Exercise 7.2 — Properties of Triangles

Some more properties of triangles.

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Class 9 Mathematics · Chapter 7

Triangles — Exercise 7.2

Isosceles triangles, congruence rules, and angle-side relationships — step-by-step solutions for CBSE, Telangana & Andhra Pradesh board exams.

The Isosceles Triangle Theorem — Foundation of Exercise 7.2

Before diving into the exercise questions, it is essential to understand the core theorem that drives every single problem in Exercise 7.2. This theorem appears in Theorem 7.2 of your Class 9 NCERT / Telangana / Andhra Pradesh textbook.

Theorem 7.2

Angles opposite to equal sides of an isosceles triangle are equal.
Conversely: If two angles of a triangle are equal, the sides opposite to them are also equal.

How the Theorem is Proved

In isosceles △ABC where AB = AC, the goal is to prove that ∠B = ∠C. The construction step is key: draw the angle bisector AD of ∠A so that it meets BC at D. This creates two smaller triangles — △ABD and △ACD — which can be compared.

Given: △ABC with AB = AC To Prove: ∠B = ∠C Construction: Draw angle bisector AD of ∠A, meeting BC at D   In △ABD and △ACD:   AB = AC (given)   ∠BAD = ∠CAD (AD is the angle bisector — construction)   AD = AD (common side) SAS Rule → △ABD ≅ △ACD ∠ABD = ∠ACD (CPCT) ∴ ∠B = ∠C ✓
A B C D ∠B ∠C bisector AB = AC
Figure: Isosceles △ABC with angle bisector AD. The bisector divides the triangle into two congruent triangles △ABD ≅ △ACD, proving ∠B = ∠C.
📌 Remember the Converse too: If ∠B = ∠C in △ABC, then AB = AC. This converse is used in several questions below to identify isosceles triangles.

Congruence Rules Used in This Exercise

All five questions in Exercise 7.2 rely on either the SAS rule or the AAS rule to prove triangle congruence. Understanding which rule to apply is the key skill tested in board exams.

Rule Full Name What Must Match Used In
SAS Side–Angle–Side Two sides & the included angle between them Theorem proof, Q1, Q2
AAS Angle–Angle–Side Two angles & a non-included side Q3, Q4
CPCT Corresponding Parts of Congruent Triangles Used after proving congruence to conclude equal sides/angles All questions
💡 Board Exam Tip: Always state the congruence rule explicitly in your proof. Writing "△ABD ≅ △ACD (by SAS)" earns full marks. Skipping the rule name can cost you 1 mark in Telangana and AP board exams.

Question 1 — Angle Bisectors of an Isosceles Triangle

Question 1
In isosceles triangle ABC, 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
Understanding the Setup

Since AB = AC, the base angles are equal: ∠B = ∠C. The bisectors of ∠B and ∠C each cut these equal angles in half, creating two new equal half-angles. Point O is where these bisectors meet.

Part (i) — Prove OB = OC
In △ABC: AB = AC (given) ∠B = ∠C (angles opposite to equal sides) ½∠B = ½∠C i.e., ∠OBC = ∠OCB (BO and CO are bisectors of ∠B and ∠C) In △OBC: ∠OBC = ∠OCB ⇒ OB = OC ✓ (sides opposite to equal angles)
Part (ii) — Prove AO bisects ∠A
In △OAB and △OAC:   OB = OC (proved in part i)   ∠OBA = ∠OCA (½∠B = ½∠C, since ∠B = ∠C)   AB = AC (given) SAS Rule → △OAB ≅ △OAC ∠OAB = ∠OAC (CPCT) ∴ AO bisects ∠A ✓
Key Insight: This question beautifully demonstrates that in an isosceles triangle, the point where the base-angle bisectors meet lies on the axis of symmetry of the triangle.

Question 2 — Perpendicular Bisector Proves Isosceles Triangle

Question 2
In △ABC, AD is the perpendicular bisector of BC. Show that △ABC is an isosceles triangle in which AB = AC.

Here the direction is reversed — instead of starting with an isosceles triangle, we are given a perpendicular bisector and asked to prove the triangle is isosceles. The perpendicular bisector gives us two facts: AD ⊥ BC (right angles) and BD = DC (D is the midpoint of BC).

In △ADB and △ADC:   DB = DC (AD is the perpendicular bisector of BC — given)   ∠ADB = ∠ADC = 90° (AD ⊥ BC — given)   AD = AD (common side) SAS Rule → △ADB ≅ △ADC AB = AC (CPCT) ∴ △ABC is isosceles ✓
A B C D BD = DC AD ⊥ BC
AD is the perpendicular bisector of BC. The two right-angled triangles △ADB and △ADC are congruent by SAS, so AB = AC.
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Question 3 — Equal Altitudes Prove Equal Sides

Question 3
ABC is an isosceles triangle in which altitudes BD and CE are drawn to equal sides AC and AB respectively. Show that these altitudes are equal (BD = CE).

An altitude is a perpendicular from a vertex to the opposite side. Since AB = AC, we already know ∠B = ∠C. The altitudes are drawn to the equal sides, not to the base — so we compare triangles that share BC as a common side.

In △ABC: AB = AC (given) ∠EBC = ∠DCB (angles opposite to equal sides AB = AC)   In △BEC and △CDB:   ∠BEC = ∠CDB = 90° (CE ⊥ AB and BD ⊥ AC — altitudes)   BC = CB (common side)   ∠EBC = ∠DCB (proved above) AAS Rule → △BEC ≅ △CDB ⇒ CE = BD ✓ (CPCT)
📐 Why AAS and not SAS here? We know two angles and a non-included side (BC). We do not know the side between the two angles, so SAS cannot be applied. AAS is the correct choice whenever the known side is not between the two known angles.

Question 4 — Equal Altitudes Prove the Triangle is Isosceles

Question 4
ABC is a triangle in which altitudes BD and CE to sides AC and AB are equal. Show that: (i) △ABD ≅ △ACE    (ii) AB = AC (i.e., ABC is isosceles)

This is the reverse of Question 3 — we are now given that the two altitudes are equal and asked to conclude the triangle is isosceles. Notice both triangles share the common angle ∠A.

Part (i) — Prove △ABD ≅ △ACE
In △ABD and △ACE:   ∠BAD = ∠CAE (same angle ∠A — common to both triangles)   ∠ADB = ∠AEC = 90° (BD ⊥ AC and CE ⊥ AB — altitudes)   BD = CE (given) AAS Rule△ABD ≅ △ACE ✓
Part (ii) — Prove AB = AC
Since △ABD ≅ △ACE (proved above) AB = AC (CPCT) ∴ △ABC is an isosceles triangle ✓
💡 Exam Strategy: For Part (i), always identify the common angle first — it is the easiest element of congruence to spot and is often overlooked. In this problem, angle A appears in both △ABD and △ACE, making it the anchor for the AAS proof.

Question 5 — Two Isosceles Triangles on the Same Base

Question 5
△ABC and △DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

Both triangles sit on base BC, with A above BC and D below (or both on the same side). Since each is isosceles, we get two pairs of equal base angles. The trick is to add them together.

A B C D AB=AC DB=DC
△ABC (upper, with AB = AC) and △DBC (lower, with DB = DC) share the same base BC, forming a kite shape.
In △ABC: AB = AC (given) ∠ABC = ∠ACB    …(1) (angles opposite to equal sides)   In △DBC: DB = DC (given) ∠DBC = ∠DCB    …(2) (angles opposite to equal sides)   Adding equations (1) and (2): ∠ABC + ∠DBC = ∠ACB + ∠DCB ⇒ ∠ABD = ∠ACD ✓
Elegant Approach: No congruence rule is needed here at all! Simply applying the isosceles triangle theorem to each triangle and then adding the resulting angle equations gives the result directly. This type of "equation approach" is a common pattern in Class 9 geometry.

Quick Reference — All 5 Questions at a Glance

📐
Question 1
SAS Rule
OB = OC; AO bisects ∠A
Question 2
SAS Rule
Perp. bisector → AB = AC
Question 3
AAS Rule
Equal sides → equal altitudes
🔄
Question 4
AAS Rule
Equal altitudes → isosceles
Question 5
No congruence needed
Adding angle equations
Question Given To Prove / Show Method Key Step
Q1(i) AB = AC; bisectors of ∠B, ∠C meet at O OB = OC Isosceles theorem + converse ∠OBC = ∠OCB → OB = OC
Q1(ii) Above + OB = OC AO bisects ∠A SAS → CPCT △OAB ≅ △OAC → ∠OAB = ∠OAC
Q2 AD ⊥ BC; BD = DC AB = AC SAS → CPCT △ADB ≅ △ADC → AB = AC
Q3 AB = AC; BD ⊥ AC; CE ⊥ AB BD = CE AAS → CPCT △BEC ≅ △CDB → CE = BD
Q4 BD ⊥ AC; CE ⊥ AB; BD = CE AB = AC AAS → CPCT △ABD ≅ △ACE → AB = AC
Q5 AB = AC; DB = DC; same base BC ∠ABD = ∠ACD Adding angle equations ∠ABC + ∠DBC = ∠ACB + ∠DCB

Common Mistakes to Avoid in Exercise 7.2

  • Not stating the congruence rule: Writing "△ABD ≅ △ACD" without specifying "(by SAS)" is incomplete and loses marks in board exams. Always name the rule — SAS, AAS, SSS, ASA, or RHS.
  • Confusing SAS and AAS: SAS requires the angle to be between the two sides. If the equal side is not between the two angles, it is AAS — not SAS. Getting this wrong changes the validity of the proof.
  • Forgetting to justify the "common side": In every proof where AD = AD or BC = CB, write the reason "(common side)" explicitly. Examiners check for this.
  • Applying the isosceles theorem without mentioning which sides are equal: Always clearly state "AB = AC (given)" before writing "∠B = ∠C". The theorem requires the given condition to be written first.
  • Skipping the construction in Q1: The join AO must be stated before using it in the proof. Never use a line in a proof that has not been constructed or given.
Most Frequent Board Exam Error: In Q3 and Q4, students often apply SAS instead of AAS. Remember — the right angle (90°) and the common or given side are not in the SAS arrangement here. Always sketch the triangles and label the matching parts before choosing the rule.

What Exercise 7.2 Prepares You For

The isosceles triangle properties proved in Exercise 7.2 are foundational for several topics in Class 9 and beyond. In Exercise 7.3, you use angle-side relationships to compare sides and angles of unequal triangles — a direct extension of the isosceles triangle ideas here.

For CBSE, Telangana, and Andhra Pradesh board examinations, questions from Exercise 7.2 regularly appear as 3-mark or 4-mark proofs. The most commonly tested problems are Q1, Q3, and Q5. Mastering the step-by-step format — Given → To Prove → Construction → Proof → Conclusion — is as important as knowing the content itself.

In Class 10, these congruence techniques reappear in Chapter 6 (Similar Triangles) and in coordinate geometry proofs. The ability to identify which congruence rule applies — a skill sharpened by Exercise 7.2 — is tested throughout Classes 9 and 10.

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