Class 9Maths · GeometryFull chapter

Triangles

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
Congruence of TrianglesCongruence Criteria: SAS, ASA and AASCongruence Criteria: SSS and RHSProperties of a Triangle: Isosceles TrianglesInequalities in a Triangle

Congruence of Triangles

Quick answer Two triangles are congruent if they have exactly the same shape and size, so every corresponding side and angle is equal (CPCT).

Congruence means two figures have exactly the same shape and the same size — if you place one on top of the other, they match perfectly. Two triangles are congruent if all three pairs of corresponding sides and all three pairs of corresponding angles are equal.

We write this using the symbol ≅. If ΔABC ≅ ΔDEF, it means vertex A matches vertex D, B matches E and C matches F — in that exact order. This matching is called correspondence, and once it is fixed, every equal pair follows automatically. This rule is called CPCT — Corresponding Parts of Congruent Triangles (are equal).

Worked Example: If ΔABC ≅ ΔPQR, AB = 6 cm, BC = 8 cm, ∠A = 50° and ∠C = 70°, find PQ, QR and ∠P.

Since the correspondence is A ↔ P, B ↔ Q, C ↔ R, by CPCT: PQ = AB = 6 cm, QR = BC = 8 cm, and ∠P = ∠A = 50°.

Note carefully: writing ΔABC ≅ ΔQRP is a completely different statement from ΔABC ≅ ΔPQR — the order of letters tells you exactly which sides and angles are equal, so it must never be changed carelessly.

CPCT ΔABC ≅ ΔDEF ⇒ AB=DE, BC=EF, CA=FD, ∠A=∠D, ∠B=∠E, ∠C=∠F Corresponding Parts of Congruent Triangles are Equal
Remember
  • Congruent triangles have identical shape and size — every corresponding side and angle is equal.
  • The order of vertices in ΔABC ≅ ΔDEF fixes the correspondence: A↔D, B↔E, C↔F.
  • CPCT (Corresponding Parts of Congruent Triangles) lets us find unknown sides/angles once congruence is proved.
  • Congruent triangles always have equal area, but two triangles with equal area need not be congruent.
  • Congruence of triangles behaves as an equivalence relation: every triangle is congruent to itself, congruence can be stated in either order, and it carries through a chain of triangles.

Congruence Criteria: SAS, ASA and AAS

Quick answer Two triangles are congruent if two sides and the included angle match (SAS), two angles and the included side match (ASA), or two angles and any one side match (AAS).

To prove two triangles congruent, we do not need to check all six measurements (3 sides + 3 angles) — a few carefully chosen ones are enough.

SAS (Side-Angle-Side): If two sides and the angle included between them in one triangle are equal to the corresponding two sides and the included angle of another triangle, the triangles are congruent. This is taken as a basic axiom.

ASA (Angle-Side-Angle): If two angles and the side included between them in one triangle are equal to the corresponding two angles and included side of another triangle, the triangles are congruent.

AAS (Angle-Angle-Side): If any two angles and one side (not necessarily the included side) of one triangle are equal to the corresponding two angles and side of another triangle, the triangles are congruent. This reduces to ASA, because once two angles of a triangle are known, the third angle is fixed by the angle sum property (180°).

Worked Example: In ΔABC and ΔDEF, ∠B = ∠E = 70°, ∠C = ∠F = 50°, and BC = EF = 5 cm. Show the triangles are congruent and find ∠A.

Here BC is included between ∠B and ∠C (similarly EF between ∠E and ∠F), so by ASA, ΔABC ≅ ΔDEF. By CPCT, AB = DE, AC = DF and ∠A = ∠D. Also ∠A = 180° − ∠B − ∠C = 180° − 70° − 50° = 60°.

SAS Congruence AB=DE, ∠A=∠D (included), AC=DF ⇒ ΔABC≅ΔDEF
ASA Congruence ∠B=∠E, BC=EF (included side), ∠C=∠F ⇒ ΔABC≅ΔDEF
AAS Congruence ∠A=∠D, ∠B=∠E, BC=EF (non-included side) ⇒ ΔABC≅ΔDEF
Remember
  • SAS: two sides + the included angle equal ⇒ congruent.
  • ASA: two angles + the included side equal ⇒ congruent.
  • AAS: two angles + any one side equal ⇒ congruent (reduces to ASA via angle sum property).
  • SAS is taken as a basic assumption (axiom); ASA and AAS can be proved using SAS.
  • Always identify which side is 'included' between the two known angles before naming the criterion.

Congruence Criteria: SSS and RHS

Quick answer Two triangles are congruent if all three sides are equal (SSS), or, for right triangles, the hypotenuse and one side are equal (RHS).

SSS (Side-Side-Side): If all three sides of one triangle are equal to the corresponding three sides of another triangle, the two triangles are congruent — no angle information is even needed.

RHS (Right angle-Hypotenuse-Side): This criterion applies only to right triangles. If the hypotenuse and one side of a right triangle are equal to the hypotenuse and corresponding side of another right triangle, the two triangles are congruent.

Worked Example (SSS): In ΔABC and ΔPQR, AB = PQ = 4 cm, BC = QR = 5.5 cm and CA = RP = 6 cm. Are the triangles congruent?

All three pairs of corresponding sides are equal, so by SSS, ΔABC ≅ ΔPQR. By CPCT, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R.

Worked Example (RHS): ΔABC is right-angled at B and ΔPQR is right-angled at Q. If the hypotenuse AC = PR = 10 cm and AB = PQ = 6 cm, show the triangles are congruent, and find BC and QR.

∠B = ∠Q = 90°, hypotenuse AC = PR = 10 cm, and side AB = PQ = 6 cm, so by RHS, ΔABC ≅ ΔPQR. By CPCT, BC = QR. Using Pythagoras: BC = √(AC2 − AB2) = √(102 − 62) = √(100 − 36) = √64 = 8 cm. So BC = QR = 8 cm.

SSS Congruence AB=DE, BC=EF, CA=FD ⇒ ΔABC≅ΔDEF
RHS Congruence ∠B=∠Q=90°, AC=PR (hypotenuse), AB=PQ ⇒ ΔABC≅ΔPQR
Remember
  • SSS: all three sides equal ⇒ congruent (no angle needed).
  • RHS applies only to right triangles: hypotenuse + one side equal ⇒ congruent.
  • In RHS, the right angle is common to both triangles, so only two more measurements (hypotenuse and one side) need to be checked.
  • SSA (two sides and a non-included angle) is NOT a valid congruence criterion in general.
  • Once SSS or RHS is proved, CPCT gives all remaining equal angles/sides.

Properties of a Triangle: Isosceles Triangles

Quick answer In an isosceles triangle, the angles opposite the two equal sides are equal, and conversely equal angles have equal opposite sides.

An isosceles triangle has two equal sides. The angles opposite these equal sides are called base angles, and they turn out to be equal too.

Theorem: If two sides of a triangle are equal, the angles opposite them are equal. That is, if AB = AC in ΔABC, then ∠B = ∠C. This is proved by drawing the bisector AD of ∠A and showing ΔABD ≅ ΔACD by SAS (AB = AC, ∠BAD = ∠CAD, AD = AD common); then ∠B = ∠C by CPCT.

Converse Theorem: If two angles of a triangle are equal, the sides opposite them are also equal. That is, if ∠B = ∠C in ΔABC, then AB = AC.

Worked Example: In ΔABC, AB = AC = 7 cm and ∠B = 55°. Find ∠A and ∠C.

Since AB = AC, the base angles are equal: ∠C = ∠B = 55°. Using the angle sum property, ∠A = 180° − ∠B − ∠C = 180° − 55° − 55° = 70°.

A useful special case: in an equilateral triangle all three sides are equal, so applying the theorem to each pair of equal sides shows all three angles are equal, and since they add up to 180°, each angle of an equilateral triangle is exactly 60°.

Isosceles Triangle Theorem AB=AC ⇒ ∠B=∠C Angles opposite equal sides are equal
Converse ∠B=∠C ⇒ AB=AC Sides opposite equal angles are equal
Angle Sum Property ∠A+∠B+∠C=180°
Remember
  • Isosceles triangle: two equal sides ⇒ base angles opposite them are equal.
  • Converse also holds: two equal angles ⇒ the sides opposite them are equal.
  • The bisector of the vertex angle in an isosceles triangle also bisects the base and is perpendicular to it.
  • In an equilateral triangle, every angle equals 60°.
  • These theorems are proved using SAS congruence, not by direct measurement.

Inequalities in a Triangle

Quick answer In any triangle, the angle opposite the longer side is larger, and the sum of any two sides is always greater than the third side.

Just as sides and angles can be equal, they can also be compared when they are unequal — these are called inequalities in a triangle.

Theorem 1: If two sides of a triangle are unequal, the angle opposite the longer side is larger. That is, if AC > AB in ΔABC, then ∠B > ∠C.

Theorem 2 (Converse): If two angles of a triangle are unequal, the side opposite the larger angle is longer. That is, if ∠B > ∠C, then AC > AB.

Theorem 3 (Triangle Inequality): The sum of the lengths of any two sides of a triangle is always greater than the length of the third side. Equivalently, the difference between any two sides is always less than the third side.

Worked Example 1: In ΔABC, AB = 4 cm, BC = 6 cm and CA = 9 cm. Which is the largest angle?

The longest side is CA = 9 cm, and the angle opposite CA is ∠B. So ∠B is the largest angle of the triangle.

Worked Example 2: Can a triangle have sides 5 cm, 6 cm and 12 cm?

Check the triangle inequality with the two smaller sides: 5 + 6 = 11 cm, which is not greater than 12 cm. Since the sum of two sides is not greater than the third side, such a triangle cannot exist.

Angle-Side Inequality AC>AB ⇒ ∠B>∠C Angle opposite longer side is larger
Side-Angle Inequality (converse) ∠B>∠C ⇒ AC>AB
Triangle Inequality AB+BC>CA, BC+CA>AB, CA+AB>BC
Remember
  • Angle opposite the longer side is larger; side opposite the larger angle is longer.
  • Sum of any two sides of a triangle must be greater than the third side.
  • Equivalently, the difference of any two sides must be less than the third side.
  • These inequality checks are used to test whether three given lengths can actually form a triangle.
  • The perpendicular distance from a point to a line is the shortest distance from that point to the line.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

ΔABC ≅ ΔDEF ⇒ AB=DE, BC=EF, CA=FD, ∠A=∠D, ∠B=∠E, ∠C=∠F
CPCT
AB=DE, ∠A=∠D (included), AC=DF ⇒ ΔABC≅ΔDEF
SAS Congruence
∠B=∠E, BC=EF (included side), ∠C=∠F ⇒ ΔABC≅ΔDEF
ASA Congruence
∠A=∠D, ∠B=∠E, BC=EF (non-included side) ⇒ ΔABC≅ΔDEF
AAS Congruence
AB=DE, BC=EF, CA=FD ⇒ ΔABC≅ΔDEF
SSS Congruence
∠B=∠Q=90°, AC=PR (hypotenuse), AB=PQ ⇒ ΔABC≅ΔPQR
RHS Congruence
AB=AC ⇒ ∠B=∠C
Isosceles Triangle Theorem
∠B=∠C ⇒ AB=AC
Converse
∠A+∠B+∠C=180°
Angle Sum Property
AC>AB ⇒ ∠B>∠C
Angle-Side Inequality
∠B>∠C ⇒ AC>AB
Side-Angle Inequality (converse)
AB+BC>CA, BC+CA>AB, CA+AB>BC
Triangle Inequality

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Congruence of Triangles easy

If ΔABC ≅ ΔPQR, which of the following is correct by CPCT?

Q2 SAS Criterion easy

Two triangles are congruent by the SAS criterion if:

Q3 ASA and AAS medium

In ΔABC and ΔDEF, ∠B = ∠E, ∠C = ∠F and AB = DE. Which congruence criterion proves ΔABC ≅ ΔDEF?

Q4 RHS Criterion easy

The RHS congruence criterion can be applied only when:

Q5 SSS Criterion medium

In ΔABC and ΔPQR, AB = PQ = 4 cm, BC = QR = 5 cm and CA = RP = 6 cm. By which criterion are the triangles congruent?

Q6 Isosceles Triangle medium

In an isosceles triangle ABC with AB = AC, if ∠B = 65°, what is ∠A?

Q7 Converse of Isosceles Theorem easy

If in ΔABC, ∠B = ∠C, then which of the following must be true?

Q8 Triangle Inequality medium

Which of the following sets of lengths can be the sides of a triangle?

Q9 Angle Opposite Longer Side medium

In ΔABC, AB = 5 cm, BC = 7 cm and CA = 9 cm. Which angle is the largest?

Q10 Side Opposite Larger Angle medium

In ΔPQR, ∠P = 50°, ∠Q = 60° and ∠R = 70°. Which side is the longest?

Q11 Triangle Inequality hard

The lengths of two sides of a triangle are 8 cm and 15 cm. Which of these cannot be the length of the third side?

Q12 Congruence Correspondence hard

If ΔABC ≅ ΔFED, which of the following is correct?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (PA = PB and QA = QB). Show that the line PQ is the perpendicular bisector of AB.SSS and SAS Congruence

Given: PA = PB and QA = QB.

Step 1: In ΔPAQ and ΔPBQ: PA = PB (given), QA = QB (given), PQ = PQ (common). So ΔPAQ ≅ ΔPBQ by SSS. By CPCT, ∠APQ = ∠BPQ, i.e. PQ bisects ∠APB.

Step 2: Let PQ intersect AB at C. In ΔPAC and ΔPBC: PA = PB (given), ∠APC = ∠BPC (proved above), PC = PC (common). So ΔPAC ≅ ΔPBC by SAS.

Step 3: By CPCT, AC = BC, so C is the midpoint of AB, and ∠ACP = ∠BCP. Since ∠ACP + ∠BCP = 180° (linear pair on line AB), each equals 90°.

Hence PQ ⊥ AB and PQ bisects AB, so PQ is the perpendicular bisector of AB.

2 In ΔABC, AB = AC. The bisectors of ∠B and ∠C intersect each other at O. Show that (i) OB = OC and (ii) AO bisects ∠A.Isosceles Triangle + SAS

Given: AB = AC, BO and CO bisect ∠B and ∠C respectively.

Step 1: Since AB = AC, ∠ABC = ∠ACB (angles opposite equal sides). As O lies on the bisectors, ∠OBC = ½∠ABC and ∠OCB = ½∠ACB, so ∠OBC = ∠OCB.

Step 2: In ΔOBC, since ∠OBC = ∠OCB, the sides opposite them are equal: OB = OC. This proves (i).

Step 3: In ΔABO and ΔACO: AB = AC (given), ∠ABO = ∠ACO (halves of equal angles), OB = OC (proved). So ΔABO ≅ ΔACO by SAS.

By CPCT, ∠BAO = ∠CAO, i.e. AO bisects ∠A. This proves (ii).

3 AD is a line from A meeting BC at D, such that D is the midpoint of BC and AD ⊥ BC. Show that ΔABC is isosceles, with AB = AC.Isosceles Triangle via SAS

Given: BD = DC (D is midpoint of BC) and AD ⊥ BC, so ∠ADB = ∠ADC = 90°.

Step 1: In ΔADB and ΔADC: BD = DC (given), ∠ADB = ∠ADC = 90° (given), AD = AD (common). So ΔADB ≅ ΔADC by SAS.

Step 2: By CPCT, AB = AC.

Hence ΔABC is isosceles with AB = AC.

4 Show that the angles of an equilateral triangle are 60° each.Isosceles Triangle Theorem

Given: ΔABC is equilateral, so AB = BC = CA.

Step 1: Since AB = BC, the angles opposite them are equal: the angle opposite AB is ∠C, and the angle opposite BC is ∠A. So ∠A = ∠C.

Step 2: Since BC = CA, similarly the angle opposite BC (∠A) equals the angle opposite CA (∠B). So ∠A = ∠B.

Step 3: Combining, ∠A = ∠B = ∠C. By the angle sum property, ∠A + ∠B + ∠C = 180°, so 3∠A = 180°, giving ∠A = 60°.

Hence each angle of an equilateral triangle is 60°.

5 ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (BE = CF). Show that (i) ΔABE ≅ ΔACF and (ii) AB = AC, i.e. ABC is isosceles.AAS Congruence

Given: BE ⊥ AC at E, CF ⊥ AB at F, and BE = CF.

Step 1: In ΔABE and ΔACF: ∠AEB = ∠AFC = 90° (altitudes), ∠A = ∠A (common angle at vertex A), BE = CF (given).

Step 2: Here BE is the side opposite ∠A in ΔABE (not the side included between the two known angles), so by the AAS criterion, ΔABE ≅ ΔACF. This proves (i).

Step 3: By CPCT, AB = AC. This proves (ii), so ΔABC is isosceles.

6 In the given figure, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.Inequalities in a Triangle

Given: AB is extended to P, AC is extended to Q, and ∠PBC < ∠QCB.

Step 1: ∠ABC and ∠PBC form a linear pair, so ∠ABC = 180° − ∠PBC. Similarly, ∠ACB and ∠QCB form a linear pair, so ∠ACB = 180° − ∠QCB.

Step 2: Since ∠PBC < ∠QCB, subtracting from 180° reverses the inequality: 180° − ∠PBC > 180° − ∠QCB, i.e. ∠ABC > ∠ACB.

Step 3: Since ∠ABC > ∠ACB, the side opposite the larger angle is longer. The side opposite ∠ABC is AC, and the side opposite ∠ACB is AB.

Hence AC > AB.

Previous-year board questions 4

Q1 In ΔABC, AB = AC and AD is the bisector of ∠BAC meeting BC at D. Prove that (i) ΔABD ≅ ΔACD, and (ii) AD ⊥ BC and D is the midpoint of BC. CBSE 2023 3 marks

Given: AB = AC, and AD bisects ∠BAC (∠BAD = ∠CAD).

Step 1: In ΔABD and ΔACD: AB = AC (given), ∠BAD = ∠CAD (AD bisects ∠A), AD = AD (common). So by SAS, ΔABD ≅ ΔACD. This proves (i).

Step 2: By CPCT, BD = DC, so D is the midpoint of BC. Also by CPCT, ∠ADB = ∠ADC.

Step 3: Since ∠ADB and ∠ADC form a linear pair (B, D, C are collinear), ∠ADB + ∠ADC = 180°. As ∠ADB = ∠ADC, each equals 90°.

Hence AD ⊥ BC, and D is the midpoint of BC. This proves (ii).

Q2 In ΔPQR, ∠P = 70° and ∠Q = 45°. Which is the smallest side of the triangle? Give reason. CBSE 2022 2 marks

Step 1: By the angle sum property, ∠R = 180° − ∠P − ∠Q = 180° − 70° − 45° = 65°.

Step 2: The three angles are ∠P = 70°, ∠Q = 45°, ∠R = 65°. The smallest angle is ∠Q = 45°.

Step 3: The side opposite the smallest angle is the smallest side. The side opposite ∠Q is PR.

Hence PR is the smallest side of ΔPQR.

Q3 D and E are points on side BC of ΔABC such that BD = EC and AD = AE. Show that ΔABD ≅ ΔACE. CBSE Periodic Test 2021 5 marks

Given: BD = EC and AD = AE, with D and E lying on BC (order B, D, E, C).

Step 1: Since AD = AE, ΔADE is isosceles, so the base angles are equal: ∠ADE = ∠AED.

Step 2: ∠ADB and ∠ADE form a linear pair on line BC at D, so ∠ADB = 180° − ∠ADE. Similarly, ∠AEC and ∠AED form a linear pair at E, so ∠AEC = 180° − ∠AED.

Step 3: Since ∠ADE = ∠AED, it follows that ∠ADB = ∠AEC.

Step 4: In ΔABD and ΔACE: BD = EC (given), ∠ADB = ∠AEC (proved), AD = AE (given). So by SAS, ΔABD ≅ ΔACE.

Q4 In ΔABC, AB = 5 cm, BC = 6 cm and AC = 7 cm. State which angle of the triangle is the largest and which is the smallest, giving reasons. CBSE 2020 2 marks

Step 1: The three sides are AB = 5 cm, BC = 6 cm, AC = 7 cm. The longest side is AC = 7 cm, and the shortest side is AB = 5 cm.

Step 2: The angle opposite the longest side is the largest angle, and the angle opposite the shortest side is the smallest angle. The side opposite ∠B is AC, and the side opposite ∠C is AB.

Hence ∠B is the largest angle (opposite AC = 7 cm), and ∠C is the smallest angle (opposite AB = 5 cm).

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