In this chapter we will prove universal concept of triangle that ” difference of any two side is always less than the third side “.

The below proof is important and is directly asked in the exams so make sure you understand the steps.

**Given;**

Given above is the triangle ABC.**Construction;**

Construct line BM that meet side AC such that side **BA = AM****Solution**

Consider **triangle ABM.**

BA = AM { construction }

We know that in triangle, equal sides have equal opposite angles.

**∠ABM = ∠AMB = ∠1**

**Applying exterior angle theorem on triangle ABM.**

∠ BMC = ∠ MAB + ∠ ABM

∠ 3 = ∠ 2 + ∠ 1

From the above expression we can conclude that ∠ 3 is greater than ∠ 1.

Hence, **∠ 3 > ∠ 1** – – – eq(1)

Now **apply exterior angle theorem to triangle BMC**

∠ 1 = ∠ MBC + ∠ MCB

∠ 1 = ∠ MBC + ∠ 4

From the above expression we conclude that ∠1 is greater than ∠ MBC.

Hence, **∠1 > ∠ MBC ** – – – eq(2)

Combining equation (1) and equation (2).

∠ 3 > ∠ 1 and ∠1 > ∠ MBC

we can say that, ** ∠3 > ∠MBC**

Now take triangle MBC.

We know that in triangle,** side opposite to greater angle is larger.**

Since ∠3 > ∠MBC;

We can say that **BC > MC**.

BC > MC

BC > AC – AM

As AM = AB, the above expression can be written as;**BC > AC – AB**

Hence, we proved that difference of any two side of triangle is always less than third side.

**Next chapter** : **Prove that sum of two sides of triangle is greater than twice the median**