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).

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