Triangles Chapter 7 Class 9 CBSE Maths Revision Notes

This comprehensive guide to Triangles for Class 9 Maths covers every concept you need to master this chapter from types of triangles and congruence criteria (SAS, ASA, AAS, SSS, RHS) to key theorems, inequality relations, and step-by-step solved examples. Aligned with the CBSE syllabus and written with clarity for students, teachers, and competitive exam aspirants.

What Is a Triangle?

A triangle is a plane (flat) figure bounded by exactly three straight line segments in a plane. Every triangle has three sides and three interior angles. For triangle ABC, the sides are AB, BC, and CA, and the angles are ∠A, ∠B, and ∠C.

One of the most fundamental results in geometry states that the sum of all interior angles of any triangle is always 180°. This is known as the Angle Sum Property and applies to every type of triangle without exception.

Types of Triangles

A. Based on Sides

Type 1: Scalene Triangle

A triangle in which no two sides are equal. All three angles are also different from each other.

Type 2: Isosceles Triangle

A triangle with exactly two sides equal. The angles opposite the equal sides are also equal (Base Angle Theorem).

Type 3: Equilateral Triangle

A triangle where all three sides are equal. Consequently, each interior angle measures exactly 60°.

B. Based on Angles

Angle Type 1: Right Triangle

Contains one angle equal to 90°. The side opposite the right angle is the longest side, called the hypotenuse.

Angle Type 2: Acute Triangle

All three angles are less than 90° (acute). An equilateral triangle is a special case of an acute triangle.

Angle Type 3: Obtuse Triangle

Contains one angle greater than 90° (obtuse). Only one obtuse angle is possible in any triangle.

Congruent Figures

Two geometric figures are called congruent if they have exactly the same shape and the same size. In practical terms, one figure can be placed (superimposed) on the other so that it covers it completely. Congruent figures have equal length, equal width, and equal height.

Think of two identical postage stamps, or two coins of the same denomination these are real-world examples of congruent figures. The symbol used to denote congruence is ≅.

Congruent Triangles

Two triangles are called congruent if and only if one can be superimposed on the other so as to cover it exactly. When △ABC ≅ △DEF, there is a precise one-to-one correspondence between vertices, giving us six equalities:

Sides: AB = DE | BC = EF | AC = DF

Angles: ∠A = ∠D | ∠B = ∠E | ∠C = ∠F

Important: The order of vertices matters. Writing △ABC ≅ △DEF is different from △ABC ≅ △DFE. The correspondence A↔D, B↔E, C↔F must be preserved.

A key principle used repeatedly is CPCTC (Corresponding Parts of Congruent Triangles are Congruent). Once two triangles are proved congruent, all their remaining corresponding parts are automatically equal.

Congruence Criteria

You do not need all six equalities to establish congruence. The following five criteria are sufficient conditions any one of them is enough to prove two triangles congruent.

SAS

Side–Angle–Side

Two sides and the included angle of one triangle equal the corresponding parts of another.

ASA

Angle–Side–Angle

Two angles and the included side of one triangle equal those of another.

AAS

Angle–Angle–Side

Two angles and a non-included side of one triangle equal the corresponding parts of another.

SSS

Side–Side–Side

All three sides of one triangle are equal to the three sides of another.

RHS

Right–Hypotenuse–Side

For right triangles only: the hypotenuse and one side of one triangle equal those of another.

Note: AAA (three equal angles) is not a congruence criterion — it only establishes similarity, not congruence. Similarly, SSA is not a valid criterion.

Theorems

Triangles Class 9 Theorem 1

Angles opposite to equal sides of an isosceles triangle are equal.

Given: △ABC with AB = AC

To Prove: ∠B = ∠C

Construction: Draw the bisector AD of ∠A, meeting BC at D.

• In △ABD and △ACD: AB = AC (given); ∠BAD = ∠CAD (AD bisects ∠A); AD = AD (common side)
• By SAS criterion: △ABD ≅ △ACD
• Therefore ∠B = ∠C [by CPCTC] ∎

Triangles Class 9 Theorem 2

If two angles of a triangle are equal, then the sides opposite to them are also equal.

Given: △ABC with ∠B = ∠C 

To Prove: AB = AC

Construction: Draw bisector of ∠A meeting BC at D.

• In △ABD and △ACD: ∠B = ∠C (given); ∠BAD = ∠CAD (AD bisects ∠A); AD = AD (common)
• By AAS criterion: △ABD ≅ △ACD
• Therefore AB = AC [by CPCTC] ∎

Triangles Class 9 Theorem 3

If the bisector of the vertical angle bisects the base, the triangle is isosceles.

Given: △ABC with AD bisecting ∠A, meeting BC at D, and BD = CD.

Construction: Produce AD to E such that AD = DE; join EC.

• In △ADB and △EDC: AD = DE (construction); ∠ADB = ∠CDE (vertically opposite); BD = DC (given)
• By SAS: △ADB ≅ △EDC, so AB = EC and ∠BAD = ∠CED (CPCTC)
• Since ∠BAD = ∠CAD, we get ∠CAD = ∠CED, so AC = EC (sides opposite equal angles)
• Therefore AC = AB — triangle is isosceles. ∎

Important Result in Triangles Class 9

Each angle of an equilateral triangle = 60°

• In equilateral △ABC: AB = BC = CA
• AB = BC ⟹ ∠C = ∠A (angles opp. equal sides)
• BC = CA ⟹ ∠A = ∠B
• So ∠A = ∠B = ∠C. Since ∠A + ∠B + ∠C = 180°, each angle = 60°. ∎

Triangles Class 9 Solved Examples

Solved Example 1

If D is the mid-point of hypotenuse AC of right △ABC, prove that BD = ½ AC.

Approach: Produce BD to E such that BD = DE; join EC.

In △ADB and △CDE: AD = DC (D is midpoint); BD = DE (construction); ∠ADB = ∠CDE (vertically opposite).

By SAS: △ADB ≅ △CDE ⟹ AB = EC and ∠CED = ∠ABD (CPCTC).

Since ∠CED and ∠ABD are alternate interior angles, CE ∥ AB, so ∠ABC + ∠ECB = 180°, giving ∠ECB = 90°.

In △ABC and △ECB: AB = EC; BC = BC (common); ∠ABC = ∠ECB = 90°. By SAS: △ABC ≅ △ECB, so AC = EB.

∴ BD = ½ BE = ½ AC.

Hence Proved.

Solved Example 2

In a right triangle, one acute angle is double the other. Prove the hypotenuse is double the smallest side.

Setup: ∠B = 90°, ∠ACB = 2·∠CAB = 2x. Produce CB to D so that BD = CB; join AD.

In △ABD and △ABC: BD = BC; AB = AB; ∠ABD = ∠ABC = 90°. By SAS: △ABD ≅ △ABC.

So AD = AC and ∠DAB = ∠CAB = x. Therefore ∠DAC = 2x = ∠ACD, giving DC = AD (sides opp. equal angles).

DC = 2BC (since BD = BC), and AD = AC, so 2BC = AC. Hence Proved.

Solved Example 3

In △PQR, PQ = PR. Show that PS > PQ (where S is on QR).

PQ = PR ⟹ ∠PRQ = ∠PQR (angles opposite equal sides) … (i)

In △PSQ, ∠PQR is an exterior angle, so ∠PQR > ∠PSQ … (ii)

From (i) and (ii): ∠PRQ > ∠PSQ ⟹ ∠PRS > ∠PSR.

So PS > PR (side opposite larger angle is longer). Since PR = PQ, we get PS > PQ. Hence Proved.

Triangles Class 9 Triangle Inequality Rules and Properties

These five properties govern how sides and angles relate when they are unequal:

• If two sides are unequal, the longer side has the greater angle opposite to it. (In △ABC, if AB > AC then ∠C > ∠B.)
• Conversely, the greater angle has the longer side opposite it. (If ∠A > ∠B then BC > AC.)
• The sum of any two sides of a triangle is always greater than the third side. (AB + BC > AC, etc.)
• Of all line segments drawn from a point to a given line, the perpendicular is the shortest.
• The difference of any two sides is always less than the third side. (|AB − BC| < AC.)

Triangles Class 9 Congruence Criteria Revision Table