Chapter 7

Triangles

Maths has several chapters that hold utmost significance, like Simplification, Number System, Ratio & Proportion, Profit & Loss, Maths Average Topic Notes, Power Indices, and many other aspects. 

The word triangle is self-descriptive, the word 'tri' means three, and 'angle' is the intersection of 2 lines. So, it can be stated that a triangle is a plane figure with three straight sides and three angles. As simple as that! For Class 9th maths chapter 7, you will learn everything about triangles, starting from their types, properties, and inequalities.

NCERT solutions for class 9 maths chapter 7 also deal with various essential theorems, formulas and crucial questions related to triangles. This chapter will further brief the students about the congruence of triangles and the rules of congruence. 

At msvgo, we understand that mathematical knowledge plays a vital role in understanding the concepts of other subjects such as social studies, science and even art. Students can download the Class 9th chapter 7 maths content solved by proficient teachers from msvgo and quickly learn the concepts related to chapter 7 maths class 9. We consistently help students build their discipline and enhance their mental rigour and logical reasoning.

Topics covered in this chapter 7 class 9th maths include: 

  • Congruence of triangles
  • Properties of triangle and theories related to it
  • SSS and RHS congruence rules
  • Concept of Inequalities in triangles

1. Congruence of triangles

It is imperative to study the congruence of triangles as the theory is used in the architectural design and the construction of large buildings. You must have observed congruent triangles in the carpeting designs, architectural designs, geometric art, etc. In real life, it is rare to find two congruent triangles. Congruence is required to build even surfaces, and many mathematicians find the triangle to be a stable shape. Thus, comprehending the concept of congruence is significant.

2. Properties of triangles

Once students start grasping the concepts by comparing the properties of triangles with a live example, they’ll find this topic intriguing. This topic in Class 9 maths chapter 7 will brief you about all the properties of triangles. 

3. SSS and RHS congruence rule

If three sides of one triangle are equal to another, the triangles are congruent, as per the SSS criteria for congruence. Meanwhile, the RHS criteria for congruence state that all right triangles are congruent, i.e. if one side and the hypotenuse of the triangle are the same as another triangle's hypotenuse and side (any one of the other two sides). 

4. Concept of Inequalities in triangles

Why is learning about the inequalities in triangles necessary? It is because you will be able to differentiate one triangle from another. Understanding the theories related to inequalities will help you compare the two triangles that are different from each other.

Exercise 7.1

1. In quadrilateral ACBD, AC = AD and AB bisect ∠A (see Fig. 7.16). Show that ΔABC\( \cong \) ΔABD. What would BC and BD be determined as?

Solution:

We know that AC = AD, as AC and AD are equal, and the line segment AB bisects ∠A.

Now, we need to prove that the two triangles ABC and ABD are similar, ΔABC \( \cong \) ΔABD

Proof:

Consider the triangles ΔCAB and ΔDAB,

(i) AC = AD (Given)

(ii) AB = AB (Common)

(iii) ∠CAB = ∠DAB (Since AB is the bisector of angle A)

So, by SAS congruence criterion, ΔABC \( \cong \) ΔABD.

BC and BD are of equal lengths by the rule of CPCT for the 2nd part of the question.

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that (see Fig. 7.17)

(i) ΔABD ≅ ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

Solution:

It is given that ∠DAB = ∠CBA and AD = BC.

(i) ΔABD and ΔBAC are similar by SAS congruence as

AB = BA (It is the common arm)

∠DAB = ∠CBA and AD = BC (Given)

So, triangles ABD and BAC are similar, i.e. ΔABD \( \cong \) ΔBAC. 

(Hence proved).

(ii) It is now known that ΔABD \( \cong \) ΔBAC so,

BD = AC (by the rule of CPCT).

(iii) Since ΔABD \( \cong \) ΔBAC so,

∠ABD = ∠BAC (by the rule of CPCT).

1. ABC is an isosceles triangle, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

(i) OB = OC (ii) AO bisects ∠A

Solution:

Given:

AB = AC and

the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

∠B = ∠C

½ ∠B = ½ ∠C

\( \Rightarrow \) ∠OBC = ∠OCB (Angle bisectors)

\( \therefore \) OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given)

AO = AO (Common arm)

OB = OC (Already Proven)

So, ΔAOB \( \cong \) ΔAOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects ∠A.

2. AD is the perpendicular bisector of BC in ΔABC. Prove that ABC is an isosceles triangle in which AB = AC. (see Fig. 7.30)

Solution:

It is given that AD is the perpendicular bisector of BC

To prove:

AB = AC

Proof:

In ΔADB and ΔADC,

AD = AD (F Common arm)

∠ADB = ∠ADC

BD = CD (Since AD is the perpendicular bisector)

So, ΔADB \( \cong \) ΔADC by SAS congruence criterion.

Thus,

AB = AC (by CPCT)

1. ΔDBC and ΔABC are two isosceles triangles on the same base BC, and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, prove that

(i) ΔABD \( \cong \) ΔACD

(ii) ΔABP \( \cong \) ΔACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

Solution:

ΔABC and ΔDBC are equal, given that they are two isosceles triangles.

(i) ΔABD and ΔACD are similar by SSS congruence because:

AD = AD (It is the common arm)

AB = AC (Since ΔABC is isosceles)

BD = CD (Since ΔDBC is isosceles)

\( \therefore \) ΔABD \( \cong \) ΔACD.

(ii) ΔABP and ΔACP are the same:

AP = AP (common side)

∠PAB = ∠PAC (by CPCT since ΔABD \( \cong \) ΔACD)

AB = AC (Since ΔABC is isosceles)

So, ΔABP \( \cong \) ΔACP by SAS congruence condition.

(iii) ∠PAB = ∠PAC by CPCT as ΔABD \( \cong \) ΔACD.

AP bisects ∠A. — (i)

Also, ΔBPD and ΔCPD are similar by SSS congruence as

PD = PD (common side)

BD = CD (Since ΔDBC is isosceles.)

BP = CP (by CPCT as ΔABP \( \cong \) ΔACP)

So, ΔBPD \( \cong \) ΔCPD.

Therefore, ∠BDP = ∠CDP by CPCT. — (ii)

By comparing (i) and (ii), we can conclude that AP bisects ∠A and ∠D.

(iv) ∠BPD = ∠CPD (by CPCT as ΔBPD \( \cong \) ΔCPD)

and BP = CP — (i)

also,

∠BPD +∠CPD = 180° (Since BC is a straight line.)

\( \Rightarrow \) 2∠BPD = 180°

\( \Rightarrow \) ∠BPD = 90° —(ii)

From equations (i) and (ii), it can be determined that,

AP is the perpendicular bisector of BC.

What are the properties of the triangle?

Listed below are the six properties of triangles:

  • The sum of all the sides of every triangle is equal to 180 degrees. 
  • Every angle measures to be 60 degrees of an equilateral triangle. 
  • The angle opposite to the longer side is bigger than the other ones in a triangle.
  • The side opposite to the greater angle is longer.
  • The third side of the triangle always remains less than the sum of the other two sides of the triangle. 
  • The sum of all three sides of the triangle is equal to the parameter of the triangle.

What is the meaning of congruence of triangles?

Two triangles are said to be congruent when one triangle matches perfectly with the other triangle. It happens when all three corresponding sides and angles of both the triangles are equal.

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