In this chapter we will try to prove that angle opposite to greater side is greater than the other angle.
Given above is triangle ABC in which side AC > AB.
Draw line BM intersecting at side AC such that AB = AM
Since AC > AB, prove that angle opposite to greater side is larger.
i.e. ∠ABC > ∠ACB
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