Let's prove logically step-by-step that the difference of the lengths of any two sides of a triangle is less than the length of the third side.

Let ΔABC in which AB, BC and CA are its sides.

To Prove: AB – BC < CA or BC – CA < AB or AC – AB < AB

Construction: Take a point D on AC such that AD = AB and Join B to D

Proof: In ΔABD,

∠3 is an exterior angle.

So, ∠3 > ∠1 …(i)

[Exterior angle of a triangle is greater than one of its opposite interior angle]

Since, AB = AD

⇒ ∠2 = ∠1 …(ii)

[Angle opposite to equal sides are equal]

⇒ ∠3 > ∠2 …(iii)

[from (i) and (ii)]

Now, In ΔBDC,

∠2 is an exterior angle.

So, ∠2 > ∠4 …(iv)

[Exterior angle of a triangle is greater than one of its opposite interior angle]

⇒ ∠3 > ∠4 [from (iii) and (iv)]

⇒ BC > CD

[greater angle has a longer side opposite to it]

or CD < BC

or AC – AD < BC [∵, AD + CD = AC ]

or AC – AB < BC [∵ by construction, AB = AD]

Hence, the difference of the lengths of any two sides of a triangle is less than the length of the third side.