Triangles Class 9 ICSE

Written by Greya LakshmiPublished on 12 October 2026

The isosceles triangle chapter in triangles class 9 ICSE starts with one definition: a triangle that has at least two equal sides. The chapter then proves two theorems, states four standard results, and applies them to angle calculations and congruence proofs. Every proof in this chapter follows the same layout of Given, To Prove, Construction and Proof. Exercise 10(A) builds the skills with angle hunting, while Exercise 10(B) turns them into full proofs.

Download Isosceles Triangles Class 9 ICSE Mathematics Chapter 10 PDF

Download the Isosceles Triangles Class 9 ICSE Mathematics chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 11 pages, 1 sections, 45 questions.

What is inside Chapter 10

11 pages · 1 sections · 9 worked examples · 45 exercise questions

Section	Title
10.1	INTRODUCTION

EXERCISE 10(A): 18 questions
EXERCISE 10(B): 27 questions

10.1 INTRODUCTION

What the chapter defines

The chapter opens with three statements that fix the vocabulary. An isosceles triangle has at least two sides equal. An equilateral triangle has all three sides equal.

The third statement is the edge case worth remembering. An equilateral triangle satisfies every property of an isosceles triangle. An isosceles triangle becomes equilateral only when its third angle is also equal.

The chapter also introduces the four labels that shape every theorem: Given, To Prove, Construction and Proof. Write a proof under these four heads and you will not miss a step.

Theorem 1: equal sides give equal angles

Theorem 1 says that if two sides of a triangle are equal, the angles opposite them are equal. In triangle ABC, if $AB = AC$ , then $\angle B = \angle C$ . The equality holds because each of these angles sits opposite one of the equal sides.

The proof uses the chapter’s layout. Given: triangle ABC with $AB = AC$ . To Prove: $\angle B = \angle C$ . Construction: draw $AD \perp BC$ .

Proof: in triangles ABD and ACD, $AB = AC$ by the given condition, $AD = AD$ as a common side, and $\angle ADB = \angle ADC = 90^\circ$ because AD is perpendicular to BC. The two right triangles match in hypotenuse and one leg, so they are congruent by RHS.

$\Delta ABD \cong \Delta ACD \Rightarrow \angle B = \angle C$

Congruent triangles have every matching part equal, so $\angle B = \angle C$ by C.P.C.T.C. The perpendicular worked because it split the isosceles triangle into two right triangles sharing AD, with equal hypotenuses $AB$ and $AC$ .

Theorem 2: equal angles give equal sides

Theorem 2 is the converse of Theorem 1. If two angles of a triangle are equal, the sides opposite them are equal. Given $\angle B = \angle C$ in triangle ABC, the conclusion is $AB = AC$ .

The construction repeats: draw $AD \perp BC$ . In triangles ABD and ACD, $\angle B = \angle C$ by the given condition, $\angle ADB = \angle ADC = 90^\circ$ because AD is perpendicular to BC, and AD is a common side. Two angles plus one side fix the triangles, so they are congruent by AAS.

$\Delta ABD \cong \Delta ACD \Rightarrow AB = AC$

Notice the congruence map the chapter follows. RHS proves Theorem 1, AAS proves Theorem 2, and the worked examples then use ASA, SAS and SSS. When you reach the step where you declare the triangles congruent, name the rule you used.

The four results you prove once

After the theorems, the chapter lists four results that follow directly from them. Each expects a proof once, and each returns inside Exercise 10(B):

The bisector of the vertex angle cuts the base into two equal parts at a right angle.
Extending the two equal sides produces a pair of exterior angles that are equal.
The perpendicular bisector of the base passes through the vertex of the triangle.
The line from the midpoint of the base to the opposite vertex is perpendicular to the base and bisects the vertex angle.

Each proof works by splitting the isosceles triangle into two congruent right triangles, exactly as Theorem 1 does, or by applying the two theorems to the triangles produced. Mark the equal sides before you start.

Angle chase with new numbers

The worked examples join the two theorems to the exterior angle rule: an exterior angle of a triangle equals the sum of the two interior opposite angles. Try the pattern with new numbers.

Take triangle ABC with $AB = BC$ , and point D on the extension of BC such that $AC = CD$ . Let $\angle ADB = 35^\circ$ .

In triangle ACD, $AC = CD$ , so $\angle CAD = \angle CDA = 35^\circ$ . The exterior angle at C gives $\angle ACB = \angle CAD + \angle CDA = 35^\circ + 35^\circ = 70^\circ$ .

In triangle ABC, $AB = BC$ , so the angles opposite them match: $\angle BAC = \angle ACB = 70^\circ$ . The full angle at A follows as $\angle BAD = \angle BAC + \angle CAD = 70^\circ + 35^\circ = 105^\circ$ .

Run the same chase with a variable x instead of 35 degrees, and you get the ratio $\angle BAD : \angle ADB = 3 : 1$ , the ratio this chapter asks you to prove. The chain of equal angles does the work, and the exterior angle rule closes the triangle.

The common mistake is pairing an equal side with the wrong angle. The check that works: name the two equal sides first, because the equal angles are always opposite them. This proof style builds on the congruence rules of the earlier chapter on triangles.

What to practise from Triangles Class 9 ICSE Chapter 10

Exercise 10(A) carries 18 questions. It mixes angle calculations of the find-a-b-c type, including parallel line angle pairs, with short proofs that use the two theorems.

Exercise 10(B) carries 27 questions and is proof heavy. The four results from the introduction reappear here as full proofs, joined by questions on medians, bisectors, altitudes and equilateral triangles. Finishing 10(A) first builds the angle facts and congruence habits the longer proofs need.

EXERCISE 10(A) — 18 questions: angle calculations and short proofs using equal sides and equal angles.
EXERCISE 10(B) — 27 questions: proof-heavy questions applying both theorems and the four standard results.

The other chapters sit beside this one in the Class 9 Maths book. The exercises follow the geometry syllabus published on the official CISCE website.

Triangles Class 9 ICSE frequently asked questions

What are the theorems in triangles class 9 ICSE chapter 10?

Chapter 10 proves two theorems. Theorem 1 says that if two sides of a triangle are equal, the angles opposite them are equal. Theorem 2 is the converse: if two angles are equal, the opposite sides are equal.

How do I prove that the angles opposite the equal sides of an isosceles triangle are equal?

Draw the perpendicular AD from the vertex A to the base BC. In triangles ABD and ACD, use RHS because $AB = AC$ , AD is common, and both angles at D are $90^\circ$ . The congruence gives $\angle B = \angle C$ by C.P.C.T.C.

What is the difference between an isosceles triangle and an equilateral triangle?

An isosceles triangle has at least two equal sides. An equilateral triangle has all three sides equal. An equilateral triangle fits every isosceles property, but an isosceles triangle needs its third angle equal to become equilateral.

Which congruence rule do I use for isosceles triangle proofs in class 9 ICSE?

Use RHS for Theorem 1 and AAS for Theorem 2. The chapter’s worked examples then apply ASA, SAS and SSS. Always name the rule at the step where you declare the triangles congruent.

How many questions are in Exercise 10(A) and Exercise 10(B) of this chapter?

Exercise 10(A) has 18 questions. Exercise 10(B) has 27 questions. The first exercise trains angle calculations and short proofs, while the second is mostly full proofs.

Reference: ICSE Class 9 Mathematics textbook material, chapter 10: Isosceles Triangles — CISCE-aligned.

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