Class 9 Inequalities in a Triangle for CBSE-NCERT

### Topic covered

color{red} ♦Inequalities in a Triangle

### Inequalities in a Triangle

So far, you have been mainly studying the equality of sides and angles of a triangle or triangles. Sometimes, we do come across unequal objects, we need to compare them.

For example, line-segment AB is greater in length as compared to line segment CD in Fig. 7.41 (i) and ∠ A is greater than ∠ B in Fig 7.41 (ii).

Let us now examine whether there is any relation between unequal sides and unequal angles of a triangle. For this, let us perform the following activity:

color {blue} text(Activity) : Fix two pins on a drawing board say at B and C and tie a thread to mark a side BC of a triangle.

Fix one end of another thread at C and tie a pencil at the other (free) end . Mark a point A with the pencil and draw Delta ABC (see Fig 7.42). Now, shift the pencil and mark another point A′ on CA  beyond A (new position of it)

So, A′C > AC (Comparing the lengths)

Join A′ to B and complete the triangle A′BC.

What can you say about ∠ A′BC and ∠ ABC?

Compare them. What do you observe?

Clearly, ∠ A′BC > ∠ ABC

Continue to mark more points on CA (extended) and draw the triangles with the side BC and the points marked.

You will observe that as the length of the side AC is increased (by taking different positions of A), the angle opposite to it, that is, ∠ B also increases.

Let us now perform another activity :

color{blue} text( Activity) :

Construct a scalene triangle (that is a triangle in which all sides are of different lengths). Measure the lengths of the sides.

Now, measure the angles. What do you observe?

In Delta ABC of Fig 7.43, BC is the longest side and AC is the shortest side.

Also, ∠ A is the largest and ∠ B is the smallest.

Repeat this activity with some other triangles.

We arrive at a very important result of inequalities in a triangle. It is stated in the form of a theorem as shown below:

color{blue} text( Theorem 7.6) :

If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).

You may prove this theorem by taking a point P on BC such that CA = CP in Fig. 7.43.

Now, let us perform another activity :

color{blue} text( Activity) :

Draw a line-segment AB. With A as centre and some radius, draw an arc and mark different points say P, Q, R, S, T on it.

Join each of these points with A as well as with B (see Fig. 7.44). Observe that as we move from P to T, ∠ A is becoming larger and larger. What is happening to the length of the side opposite to it? Observe that the length of the side is also increasing; that is ∠ TAB > ∠ SAB > ∠ RAB > ∠ QAB > ∠ PAB and TB > SB > RB > QB > PB.

Now, draw any triangle with all angles unequal to each other. Measure the lengths of the sides (see Fig. 7.45).

Observe that the side opposite to the largest angle is the longest. In Fig. 7.45, ∠ B is the largest angle and AC is the longest side.

Repeat this activity for some more triangles and we see that the converse of Theorem 7.6 is also true. In this way, we arrive at the following theorem:

 color{blue} text( Theorem 7.8) :

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

In Fig. 7.46, observe that the side BA of Delta ABC has been produced to a point D such that AD = AC. Can you show that ∠ BCD > ∠ BDC and BA + AC > BC ? Have you arrived at the proof of the above theorem.

Let us now take some examples based on these results.
Q 3250178014

AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (see Fig. 7.37). Show that the line PQ is the perpendicular bisector of AB.
Class 9 Chapter 7 Example 7
Solution:

You are given that PA = PB and

QA = QB and you are to show that PQ ⊥ AB and

PQ bisects AB. Let PQ intersect AB at C.

Can you think of two congruent triangles in this figure?

Let us take Delta PAQ and Delta PBQ.

In these triangles,

AP = BP (Given)

AQ = BQ (Given)

PQ = PQ (Common)

So, Delta PAQ ≅ Delta PBQ (SSS rule)

Therefore, ∠ APQ = ∠ BPQ (CPCT).

Now let us consider Delta PAC and Delta PBC.

You have : AP = BP (Given)

∠ APC = ∠ BPC (∠ APQ = ∠ BPQ proved above)

PC = PC (Common)

So, Delta PAC ≅ Delta PBC (SAS rule)

Therefore, AC = BC (CPCT) (1)

and ∠ ACP = ∠ BCP (CPCT)

Also, ∠ ACP + ∠ BCP = 180°

So, 2 ∠ ACP = 180°

or, ∠ ACP = 90° (2)

From (1) and (2), you can easily conclude that PQ is the perpendicular bisector of AB.
[Note that, without showing the congruence of Delta PAQ and Delta PBQ, you cannot show
that Delta PAC ≅ Delta PBC even though AP = BP (Given)

PC = PC (Common)

and ∠ PAC = ∠ PBC (Angles opposite to equal sides in
Delta APB)

It is because these results give us SSA rule which is not always valid or true for congruence of triangles. Also the angle is not included between the equal pairs of sides.]
Q 3200178018

P is a point equidistant from two lines l and m intersecting at point A (see Fig. 7.38). Show that the line AP bisects the angle between them.
Class 9 Chapter 7 Example 8
Solution:

You are given that lines l and m intersect each other at A. Let PB ⊥ l,

PC ⊥ m. It is given that PB = PC.

You are to show that ∠ PAB = ∠ PAC.

Let us consider D PAB and D PAC. In these two
triangles,

PB = PC (Given)

∠ PBA = ∠ PCA = 90° (Given)

PA = PA (Common)

So, Delta PAB ≅ Delta PAC (RHS rule)

So, ∠ PAB = ∠ PAC (CPCT)