1) above, we get;

∠ABM > ∠MCB

This expression can also be written as;

**∠ABM > ∠ACB ** – – > eq(2)

Observe the above figure, ∠ABC can be written as;

∠ABC = ∠ABM + ∠MBC

It means that **∠ABC > ∠ABM** – -> eq(3)

Combining eq(2) and eq(3), we get;

∠ ABC > ∠ABM > ∠ACB

Hence we get, ∠ ABC > ∠ ACB

Here we proved that in triangle, the angle opposite to greater side is larger than other angle.

**Next chapter :** **Prove that side opposite to greater angle is longer**