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

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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.

Inequalities in Triangles

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.

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

             Statement

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)

            Reason

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).

9th Grade Math

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