# The Sum of any Two Sides of a Triangle is Greater than the Third Side

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.

 Statement1. ∠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) Reason1. 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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