**Congruence of Triangles – Maharashtra Board Class 9 Solutions for Geometry**

AlgebraGeometryScience and TechnologyHindi

**Exercise – 3.1**

**Solution 1:**

Congruent sides:

side AB ≅ side MN

side BC ≅ side NP

side AC ≅ side MP

Congruent angles:

∠A ≅ ∠M

∠B ≅ ∠N

∠C ≅ ∠P

**Solution 2:**

Seg PQ ≅ seg QR

∴∠R ≅ ∠P ….(Isosceles triangle theorem)

m∠P = 70° …(Given)

∴m∠R = 70° ….(1)

Now, the sum of the measures of the angles of a triangle is 180°.

∴ m∠P + m∠Q + m∠R = 180°

∴70° + m∠Q + 70° = 180° …[Given and from (1)]

∴140° + m∠Q = 180°

∴m∠Q = 180° – 140°

∴m∠Q = 40°

The measures of the remaining angles are m∠R = 70° and m∠Q = 40°.

**Solution 3:**

In ∆PST, seg PS ≅ seg PT. ….(Given)

∴ By Isosceles Triangle Theorem,

∠PTS ≅ ∠PST ….(1)

The sum of the measures of the angles of a triangle is 180°.

∴m∠P + m∠PTS + m∠PST = 180° ….(2)

∴m∠P + 2m∠PTS = 180° ….. [From (1) and (2)] …..(3)

Seg PS ≅ seg PT and seg SQ ≅ seg TR …(Given)

∴ PS = PT and SQ = TR

∴ PS + SQ = PT + TR

∴ PQ = PR ….(P-S-Q and P-T-R)

∴ seg PQ ≅ seg PR.

In ∆PQR, seg PQ ≅ seg PR ….(Proved)

∴ ∠PRQ ≅ ∠PQR ….(4)

m∠P + m∠PQR + m∠PRQ = 180° …..(5)

∴m∠P + 2m∠PQR = 180° ……[From (4) and (5)]….(6)

From (3) and (6), ∠PTS = ∠PRQ

∴ seg ST ∥ seg QR ….(corresponding angles test for parallel lines)

i.e. Side ST ∥ Side QR.

**Solution 4:**

Proof with justification:

In ∆POR and ∆SOQ,

seg OP ≅ seg OS and seg OR ≅ seg OQ …..(Given)

∠POR ≅ ∠SOQ ….(Vertically opposite angles)

∴ ∆POR ≅ ∆SOQ ….(SAS test)

∴ seg PR ≅ seg SQ ….(c.s.c.t.)

Also, ∠P ≅∠S and ∠R ≅∠Q …..(c.a.c.t.)

For seg PR and seg SQ, PQ is the transversal.

∴ For seg PR and seg SQ to be parallel, the alternate angles, ∠ P and ∠Q, should be congruent.

But these angles are not congruent.

∴ seg PR and seg SQ are not parallel.

Also, taking RS as the transversal, we can show that seg PR and seg SQ are not parallel.

**Solution 5:**

∠ABP ≅ ∠CBQ …..(Given)

∴ ∠ABP = ∠CBQ

Adding ∠PBC to both the sides

∠ABP + ∠PBC = ∠CBQ + ∠PBC

∠ABC ≅ ∠PBQ ….(1)

seg AB = seg PB and seg BC ≅ seg BQ …..(Given)

∴∆ABC ≅ ∆PBQ …..(SAS test)

∴ seg AC ≅ seg PQ ….(c.s.c.t.)

**Solution 6:**

Solution with justification:

In ∆ ABC, seg AB ≅ seg AC ……(Given)

∴ by Isosceles Triangle Theorem,

∠ABC ≅ ∠ACB.

i.e. ∠ABC = ∠ACB …(1)

The sum of the measures of the angles of a triangle is 180°

∴m∠A + m∠ABC + m∠ACB = 180°

∴m∠A + m∠ABC + m∠ABC = 180° …..[From (1)]

∴m∠A + 2m∠ABC = 180°

∴40° + 2m∠ABC = 180° …[Given: m∠A = 40°]

∴2m∠ABC = 180° – 40° = 140°

∴m∠ABC = 70°

∴m∠ABC = m∠ACB = 70° …(2)

In ∆ABP, seg AB ≅ seg PB …(Given)

∴∠P ≅ ∠PAB i.e. ∠P = ∠PAB …(3)

By Remote Interior Angle Theorem,

∠ABC = ∠P + ∠PAB

From (2) and (3), 70° = m∠P + m∠P

∴m∠P = m∠PAB = 35° ….(4)

Similarly, we can prove that

m∠Q = m∠QAC = 35° ….(5)

In ∆APQ, m∠PAQ + m∠P + m∠Q = 180°

∴m∠PAQ + 35° + 35° = 180° ….[From (4) and (5)]

∴m∠PAQ = 180° – 70°

∴m∠PAQ = 110° …..(6)

So, m∠P = m∠Q = m∠PAB = m∠QAC = 35°

PC = PB + BC and QB = QC + BC …..(7)

But PB = QC ….(Given) ….(8)

∴ PC = QB ….[From (7) and (8)]

In ∆PAC and ∆QAB,

PC = QB …..(Proved)

AC = AB ….(Given)

∠ACB = ∠ABC …..[From (2)]

i.e. ∠ACP = ∠ABQ

∴∆PAC ≅ ∆QAB …..(SAS test)

Similarly, we can prove that

∆PAB ≅ ∆QAC.

Hence, m∠PAQ = 110°; Also ∆PAC ≅ ∆QAB and ∆PAB ≅ ∆QAC are the congruent triangles.

**Exercise – 3.2**

**Solution 1:**

In ∆ACO and ∆DBO,

∠ACO ≅ ∠DBO ….(Given)

seg CO ≅ seg BO …(Given)

∠AOC ≅ ∠DOB …..(Vertically opposite angles)

∴∆ACO ≅ ∆DBO ….(ASA test)

The remaining congruent parts:

∠A ≅ ∠D, seg AC ≅ seg DB, and seg AO ≅ seg DO

**Solution 2(i):**

∆BDA ≅ ∆BDC

1. Test of congruence: SAA test

2. Correspondence: BDA ↔ BDC

3. Congruent Parts:

∠ABD ≅ ∠CBD

seg AB ≅ seg CB

seg AD ≅ seg CD

**Solution 2(ii):**

∆PQR ≅ ∆SRQ

1. Test of congruence: SAS test

2. Correspondence: PQR ↔ SRQ

3. Congruent Parts:

seg PR ≅ seg SQ

∠QPR ≅ ∠RSQ

∠PRQ ≅ ∠SQR

**Solution 2(iii):**

∆MQN ≅ ∆PQN

1. Test of congruence: SAS test

2. Correspondence: MQN ↔ PQN

3. Congruent Parts:

seg MN ≅ seg PN

∠QMN ≅ ∠QPN

∠QNM ≅ ∠QNP

**Solution 2(iv):**

∆BPA ≅ ∆BPC

1. Test of congruence: ASA test

2. Correspondence: BPA ↔ BPC

3. Congruent Parts:

∠BAP ≅ ∠BCP

seg BA ≅ seg BC

seg PA ≅ seg PC

**Solution 2(v):**

∆ ABP ≅ ∆CBP

1. Test of congruence: SSS test

2. Correspondence: ABP ↔ CBP

3. Congruent Parts:

∠PBA ≅ ∠PBC

∠BAP ≅ ∠BCP

∠BPA ≅ ∠BPC

**Solution 2(vi):**

BD + DC = FC + DC

∴BC = FD

∆ ABC ≅ ∆EFD

1. Test of congruence: SAS test

2. Correspondence: ABC ↔ EFD

3. Congruent Parts:

seg AB ≅ seg EF

∠ABC ≅ ∠EFD

∠BAC ≅ ∠FED

**Solution 2(vii):**

AP = AQ

PB = QC

∴ AP + PB = AQ + QC

∴AB = AC …..(A-P-B; A-Q-C)

∆ ABQ ≅ ∆ACP

1. Test of congruence: SAS test

2. Correspondence: ABQ ↔ ACP

3. Congruent Parts:

seg BQ ≅ seg CP

∠ABQ ≅ ∠ACP

∠AQB ≅ ∠APC

**Solution 2(viii):**

∆ PSQ ≅ ∆PSR

1. Test of congruence: Hypotenuse-Side theorem

2. Correspondence: PSQ ↔ PSR

3. Congruent Parts:

seg QS ≅ seg RS

∠QPS ≅ ∠RPS

∠PQS ≅ ∠PRS

**Solution 2(ix):**

∆ APR ≅ ∆BPQ ≅ ∆CQR ≅ ∆QPR

1. Test of congruence: SSS test

2. Correspondence:

APR ↔ BPQ ↔ CQR ↔ QPR or any correspondence

3. Congruent Parts:

All the angles of ∆APR, ∆BPQ, ∆CQR and ∆QPR are 60° each.

**Solution 2(x):**

∆ PQS ≅ ∆RQS

1. Test of congruence: Hypotenuse – side theorem

2. Correspondence: PQS ↔ RQS

3. Congruent Parts:

seg PS ≅ seg RS

∠PQS ≅ ∠RQS

∠PSQ ≅ ∠RSQ

**Solution 3:**

i.

1. ∆XYZ and ∆DYZ are drawn as above.

2. Congruent parts seg XZ ≅ seg DY and

∠XZY ≅ ∠ZYD are shown in the figure by identical marks.

3. Test of congruence: SAS test

4. Congruent triangle: ∆ XYZ ≅ ∆DZY.

Correspondence: XYZ ↔ DZY

ii.

1. ∆EFG and ∆LMN are drawn as above.

2. Congruent parts seg FG ≅ seg MN,

∠G ≅ ∠N and ∠F ≅ ∠M are shown in the figure by identical marks.

3. Test of congruence: ASA test

4. Congruent triangle: ∆ EFG ≅ ∆LMN.

Correspondence: EFG ↔ LMN

iii.

1. ∆MNO and ∆CNR are drawn as above.

2. Congruent parts seg MN ≅ seg CN,

seg NO ≅ seg NR and ∠MNO ≅ ∠CNR are shown in the figure by identical marks.

3. Test of congruence: SAS test

4. Congruent triangle: ∆MNO ≅ ∆CNR.

Correspondence: MNO ↔ CNR

iv.

1. ∆BTR and ∆BTP are drawn as above.

1. Congruent parts seg BR ≅ seg BP,

Seg RT ≅ seg PT are shown in the figure by identical marks.

2. Test of congruence: SSS test

3. Congruent triangle: ∆BTR ≅ ∆BTP.

4. Correspondence: BTR ↔ BTP

v. In ∆LMP and ∆LPN,

∠LMP ≅ ∠LNP, ∠LPM ≅ ∠NPL.

Ans. (1) and (2)

1. ∆LMP and ∆LPN are drawn as above.

2. Congruent parts ∠LMP ≅ ∠LNP and ∠LPM ≅ ∠NPL are shown in the figure by identical marks.

3. Test of congruence: SAA test

4. Congruent triangle: ∆LMP ≅ ∆LNP.

Correspondence: LMP ↔ LNP

**Solution 4:**

Missing information required and sufficient to prove

∆ADC ≅ ∆CBA:

SAS test

seg AD ≅ seg CB

SAA test

∠D ≅ ∠B

ASA test

∠ACD ≅ ∠CAB

**Solution 5:**

Seg QS is the bisector of ∠PQR ….(Given)

∴∠PQS ≅ ∠RQS …(1)

Seg QS is the bisector of ∠PSR ….(Given)

∴∠PSQ ≅ ∠RSQ …(2)

In ∆PQS and ∆RQS,

∠PQS ≅ ∠RQS …[From (1)]

seg QS ≅ seg QS …(Common side)

∠PSQ ≅ ∠RSQ …..[From (2)]

∴ ∆PQS ≅ ∆RSQ ….(ASA test)

∴∠P ≅ ∠R ….(c.a.c.t.)

**Solution 6:**

Diagonal MP is the perpendicular bisector of diagonal NQ.

∴ seg NT ≅ seg QT and

m∠NTM = m∠QTM = m∠PTN = m∠PTQ = 90° …..(1)

Now, in ∆MNT and ∆MQT,

seg NT ≅ seg QT …[From (1)]

∠MTN ≅ ∠MTQ …[From (1)]

seg MT ≅ seg MT …(Common side)

∴ ∆MNT ≅∆MQT …(SAS test)

∴ seg MN ≅ seg MQ …(c.s.c.t.)

In ∆PNT and ∆PQT,

seg NT ≅ seg QT ….[From (1)]

∠PTN ≅ ∠PTQ ….[From (1)]

seg PT ≅ seg PT …(Common side)

∴ ∆PNT ≅∆PQT …(SAS test)

∴ seg NP ≅ seg QP …(c.s.c.t.)

In ∆NMP and ∆QMP,

seg MN ≅ seg MQ

seg NP ≅ seg QP

seg MP ≅ seg MP …(Common side)

∴ ∆PNT ≅∆PQT …(SSS test)

**Solution 7:**

i. In ∆ABC and ∆CDE,

m∠B = m∠D = 90°

hypotenuse AC ≅ hypotenuse CE ….(Given)

seg BC ≅ seg ED ….(Given)

∴ ∆ABC ≅ ∆CDE ….(Hypotenuse- side theorem)

ii. As ∆ABC ≅ ∆CDE

∴∠BAC ≅ ∠DCE ….(c.a.c.t.) …(1)

iii. In ∆ABC, m∠BAC + m∠ACB = 90° … (Acute angles of a right angled triangle)…(2)

From (1) and (2),

m∠DCE + m∠ACB = 90° ….(3)

m∠ACB + m∠ACE + m∠DCE = 180°

∴ m∠ACB + m∠DCE + m∠ACE = 180°

∴ 90° + m∠ACE = 180° ….[From (3)]

∴ m∠ACE = 90°

**Solution 8:**

i. The pairs of congruent triangles:

a. ∆ABP and ∆ACP

b. ∆ABQ and ∆ACQ

c. ∆BQP and ∆CQP

a.

In ∆ABP and ∆ACP,

seg AB ≅ seg AC …(Given)

∠BAP ≅ ∠CAP ….(AP is the bisector of ∠BAC)

seg AP ≅ seg AP …(Common side)

∴ ∆ABP ≅ ∆ACP ….(SAS test)

∴ seg BP ≅ seg CP …(c.s.c.t.)….(1)

b.

In ∆ABQ and ∆ACQ,

seg AB ≅ seg AC …(Given)

∠BAQ ≅ ∠CAQ ….(AP, i.e. AQ is the bisector of ∠BAC)

seg AQ ≅ seg AQ …(Common side)

∴ ∆ABQ ≅ ∆ACQ ….(SAS test)

∴ seg BQ ≅ seg CQ …(c.s.c.t.)….(2)

c.

In ∆BQP and ∆CQP,

seg BP ≅ seg CP …[ From (1)]

seg BQ ≅ seg CQ …[ From (2)]

seg QP ≅ seg QP …(Common side)

∴ ∆BQP ≅ ∆CQP …(SSS test)

ii ∆BQC is an isosceles triangle. ….[From (2)]

seg AB ≅ seg AC …(Given)

∆ABC is an isosceles triangle.

**Exercise – 3.3**

**Solution 1:**

m∠A + m∠B + m∠C = 180° …(Angles of a triangles)

∴40° + 80° + m∠C = 180°

∴m∠C = 180° – 120° ∴m∠C = 60°

The descending order of the measures of the angles is ∠B > ∠C > ∠A

∴ side AC > side AB > side BC ….(Sides opposite to the angles)

The shortest side is side BC and the longest side is side AC.

**Solution 2:**

AB = 5 cm, BC = 8 cm, AC = 10 cm.

∴ AC > BC > AB

∴∠B > ∠A > ∠C …..(Inequality property of a triangle)

The smallest angle is ∠C and the greatest angle is ∠B.

Their descending order is ∠B > ∠A > ∠C.

**Solution 3:**

In ∆ABC, AB = 4 cm, AC = 6 cm.

∴ AC > AB.

Angle opposite to the greater side is greater.

∴a > b

b is the exterior angle of ∆ACD.

∴ By exterior angle theorem b > d ….(2)

From (1) and (2) a > b > d ….(3)

Now, c is the exterior angle of ∆ABC.

∴ c > a ….(By exterior angle theorem) …(4)

From (3) and (4) c > a > b > d.

**Solution 4:**

In ∆ABC, m∠B = 30°, m∠C = 25°,

∴ ∠B > ∠C

∴side AC > side AB

…..(In a triangle, the side opposite to the greater angle is greater) ….(1)

m∠ADB + m∠ADC = 180° …(Angles in a linear pair)

∴m∠ADB + 70° = 180°

∴m∠ADB = 110°

In ∆ABD, m∠ADB = 110° and m∠B = 30°

∴ ∠ADB > ∠B

∴ side AB > side AD …(2)

From (1) and (2),

side AC > side AB > side AD.

**Solution 5:**

(i) In ∆ABC, side AB ≅ side BC and A-P-C. …(Given)

∴ ∠A ≅ ∠C …(Isosceles Triangle Theorem) ..(1)

∠BPC > A …(Exterior Angle Theorem) …(2)

From (1) and (2), ∠BPC > ∠C

∴ BC > BP

…(Side opposite to greater angle)

i.e. BP < BC …(3)

AB ≅ BC ….(Given)…(4)

From (3) and (4),

BP

∴ BP< congruent sides.

(ii) In ∆ABC, side AB ≅ side BC and A-C-P.

….(Given)

∴ ∠A ≅ ∠BCA ….(Isosceles Triangle Theorem) …(1)

∠BCA > ∠P …(Exterior Angle Theorem)…(2)

From (1) and (2), ∠A > ∠P

∴ BP > BA …(Side opposite to greater angle) ..(3)

side AB ≅ side BC …(Given) …(4)

From (3) and (4),

BP > BA and BP > BC.

∴ BP > congruent sides.

**Solution 6:**

∠APC > ∠B …(Exterior Angle Theorem) …(1)

∠B ≅ ∠ACB …(Isosceles Triangle Theorem) …(2)

From (1) and (2), ∠APC > ∠ACP.

∴ AC > AP …(Side opposite to greater angle) … (3)

AB = AC ….(Given) …(4)

From (3) and (4), AB > AP

i.e. AP < AB …(5)

∠ACB > ∠AQC …(Exterior Angle Theorem) ..(6)

∴ ∠ABC > ∠AQC …[From (2) and (6)]

i.e. ∠ABQ > ∠AQC

∴ in ∆ABQ, AQ > AB

…(Side opposite to greater angle)

i.e. AB < AQ …(7)

From (5) and (7), AP < AQ.

**Solution 7:**

Let the length of side PR be x cm.

The sum of the lengths of any two sides of a triangle is greater than the third side.

∴ PQ + QR > PR

∴ 4 + 6 > PR

∴ 10 > PR

∴ 10 > x …(1)

The difference between the lengths of any two sides of a triangle is less than the length of the third side.

∴ QR – PQ < PR

∴ 6 – 4 < PR

∴ 2 < PR i.e. PR > 2

∴ x > 2 …(2)

From (1) and (2), 10 > x > 2

The length of side PR is greater than 2 cm but less than 10 cm.