CBSE Notes Class 9 Maths Chapter 7 Triangles

1.0Introduction to Triangles 

A triangle is a figure with three sides, and the sum of all the angles is always 180 degrees. There are different types of triangles based on their sides and angles. 

2.0Download CBSE Notes for Class 9 Maths Chapter 7 Triangles - Free PDF

Master the concepts of triangles with our free downloadable PDF for CBSE Class 9 Maths Chapter 7 – Triangles. These notes are designed to help students understand the fundamentals of triangle properties, congruence rules, and important theorems with ease.

3.0CBSE Class 9 Maths Chapter 7 Triangles - Revision Notes

What is a Congruent Triangle? 

Congruent triangles are pairs of triangles that are exactly equal. Both triangles have sides and angles that are equal to each other. It’s like a carbon copy of an image. Like this: 

Note: While solving the question, we use CPCT, which stands for congruent triangles' corresponding parts, to show congruence in the remaining parts of congruent triangles. 

Properties Related to Congruent Triangles

1. SAS Congruence Rule: 

SAS stands for Side Angle Side. This means that while solving, if a pair of triangles has two sides and one angle between those two sides is equal, then the given pair is congruent with each other.

Remember, the equal angle must be between the two equal sides when applying this rule. 

ABC  ≅ QPR

2. ASA Congruence Rule: 

ASA means Angle Side Angle. If a pair of triangles has two angles equal angles and one equal side in between the angles, then the pair is said to be a congruent triangle with ASA congruency. 

ABC  ≅ PQR

3. AAS Congruence Rule: 

In a pair of triangles, if two angles and one side that is not in between the angles are equal, then the triangles are said to be congruent with the AAS (Angle Angle Side) congruency rule. 

ABC  ≅ QPR

4. SSS Congruence Rule: 

In Maths, If all three sides of a pair of triangles are equal to each other, then the pair is said to be a congruent triangle. Like this: 

5. RHS Congruence Rule: 

RHS stands for Right angle, Hypotenuse, and Side. If in two right triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent. 

Theorems with Proof

Theorem 1: In an isosceles triangle, angles opposite to equal sides are equal.

To Prove: ∠A = ∠ B 

Given: ABC is an isosceles triangle in which AC = BC.

Construction: Construct an angle bisector of ∠ C. 

Proof: In △ ADC and △ BDC 

AC = BC (Given) 

DC = DC (Common) 

∠ACD = ∠BCD (by construction) 

ADC ≅ BDC (By SAS rule) 

Since Triangle ADC and BDC are congruent then, 

∠ A = ∠ B (CPCT) 

Theorem 2: The sides opposite to equal angles of a triangle are equal. 

To prove: AB = AC 

Given: ∠ B = ∠ C 

Construction: construct an angle bisector of angle A. 

Proof: In $△ABD$ and $△ACD$

∠ABD = ∠ACD (given) 

∠BAD = ∠CAD (By construction) 

AD = AD (Common) 

$△ABD\simeq △ACD(ASArule)$

AB = AC (CPCT)

Also Read: Triangles and Its Properties

4.0Solved Questions

Example 1: ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

Solution: In ABD and ACD,

AB = AC (Side) (given)

BD = CD (Side) (given)

BC = BC (Common Side)

$△ABD\cong △ACD(SSSRule)$

Therefore, ∠BAD = ∠CAD (CPCT) ……. (1)

In ABO and ACO,

AB=AC (Side) (given)

∠BAO = ∠CAO (Angle) (from equation (1))

AO = AO (Common Side)

$△ABO\cong △ACO(SASRule)$

∠AOB = ∠AOC (CPCT) ....... (2)

BO = OC (CPCT) ……. (3)

We know that $\angle$AOB and $\angle$AOC are a linear pair (sums up to 180 degrees) on line BOC.

So, ∠AOB + ∠AOC = 180

Using equation (2), ∠AOB = ∠AOC = 90 degrees, making AD perpendicular to BC. Moreover, using equation (3), AD becomes the perpendicular bisector of BC.

Example 2: ABC is a right triangle and right-angled at B such that ∠BCA = 2 ∠BAC. Show that hypotenuse AC = 2 BC.

To prove: AC = 2BC

Given: ∠BCA = 2 ∠BAC

Construction: Extend CB to point D such that BC = BD and join AD.

Solution: In ∆ ABC and ∆ ABD

BC = BD (By construction)

AB = AB (Same side)

∠ABC = ∠ABD (Each of 90°)

$△ABC\cong △ABD(SASRule)$

∠CAB = ∠DAB (CPCT) ……(1)

AC = AD (CPCT) ……(2)

∠CAD = ∠CAB + ∠BAD = x + x = 2x [From (1)] (3)

and ∠ACD = ∠ADB = 2x [From (2), AC = AD] (4)

Hence, ∆ ACD is an equilateral triangle. [From (3) and (4)]

or AC = CD, i.e., AC = 2 BC (Since BC = BD)

Example 3: If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles.

To prove: ABC is an isosceles triangle. 

Given: ∠BAD = ∠CAD, BD = DC 

Construction: Extend AD to point E, such that AD = ED, and also join point C to E. 

Solution: In $△ABD$ and $△ECD$

BD = CD (Given) 

AD = DE (Construction)

∠ADB = ∠EDC (Vertically opposite angle) 

$△ABD\cong △ECD(SASRule)$

AB = EC (CPCT) ………(1)

∠BAD = ∠CED (CPCT) 

∠BAD = ∠CAD (Given)

So,  

∠CED = ∠CAD 

AC = EC (Side opposite to equal angles are equal) ……..(2)

AB = AC (From 1 and 2) 

Hence, ABC is an isosceles triangle. 

5.0Key Features of CBSE Maths Notes for Class 9 Chapter 7

• These notes consist of a visual representation of every rule related to the triangle, which makes it easy to understand its concepts.
• The notes are well aligned with the latest CBSE Maths curriculum. 
• The language used in the content is easy to understand, making it ideal for every level of students, whether they are beginner or advanced.