In this chapter we will prove one of the universal law of triangle that ” **sum of two side is always greater than third side** “.

The proof is very important as it is directly asked in mathematics exams, so make sure you understand and remember the steps;

**Given;**

Given above is the triangle ABC.**To prove:**

AB + AC > BC

**Construction:**

Extend side AB to AM such that AC = AM

**Solution**

Consider** triangle AMC** in which AM = AC.

We know that in triangle, angle opposite to equal sides are equal.

So, **∠AMC = ∠ ACM**

**Now consider triangle MBC.**

Note that ∠BCM = ∠BCA + ∠ACM

So we can say that; **∠BCM > ∠ACM.**

As ∠AMC = ∠ ACM, the expression can be written as;

**∠BCM > ∠AMC **

**Now consider triangle BMC.**

∠BCM > ∠AMC { Proved above }

We know that side opposite to greater angle is equal.

So, BM > BC

BA + AM > BC

As AM = AC, we can write the expression as;**BA + AC > BC**

Hence, we proved that in any triangle the sum of two side is always greater than the third side.

Similarly one can also prove the following expressions;

AB + BC > AC

AC + BC > AB

**Next chapter :** **Prove that in triangle difference of two side is less than third side**