Difference of any two sides of a triangle is greater than the third side

Geometry is the branch of mathematics that deals with the study of different types of shapes and figures and sizes. The branch of geometry deals with different angles, transformations, and similarities in the figures seen.

Triangle

A triangle is a closed two-dimensional shape associated with three angles, three sides, and three vertices. A triangle associated with three vertices says A, B, and C is represented as △ABC. It can also be termed as a three-sided polygon or trigon. Some of the common examples of triangles are signboards and sandwiches.

To prove: The sum of two sides of a triangle is greater than the third side, BA + AC > BC

Assume: Let us assume ABC to be a triangle.

Proof:

Extend the line segment BA to D,

Such that, AD = AC

⇒ ∠ ADC = ∠ ACD

Observing by the diagram, we obtain,

∠ DCB > ∠ ACD

⇒ ∠ DCB > ∠ ADC

⇒ BD > AB (Since the sides opposite to the larger angle is larger and the sides opposite to smaller angle is smaller)

⇒ BA + AD > BC

⇒ BA + AC > BC.

Hence proved.

Note: Similarly it can be also proved that, BA + BC > AC or AC + BC > BA

Hence, The sum of two sides of a triangle is greater than the third side.

Sample Questions

Question 1. Prove that the above property holds for the lowest positive integral value.

Solution:

Let us assume ABC to be a triangle.

Each of the sides is 1 unit.

Now,

It is an equilateral triangle where all the sides are 1 each.

Taking sum of two sides,

AB + BC ,

1 + 1 > BC

1+1 > 1

2 > 1

Question 2. Illustrate this property for a right-angled triangle

Solution:

Let us assume the sides of the right angles triangle to be 5,12 and 13.

Now,

Taking the smaller two sides, we obtain,

5 + 12 > 13

17 > 13

Hence, the property holds.

Question 3. Does this property hold for isosceles triangles?

Solution:

Let us assume a triangle with sides 2x, 2x, and x.

Now,

Taking the sum of equal two sides, we obtain,

2x + 2x = 4x

which is greater than the third side, equivalent to x.

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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.

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.

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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Answer

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Here, \[AB=AD\] we know that the angles opposite to equal sides are equal and the exterior angle is equal to the sum of opposite two angles. By using these two theorems we need to prove \[DC< BC\] which gives the required result.

Complete step-by-step solution:

Let us consider a triangle \[\Delta ABC\]Now, let us construct the side \[BD\] such that \[AB=AD\] as shown below

We know that the angles opposite to equal sides are equal.By using the above theorem we can write\[\angle ABD=\angle ADB=\angle 2\]We know that the exterior angle is equal to the sum of the opposite angles.By using the above theorem we can write\[\begin{align} & \Rightarrow \angle BDC=\angle A+\angle ABD \\ & \Rightarrow \angle 1=\angle A+\angle 2 \\ \end{align}\]Here, we can see that we need to add \[\angle A\] to \[\angle 2\] for getting \[\angle 1\]So, from the inequality we can write\[\Rightarrow \angle 1> \angle 2.....equation(i)\] Now, similarly let us go for \[\Delta DBC\] that is\[\begin{align} & \Rightarrow \angle ADB=\angle C+\angle DBC \\ & \Rightarrow \angle 2=\angle C+\angle 3 \\ \end{align}\]Here, we can see that we need to add \[\angle C\] to \[\angle 3\] for getting \[\angle 2\]So, from the inequality we can write\[\Rightarrow \angle 2 >\angle 3.....equation(ii)\]Now, by combining the equation (i) and equation (ii) we get\[\begin{align} & \Rightarrow \angle 1> \angle 2> \angle 3 \\ & \Rightarrow \angle 1> \angle 3 \\ & \Rightarrow \angle 3<\angle 1 \\ \end{align}\]We know that we can replace the sides in an inequality with their opposite angles and vice versa.By using the above theorem we can take the sides opposite to \[\angle 3\] and \[\angle 1\] from \[\Delta DBC\] to get\[\begin{align} & \Rightarrow \angle 3< \angle 1 \\ & \Rightarrow DC< BC \\ \end{align}\]Here, we have the length \[DC\] can be written as\[\Rightarrow AC-AD< BC\]Now, by substituting \[AB=AD\] in above equation we get\[\Rightarrow AC-AB< BC\]Therefore, we can see that the difference of two sides is less than the third side.Hence the required result has been proved.

Note: The explanation for the above question can be done in another way also.

We know that the definition of a triangle as the polygon having three sides such that the sum of any two sides is greater than the third side.Let us consider a triangle \[\Delta ABC\]

From the definition of triangle, we can write\[\Rightarrow AB+BC>AC\]Now, the above equation can be reformed as\[\Rightarrow AC-BC< AB\]Therefore, we can say that the difference between two sides is less than the third side.Similarly, we can say for all sides.

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