Here we will prove that the sum of any two sides of a triangle is greater than the third side.
Given: XYZ is a triangle.
To Prove: (XY + XZ) > YZ, (YZ + XZ) > XY and (XY + YZ) > XZ
Construction: Produce YX to P such that XP = XZ. Join P and Z.
1. ∠XZP = ∠XPZ.
2. ∠YZP > ∠XZP.
3. Therefore, ∠YZP > ∠XPZ.
4. ∠YZP > ∠YPZ.
5. In ∆YZP, YP > YZ.
6. (YX + XP) > YZ.
7. (YX + XZ) > YZ. (Proved)
1. XP = XZ.
2. ∠YZP = ∠YZX + ∠XZP.
3. From 1 and 2.
4. From 3.
5. Greater angle has greater side opposite to it.
6. YP = YX + XP
7. XP = XZ
Similarly, it can be shown that (YZ + XZ) >XY and (XY + YZ) > XZ.
Corollary: In a triangle, the difference of the lengths of any two sides is less than the third side.
Proof: In a ∆XYZ, according to the above theorem (XY + XZ) > YZ and (XY + YZ) > XZ.
Therefore, XY > (YZ - XZ) and XY > (XZ - YZ).
Therefore, XY > difference of XZ and YZ.
Note: Three given lengths can be sides of a triangle if the sum of two smaller lengths greater than the greatest length.
For example: 2 cm, 5 cm and 4 cm can be the lengths of three sides of a triangle (since, 2 + 4 = 6 > 5). But 2 cm, 6.5 cm and 4 cm cannot be the lengths of three sides of a triangle (since, 2 + 4 ≯ 6.5).