Prove that sum of two sides of triangle is greater than third side

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