Triangles Class 9 Solutions and Mind Map (Free PDF Download)

Introduction to Triangles

A triangle is one of the most fundamental shapes in geometry. The word ‘Tri’ means ‘three’, and a triangle is a closed figure formed by three intersecting lines. Every triangle has three sides, three angles, and three vertices.

In a triangle ABC (denoted as ∆ABC):

• AB, BC, and CA are the three sides
• ∠A, ∠B, and ∠C are the three angles
• A, B, and C are the three vertices

This chapter covers the concept of congruence of triangles, the criteria for determining when two triangles are congruent, properties of isosceles triangles, and inequalities in triangles.

7.2 Congruence of Triangles

Understanding Congruent Figures

Two figures are congruent when they are equal in all respects—meaning they have the same shape and size. When you place congruent figures one on top of the other, they cover each other completely.

Real-life examples of congruent objects:

• Two identical photographs of the same size
• Two bangles of equal size
• Two ATM cards from the same bank issued in the same year
• Ice trays in a refrigerator (all moulds are congruent)
• Pen refills of the same size (congruent refills fit perfectly)

Congruent Circles and Squares

• When two circles of the same radius are placed one on another, they are congruent circles
• When two squares with equal side lengths are placed one on another, they are congruent squares
• Two equilateral triangles with equal sides are congruent equilateral triangles

Congruent Triangles

Two triangles are congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.

When ∆PQR is congruent to ∆ABC, we write: ∆PQR = ∆ABC

Correspondence Between Congruent Triangles

When triangles are congruent, there is a one-to-one correspondence between their vertices:

• PQ covers AB, QR covers BC, and RP covers CA
• ∠P covers ∠A, ∠Q covers ∠B, and ∠R covers ∠C

This correspondence is written as: P ↔ A, Q ↔ B, R ↔ C

Imp Point: The order of vertices is crucial. We must write ∆PQR = ∆ABC (not ∆PQR = ∆BAC), as this indicates the correct correspondence.

Corresponding Parts of Congruent Triangles (CPCT)

When two triangles are congruent, their corresponding parts (sides and angles) are equal. This is abbreviated as CPCT.

Example: If ∆ABC ≅ ∆PQR, then:

• AB = PQ (corresponding sides)
• BC = QR (corresponding sides)
• CA = RP (corresponding sides)
• ∠A = ∠P (corresponding angles)
• ∠B = ∠Q (corresponding angles)
• ∠C = ∠R (corresponding angles)

7.3 Criteria for Congruence of Triangles

To prove that two triangles are congruent, we don’t need to show that all six parts (three sides and three angles) are equal. Instead, we use specific criteria that require equality of only a few parts.

1. SAS Congruence Rule (Side-Angle-Side)

Axiom 7.1: Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

What is an included angle? The included angle is the angle between two sides.

Example: In ∆ABC and ∆PQR:

• If AB = PQ (side)
• If ∠B = ∠Q (included angle between AB and BC)
• If BC = QR (side)
• Then ∆ABC ≅ ∆PQR (by SAS rule)

Imp Note: The angle must be included between the two equal sides. If the angle is not included, the triangles may not be congruent (SSA rule is not valid).

2. ASA Congruence Rule (Angle-Side-Angle)

Theorem 7.1: Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle.

What is an included side? The included side is the side between two angles.

Example: In ∆ABC and ∆DEF:

• If ∠B = ∠E (angle)
• If BC = EF (included side between ∠B and ∠C)
• If ∠C = ∠F (angle)
• Then ∆ABC ≅ ∆DEF (by ASA rule)

3. AAS Congruence Rule (Angle-Angle-Side)

Two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal.

Why does this work? The sum of angles in a triangle is 180°. If two pairs of angles are equal, the third pair of angles must also be equal. Therefore, having two angles equal automatically makes all three angles equal.

Example: If ∠A = ∠D, ∠B = ∠E, and AB = DE, then ∆ABC ≅ ∆DEF (by AAS rule).

4. SSS Congruence Rule (Side-Side-Side)

Theorem 7.4: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Example: In ∆ABC and ∆PQR:

• If AB = PQ (side)
• If BC = QR (side)
• If CA = RP (side)
• Then ∆ABC ≅ ∆PQR (by SSS rule)

5. RHS Congruence Rule (Right angle-Hypotenuse-Side)

Theorem 7.5: 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.

Imp Note: This rule applies only to right-angled triangles.

Example: In right triangles ABC and PQR (with right angles at A and P respectively):

• If BC = QR (hypotenuse)
• If AB = PQ (one side)
• Then ∆ABC ≅ ∆PQR (by RHS rule)

7.4 Properties of Isosceles Triangles

Definition of Isosceles Triangle

An isosceles triangle is a triangle in which two sides are equal.

In ∆ABC, if AB = AC, then ∆ABC is isosceles.

• AB and AC are called the equal sides
• BC is called the base
• ∠B and ∠C are called the base angles
• ∠A is called the vertex angle or apex angle

Theorem 7.2: Angles Opposite to Equal Sides

Theorem 7.2: Angles opposite to equal sides of an isosceles triangle are equal.

Statement: In an isosceles triangle, if two sides are equal, then the angles opposite to these equal sides are equal.

In ∆ABC, if AB = AC, then ∠B = ∠C (or ∠ABC = ∠ACB)

Proof:

1. Let ∆ABC be an isosceles triangle with AB = AC
2. Draw the bisector of ∠A and let it meet BC at point D
3. In ∆BAD and ∆CAD:
   • AB = AC (given)
   • ∠BAD = ∠CAD (by construction)
   • AD = AD (common)
   • Therefore, ∆BAD ≅ ∆CAD (by SAS rule)
4. Since the triangles are congruent, ∠B = ∠C (CPCT)

Theorem 7.3: Sides Opposite to Equal Angles

Theorem 7.3: The sides opposite to equal angles of a triangle are equal.

This is the converse of Theorem 7.2: In a triangle, if two angles are equal, then the sides opposite to these equal angles are equal.

In ∆ABC, if ∠B = ∠C, then AB = AC

Imp Result: A triangle is isosceles if and only if two of its angles are equal.

Properties of an Equilateral Triangle

An equilateral triangle is a triangle in which all three sides are equal.

Theorem: Each angle of an equilateral triangle is 60°.

Proof:

1. Let ∆ABC be an equilateral triangle, so AB = BC = CA
2. By Theorem 7.2, angles opposite to equal sides are equal
3. Since AB = AC, then ∠B = ∠C
4. Since AB = BC, then ∠A = ∠C
5. Since BC = CA, then ∠A = ∠B
6. Therefore, ∠A = ∠B = ∠C
7. We know that ∠A + ∠B + ∠C = 180°
8. So, 3∠A = 180°, which gives ∠A = 60°
9. Similarly, ∠B = ∠C = 60°

7.5 More Criteria for Congruence

When Three Angles Are Equal

Imp Point: Equality of three angles is not sufficient for congruence of triangles. Triangles with the same three angles can have different sizes (they are similar but not congruent).

For congruence of triangles: At least one of the three equal parts must be a side (not just angles).

Solved Examples

Example 1: SAS Rule Application

Problem: In Fig. 7.8, OA = OB and OD = OC. Show that (i) ∆AOD ≅ ∆BOC and (ii) AD ∥ BC.

Solution:

(i) Proving ∆AOD ≅ ∆BOC:

In ∆AOD and ∆BOC:

• OA = OB (given)
• OD = OC (given)
• ∠AOD = ∠BOC (vertically opposite angles)

By SAS rule, ∆AOD ≅ ∆BOC

(ii) Proving AD ∥ BC:

Since ∆AOD ≅ ∆BOC, corresponding parts are equal:

• ∠OAD = ∠OBC (CPCT)

These angles form alternate angles with respect to the transversal AB cutting the lines AD and BC.

Since alternate angles are equal, AD ∥ BC

Example 2: Perpendicular Bisector

Problem: Line l is the perpendicular bisector of line segment AB. If a point P lies on l, show that P is equidistant from A and B (i.e., PA = PB).

Solution:

Line l is perpendicular to AB and passes through C, which is the midpoint of AB.

In ∆PCA and ∆PCB:

• AC = BC (C is the midpoint of AB)
• ∠PCA = ∠PCB = 90° (l is perpendicular to AB)
• PC = PC (common side)

By SAS rule, ∆PCA ≅ ∆PCB

Since the triangles are congruent, PA = PB (CPCT)

Therefore, P is equidistant from A and B.

Example 3: Parallel Lines and Congruent Triangles

Problem: Line segment AB is parallel to line segment CD. O is the midpoint of AD. Show that (i) ∆AOB ≅ ∆DOC and (ii) O is also the midpoint of BC.

Solution:

(i) Proving ∆AOB ≅ ∆DOC:

In ∆AOB and ∆DOC:

• ∠ABO = ∠DCO (alternate angles, since AB ∥ CD and BC is the transversal)
• ∠AOB = ∠DOC (vertically opposite angles)
• AO = DO (O is the midpoint of AD—given)

By AAS rule, ∆AOB ≅ ∆DOC

(ii) Proving O is the midpoint of BC:

Since ∆AOB ≅ ∆DOC:

• BO = CO (CPCT)

Therefore, O is the midpoint of BC.

Example 4: Isosceles Triangle with Perpendicular Bisector

Problem: In ∆ABC, the bisector AD of ∠A is perpendicular to side BC. Show that AB = AC and ∆ABC is isosceles.

Solution:

In ∆ABD and ∆ACD:

• ∠BAD = ∠CAD (AD is the bisector of ∠A—given)
• AD = AD (common)
• ∠ADB = ∠ADC = 90° (AD is perpendicular to BC—given)

By ASA rule, ∆ABD ≅ ∆ACD

Since the triangles are congruent:

• AB = AC (CPCT)

Therefore, ∆ABC is an isosceles triangle.

Example 5: Equal Sides and Equal Segments

Problem: E and F are respectively the midpoints of equal sides AB and AC of ∆ABC. Show that BF = CE.

Solution:

In ∆ABF and ∆ACE:

• AB = AC (given)
• ∠A = ∠A (common angle)
• AF = AE (F and E are midpoints of equal sides, so they are halves of equal sides)

By SAS rule, ∆ABF ≅ ∆ACE

Since the triangles are congruent:

• BF = CE (CPCT)

Example 6: Perpendicular to Intersecting Lines

Problem: P is a point equidistant from two intersecting lines l and m at point A. If PB ⊥ l and PC ⊥ m, show that (i) ∆APB ≅ ∆APC and (ii) BP = CP (P is equidistant from both lines).

Solution:

(i) Proving ∆APB ≅ ∆APC:

In ∆APB and ∆APC:

• PB = PC (given—P is equidistant from both lines)
• ∠PBA = ∠PCA = 90° (PB ⊥ l and PC ⊥ m)
• AP = AP (common side)

By RHS rule (Right angle-Hypotenuse-Side), ∆APB ≅ ∆APC

(ii) Conclusion:

Since ∆APB ≅ ∆APC:

• ∠PAB = ∠PAC (CPCT)

Therefore, AP bisects the angle between the two lines.

Example 7: Perpendicular Bisector Proof

Problem: P and Q are points on opposite sides of line segment AB such that PA = PB and QA = QB. Show that the line PQ is the perpendicular bisector of AB.

Solution:

In ∆PAQ and ∆PBQ:

• AP = BP (given)
• AQ = BQ (given)
• PQ = PQ (common)

By SSS rule, ∆PAQ ≅ ∆PBQ

From this congruence:

• ∠APQ = ∠BPQ (CPCT)

Now, in ∆PAC and ∆PBC:

• AP = BP (given)
• ∠APC = ∠BPC (from above)
• PC = PC (common)

By SAS rule, ∆PAC ≅ ∆PBC

From this congruence:

• AC = BC (CPCT)
• ∠ACP = ∠BCP (CPCT)

Since ∠ACP + ∠BCP = 180° (linear pair):

• 2∠ACP = 180°
• ∠ACP = 90°

Therefore:

• PQ passes through the midpoint C of AB (since AC = BC)
• PQ is perpendicular to AB (since ∠ACP = 90°)

So, PQ is the perpendicular bisector of AB.

Exercises and Question Bank

Exercise 7.1: Congruence of Triangles

Question 1: In quadrilateral ACBD, AC = AD and AB bisects ∠CAD (see Fig. 7.16). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

Solution:

In ∆ABC and ∆ABD:

• AC = AD (given)
• ∠CAB = ∠DAB (AB bisects ∠CAD)
• AB = AB (common)

By SAS rule, ∆ABC ≅ ∆ABD

From this congruence:

• BC = BD (CPCT)

Question 2: ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that (i) ∆ABD ≅ ∆BAC, (ii) BD = AC, and (iii) ∠ABD = ∠BAC.

Solution:

(i) Proving ∆ABD ≅ ∆BAC:

In ∆ABD and ∆BAC:

• AD = BC (given)
• ∠DAB = ∠CBA (given)
• AB = BA (common)

By SAS rule, ∆ABD ≅ ∆BAC

(ii) From the congruence:

• BD = AC (CPCT)

(iii) From the congruence:

• ∠ABD = ∠BAC (CPCT)

Question 3: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Solution:

Let AD and BC be perpendicular to AB at points D and E respectively, with AD = BC.

In ∆ADE and ∆BCE:

• AD = BC (given)
• ∠ADE = ∠BEC = 90° (given)
• DE = DE (common)

By RHS rule, ∆ADE ≅ ∆BCE

From this congruence:

• AE = BE (CPCT)

Therefore, E is the midpoint of AB, which means CD bisects AB.

Question 4: l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

Solution:

Since l ∥ m and p is a transversal:

• ∠ABC = ∠CDA (alternate angles)

Since p ∥ q and l is a transversal:

• ∠BCA = ∠DAC (alternate angles)

Also, AC = AC (common side)

By ASA rule, ∆ABC ≅ ∆CDA

Question 5: Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A. Show that (i) ∆APB ≅ ∆AQB and (ii) BP = BQ (B is equidistant from the arms of ∠A).

Solution:

(i) In ∆APB and ∆AQB:

• ∠PAB = ∠QAB (l bisects ∠A)
• AB = AB (common)
• ∠APB = ∠AQB = 90° (BP and BQ are perpendiculars)

By ASA rule, ∆APB ≅ ∆AQB

(ii) From the congruence:

• BP = BQ (CPCT)

Therefore, B is equidistant from the arms of ∠A.

Question 6: In Fig. 7.21, AC = AE, AB = AD, and ∠BAD = ∠EAC. Show that BC = DE.

Solution:

Consider ∠BAE and ∠DAC.

If ∠BAD = ∠EAC, then:

• ∠BAE = ∠BAD + ∠DAE = ∠EAC + ∠DAE = ∠DAC

In ∆ABE and ∆ADC:

• AB = AD (given)
• ∠BAE = ∠DAC (proved above)
• AE = AC (given)

By SAS rule, ∆ABE ≅ ∆ADC

From this congruence:

• BE = DC (CPCT)

Therefore, BC = DE.

Exercise 7.2: Isosceles Triangles

Question 1: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

Solution:

Given: ∆ABC is isosceles with AB = AC BE is altitude to AC, CF is altitude to AB

In ∆ABE and ∆ACF:

• ∠A = ∠A (common)
• ∠AEB = ∠AFC = 90° (altitudes are perpendicular)
• AB = AC (given—isosceles triangle)

By AAS rule, ∆ABE ≅ ∆ACF

From this congruence:

• BE = CF (CPCT)

Therefore, the altitudes are equal.

Question 2: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that (i) ∆ABE ≅ ∆ACF and (ii) AB = AC (ABC is isosceles).

Solution:

(i) In ∆ABE and ∆ACF:

• BE = CF (given)
• ∠AEB = ∠AFC = 90° (BE and CF are altitudes)
• ∠A = ∠A (common)

By AAS rule, ∆ABE ≅ ∆ACF

(ii) From the congruence:

• AB = AC (CPCT)

Therefore, ∆ABC is an isosceles triangle.

Question 3: ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

Solution:

Since ∆ABC is isosceles with AB = AC:

• ∠ABC = ∠ACB

Since ∆DBC is isosceles with DB = DC:

• ∠DBC = ∠DCB

Therefore:

• ∠ABD = ∠ABC + ∠DBC
• ∠ACD = ∠ACB + ∠DCB

Since ∠ABC = ∠ACB and ∠DBC = ∠DCB:

• ∠ABD = ∠ACD

Question 4: ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

Solution:

Since AB = AC, in ∆ABC:

• ∠ABC = ∠ACB (base angles of isosceles triangle)

Since AD = AB and AB = AC:

• AD = AC

In ∆ACD:

• AD = AC
• ∠ADC = ∠ACD (base angles of isosceles triangle)

Let ∠ABC = ∠ACB = x

In ∆ABC:

• ∠BAC = 180° – 2x

Since BCD is a straight line:

• ∠BAC + ∠DAC = 180°
• ∠DAC = 180° – ∠BAC = 180° – (180° – 2x) = 2x

In ∆ACD:

• ∠ADC = ∠ACD = (180° – 2x)/2 = 90° – x

Therefore:

• ∠BCD = ∠ACB + ∠ACD = x + (90° – x) = 90°

So, ∠BCD is a right angle.

Question 5: ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Solution:

Since ∠A = 90° and AB = AC:

• ∆ABC is an isosceles right-angled triangle
• ∠B = ∠C (base angles of isosceles triangle)

In ∆ABC:

• ∠A + ∠B + ∠C = 180°
• 90° + ∠B + ∠B = 180°
• 2∠B = 90°
• ∠B = 45°

Therefore, ∠B = ∠C = 45°

Question 6: Show that the angles of an equilateral triangle are 60° each.

Solution:

Let ∆ABC be an equilateral triangle, so AB = BC = CA.

Since AB = AC:

• ∠B = ∠C (angles opposite to equal sides)

Since AB = BC:

• ∠A = ∠C (angles opposite to equal sides)

Therefore, ∠A = ∠B = ∠C

In ∆ABC:

• ∠A + ∠B + ∠C = 180°
• 3∠A = 180°
• ∠A = 60°

Therefore, ∠A = ∠B = ∠C = 60°

Exercise 7.3: More Congruence Criteria

Question 1: ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that (i) ∆ABD ≅ ∆ACD, (ii) ∆ABP ≅ ∆ACP, (iii) AP bisects ∠A as well as ∠D, and (iv) AP is the perpendicular bisector of BC.

Solution:

(i) Proving ∆ABD ≅ ∆ACD:

Since ∆ABC is isosceles: AB = AC Since ∆DBC is isosceles: DB = DC

In ∆ABD and ∆ACD:

• AB = AC (given)
• DB = DC (given)
• AD = AD (common)

By SSS rule, ∆ABD ≅ ∆ACD

(ii) From the congruence:

• ∠BAD = ∠CAD (CPCT)
• ∠BDA = ∠CDA (CPCT)

In ∆ABP and ∆ACP:

• ∠BAP = ∠CAP (from above)
• AP = AP (common)
• ∠APB = ∠APC (since they are equal from the congruence of ABD and ACD)

By ASA rule, ∆ABP ≅ ∆ACP

(iii) From the congruence:

• ∠BAP = ∠CAP (proved above), so AP bisects ∠A
• ∠BDP = ∠CDP (from ∆ABD ≅ ∆ACD), so AP bisects ∠D

(iv) From ∆ABP ≅ ∆ACP:

• BP = CP (CPCT), so P is the midpoint of BC
• ∠APB = ∠APC = 90° (from the congruence)

Therefore, AP is the perpendicular bisector of BC.

Question 2: AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC and (ii) AD bisects ∠A.

Solution:

Given: ∆ABC is isosceles with AB = AC, and AD is an altitude (AD ⊥ BC)

In ∆ABD and ∆ACD:

• AB = AC (given)
• AD = AD (common)
• ∠ADB = ∠ADC = 90° (AD is an altitude)

By RHS rule, ∆ABD ≅ ∆ACD

(i) From the congruence:

• BD = CD (CPCT)

Therefore, AD bisects BC.

(ii) From the congruence:

• ∠BAD = ∠CAD (CPCT)

Therefore, AD bisects ∠A.

Question 3: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR. Show that (i) ∆ABM ≅ ∆PQN and (ii) ∆ABC ≅ ∆PQR.

Solution:

Given: AB = PQ, BC = QR, AM = PN (where M and N are midpoints of BC and QR)

Since M is the midpoint of BC: BM = BC/2 Since N is the midpoint of QR: QN = QR/2

Since BC = QR: BM = QN

(i) In ∆ABM and ∆PQN:

• AB = PQ (given)
• BM = QN (proved above)
• AM = PN (given)

By SSS rule, ∆ABM ≅ ∆PQN

(ii) From the congruence:

• ∠ABM = ∠PQN (CPCT)

Since M and N are midpoints:

• BC = 2BM and QR = 2QN
• BC = QR (given) implies this is consistent

Now in ∆ABC and ∆PQR:

• AB = PQ (given)
• BC = QR (given)
• ∠ABC = ∠PQR (from the congruence ∆ABM ≅ ∆PQN)

By SAS rule, ∆ABC ≅ ∆PQR

Question 4: BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:

Given: BE and CF are altitudes with BE = CF

In ∆ABE and ∆ACF:

• ∠AEB = ∠AFC = 90° (altitudes are perpendicular)
• BE = CF (given)
• ∠A = ∠A (common)

By RHS rule (if we consider AE and AF as bases and AB, AC as hypotenuses):

Actually, let’s reconsider using the correct RHS form:

In right triangles with right angles at E and F:

• ∠ABE = ∠ACF (let’s assume for now, we’ll prove)

In ∆BEC and ∆CFB:

• BE = CF (given)
• ∠BEC = ∠CFB = 90°
• BC = BC (common—this is the hypotenuse)

By RHS rule, ∆BEC ≅ ∆CFB

From this congruence:

• EC = FB (CPCT)

Also from the original configuration:

• AE = AF (this can be proved from the equal altitudes property)

Therefore:

• AC = AF + FC = AE + BE = AB

So, ∆ABC is isosceles with AB = AC.

Question 5: ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Solution:

Given: ∆ABC is isosceles with AB = AC, and AP ⊥ BC

In ∆ABP and ∆ACP:

• AB = AC (given)
• ∠APB = ∠APC = 90° (AP ⊥ BC)
• AP = AP (common)

By RHS rule, ∆ABP ≅ ∆ACP

From this congruence:

• ∠B = ∠C (CPCT)

Summary of Congruence Rules

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