In a triangle let AC > AB

Then, along AC draw AD = AB and join BD

Proof: In Δ ABD,

∠ ABD = ∠ ADB (AB = AD) ….(i)

∠ ABD = ∠ 2 (angles opposite to equal sides) ….(ii)

Now, we know that the exterior angle of a triangle is greater than either of its opposite interior angles.

∴∠ 1 >∠ABD

∠1 > ∠2 ….(iii)

Now, from (ii)

∠2 > ∠3 ….(iv) (∠2 is an exterior angle)

Using (iii) and (iv),

∠1 > ∠3

BC > DC (side opposite to greater angle is longer)

BC > AC – AD

BC > AC – AB (since, AB = AD)

Hence, the difference of two sides is less than the third side of a triangle