Prove that the difference between any two sides of a triangle is less than its third side.

To prove AC - AB < BC, BC - AC < AB and BC - AB < AC

Construction Let AC > AB. Then, along AC, cut-off AD = AB and join BD.

Proof In ∆ ABD, AB=AD⇒ ∠2=∠1

[∵ angles opposite to equal sides are equal]

Side CD of ∆ BCD has been produced to A.

∠2 >∠4

[∵ exterior angle is greater than each interior opposite angle]

Again, side AD of ∆ABD has been produced to C.

∴∠3>∠1 …(iii)

[∵ exterior angle is greater than each interior opposite angle]

From Eqs. (i) and (iii), we get∠3 > ∠2

Now, ∠3 >∠2 and ∠2 > ∠4 => ∠3 > ∠4

∴BC >CD

[∵ sides opposite to greater angle is longer]

⇒ CD < BC ⇒ AC - AD < BC

Hence, AC - AB < BC [∵ AD = AB]

Similarly, BC - AC < AB and

BC - AB< AC Hence proved.