SEBA Class 9 Mathematics Chapter 7 Triangles

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SEBA Class 9 Mathematics Chapter 7 Triangles

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Triangles

Chapter – 7

Exercise 7.1

1. In quadrilateral ABCD AC = AD and AB bisects ∠A. Show that ∠ABC ≅ ∆ADB. What can you say about BC and RD?

Ans: In ∆ABC and ∆ABD 

AC = AD (Given) 

∠CAB = ∠DAB (Given) 

AB = AB  (Given)

Therefore, 

By SAS congruence condition 

∆ABC ≅ ∆ABD 

So, BC = BD (By C.P.C.T)

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠BCA Prove that

(i) ∆AB D ≅ ∆ BAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC

Ans: (i) In ∆ ABD and ∆BAC

(ii) AD = BC (Given)

(iii) ∠DAB = ∠CBA  (Given)

and AB = BA (Common)

By SAS Congruence Condition 

∆ABD ≅ ∆BAC 

(ii) BD = AC( By C.P.C.T)

(iii) ∠ABD = ∠ BAC

(Again by C.P.C.T)

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

Ans: In ∆ AOD and ∆ BOC

AD = BC (Given) 

∠OAD = ∠OBC (each 90⁰) 

∠AOD = ∠BOC 

(Vertically opposite angles) 

Therefore, by ASA congruence condition. 

So, OA = OB 

(By C.P. С.Т.)

Hence, CD bisects line segment AB.

4. I and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC = ∆CDA

Ans: We have given that l || m and p || q 

Therefore, In ∆ ABC and ∆ACDA 

(Alternate interior angles as AB || CD) 

∠ACB = ∠CAD

 (Alternate interior angles as BC II DA) 

AC = CA

So, By A-S-A congruence condition.

∆ ABC ≅ ∆CDA

5. Line 1 is the bisector of an angle ∠A and B is any point on 1. BP and BQ are perpendiculars from B to the arms of ∠A show that:

(i) ∆APB = ∆AQB

(ii) BP = BQ or B is equidistant from the arms of ∠A

Ans: (i) In ∆ABP and ∆ABQ

∠BAP = ∠BAQ   (Given)

∠APB = ∠AQB     (Each 90⁰)

 AB = AB    (Common) 

By A – A – S congruence condition. 

So, ∆ABP = ∆ABQ 

(ii) BP = BQ (By C.P.C.T.)

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

Ans: In ∆ BAC and ∆ DAE 

AB = AD   (Given)

AC = AB  (Given) 

∠BAD = ∠EAC …(1) (Given)

Adding ∠DAC both side in equation (1) 

∠BAD + ∠DAC = ∠EAC + ∠DAC 

∠BAC = ∠DAE 

Therefore by S-A-S Congruency Condition

∆BAC ≅ ∆DAE 

So, BC = DE (By C.P.C.T.)

7. AB is a line segment and P is its mid point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB show that

(i) ∆DA P ≅ EBP

Ans: In ∆DAP and ∆EBP

∠DAP = ∠EBP (Given)

∠APE = ∠DPB (Given) 

∴ ∠APE + ∠ EPD = ∠DPB + ∠EPD

(Add ∠EPD both side) 

∠APD =∠BPE

AP = BP 

(Given P is the midpoint of AB) 

By A-S-A Congruence Condition.

∆DAP = ∆EBP

(ii) AD = BE

Ans: AD = BE

(By C.P.C.T.)

8. In the right triangle ABC, right angled at C, M is the midpoint of hypotenuse AB, C is joined to M and produces a point D such that DM = CM. Point D is joined to point B. Show that

(i) ∆AMC ≅ ∆BMD

Ans: In ∆AMC and ∆BMD 

AM = BM (Given) 

CM = DM (Given)

∠AMC = ∠BMD

 (Vertically opp. angles) 

∴ By S-A-S Congruency Condition. 

∆AMC ≅ ∆BMD

(ii) ∠DBC is right angle.

Ans: ∠CAM = ∠DBM   (by C.P.C.T)

Also, ∠CAM+∠MBC = 90⁰ (Since ∠C 90⁰ )

∴ ∠DBM + ∠MBC = 90⁰ (∠CAM = ∠DBM)

 or, ∠ DBC = 90⁰

(iii) ∆DBC ≅ ∆ACB

Ans: In ∆ DBC and ∆ ACB 

BC = BC (Common)

DB = AC 

(∴ ∆BMD = ∆AMC, by C.P.M.T) 

and ∠DBC = ∠AVB (each 90 proved above) 

Therefore, by S-A-S Congruency Condition. 

∆ DBC ≅ ∠ ACB

(iv) CM = 1/2 AB

Ans: Since ∆ DBC ≅ ∠ ACB 

DC = AB 

CM = AM 

Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC the bisectors of ∠ B and ∠ C intersect each other at O. Show that:

(i) OB = OC

(ii) AO bisects ∠ A 

Ans: 

Given: In an isoscelés triangle ABC, with AB = AC the bisectors of ∠B and ∠ C intersect each other at O.

Join A to O.

To Prove: (i) Ob = OC

(ii) AO bisects A

Proof: (i) AB = AC | Given 

∴ ∠ A = ∠C     |Angles opposite to equal sides of a triangle are equal

∴ BO and CO are the bisectors of ∠B and ∠C respectively

∴ OB = OC | Sides opposite to equal angles of a triangle are equal

(ii) In ∆OAB and ∆OAC

AB = AC  | Given

OB = OC  | Proved (i) above

AB = AC  | Given 

∴ ∠ B = ∠C | Angles opposite to equal sides of a triangle are equal 

∴ ∠ ABO = ∠ ACO 

∵ BO and CO are the bisectors of ∠B and ∠C respectively

∴ ∆OAB ≅ ∆OAC  |By SAS Rule

∴ ∠OAB = ∠OAC  |c.p.c.t.

∴ AO bisects ∠ A

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

Ans: Given: In ∆ ABC, AD is the perpendicular bisector of BC. 

Το Prove: ∆ ABC is an isosceles triangle in which AB = AC 

Proof: In ∆ ADB and ∆ ADC, 

∠ ADB = ∠ ADC | Each = 90⁰

DB = DC | ∵ AD is the perpendicular bisector of BC 

AD = AD | Common

∴ ∆ ADB ≅ ∆ADC | By SAS Rule 

∴ ∆ABC is an isosceles triangle in which AB = AC

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

Ans: Given: ABC is an isosceles triangle in which altitudes BE and CF are drawn on sides AC and AB respectively.

To Prove: BE = CF 

Proof: ∵ ABC is an isosceles triangle 

∴ AB = AC

∴ ∠ ABC = ∠ ACB … (1) | Angles opposite to equal sides of a triangle are equal 

In ∆ BEC and ∆CFB

∠BEC = ∠CFB   | Each = 90⁰

BC = CB   | Common

∠ECB = ∠FBC  | From (1)

∴ ∆ BEC ≅ CFB    |By ASA Rule 

∴ BE = CE   | c.p.c.t.

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that:

(i) ∆ABE ≅ ACF

(ii) AB =AC, i.e. ∆ ABC is an isosceles triangle.

Ans: Given: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal

To Prove:

(i) ΔΑΒΕ ≅ ∆ACF

(ii) AB = AC, i.e. ∆ ABC is an isosceles triangle

Proof: (i) In ∆ABE and ∆ACF 

BE = CF | Given

∠BAC = ∠BAE = ∠CAF  | Common

∠AEB = ∠AFC   | Each = 90⁰

∴ Δ ΑΒΕ ≅ ∆ACF    | By AAS Rule

(ii) Δ ΑΒΕ ≅ ΔACF   | Proved in (i) above

∴ AB = AC  | c.p.c.t.

∴ ∆ ABC is an isosceles triangle.

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

Ans: Given: ABC and DBC are two isosceles triangles on the same base BC.

To Prove: ∠ABD = ∠ACD

Proof: ∵ ABC is an isosceles triangle on the base BC 

∴ ∠ABC = ∠ACB …(1)

∵ DBC is an isosceles triangle on the base BC

∴ ∠DBC = ∠DCB …(2)

Adding the corresponding sides of (1) and (2), we get 

∠ABC + ∠DBC = ∠ACB + ∠DCB 

⇒ ∠ABD = ∠ACD

6. ∆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.

Ans: Given: ∆ ABC is an isosceles triangle in which AB = AC 

Side BA is produced to D such that AD = AB 

To Prove: ∠ BCD is a a right angle. 

Proof: ∵ ABC is an isosceles triangle 

∴ ∠ ABC = ∠ ACB …(1) 

∵ AB = AC and AD = AB 

∴ AC = AD

∴ In ∆ACD

∠CDA = ∠ACD Angles opposite to equal sides of a triangle are equal

⇒ ∠CDB = ∠ACD  …(2) 

Adding the corresponding sides of (1) and (2), we get 

∠ABC + ∠CDB = ∠ACB +∠ACD 

⇒ ∠ABC + ∠CDB = ∠BCD   …(3)

In ∆BCD,

∠BCD + ∠DBC + ∠CDB = 180⁰  | ∵ Sum of all the angles of a triangle is 180⁰ 

⇒ ∠BCD + ∠ABC + ∠CDB = 180⁰

⇒ ∠BCD + ∠BCD = 180⁰ | Using (3) 

⇒ 2 ∠BCD = 180⁰

⇒ ∠BCD = 90⁰

⇒ ∠BCD is a right angle.

7. ABC is a right angled triangle in which ∠ A = 90⁰ and AB = AC. Find ∠ B and ∠ C.

Ans: ∵ In ∆ABC, AB = AC

∴ ∠ B = ∠C …(1)  | Angles opposite to equal sides of a triangle are equal 

In ∆ ABC,

∠A+∠B+ ∠C = 180⁰ Sum of all the angles of a triangle is 180⁰

⇒ 90⁰+∠B+∠C = 180⁰ | ∵ ∠ A = 90⁰  (given) 

⇒ ∠ B+∠C = 90⁰ …(2)

From (1) and (2), we get, ∠ B = ∠C = 45⁰

8. Show that the angles of an equilateral triangle are 60⁰ each.

Ans: Given: An equilateral triangle ABC 

To Prove: ∠ A = ∠B = ∠ C = 60⁰

Proof: ∵ ABC is an equilateral triangle 

∴ AB = BC = CA …(1) 

∵ AB = BC 

∴ ∠ A =∠C …(2)  | Angles opposite to equal sides of a triangle are equal 

∵ BC = CA 

∴ ∠A =∠B …(3)  | Angles opposite to equal sides of a triangle are equal 

From (2) and (3), we obtain 

∠ A = ∠ B = ∠C   …(4)

 In ∆ABC, ∠ A+∠B+ ∠ C = 180⁰ …(5) | Sum of all the angles of a triangle is 180⁰

From (4) and (5), we get, ∠ A = ∠B = ∠ C = 60⁰

Exercise 7.3

1. ∆ ABC and ∆DBE 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) ∆ADB ≅ ∆ACD

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects ∠ A as well as ∠ D

(iv) AP is the perpendicular bisector of BC

Ans: Given: ∆ ABC and ∆ DBE are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.

To Prove: (i) ∆AB D ≅ ACD

(ii) ∆ ABP ≅ ACP

(iii) AP bisects ∠ A as well as ∠ D

(iv) AP is the perpendicular bisector of BC

Proof:

(i) In ∆ABD and ∆ACD

AB = AC …(1)  | ∆ ABC is an isosceles triangle

BD = CD …(2) | ∵ ∆DBC is an isosceles triangle

AD = AD   …(3) | Common side SSS Rule

∴ ∆ ABD ≅ ∆ ACD 

(ii) In ∆ABP and ∆ACP, 

AB = AC …(4) | From (1) 

∠ ABP = ∠ ACP …(5) 

∵ AB = AC | From (1) 

∴ ∠ ABP =∠ACP |Angles opposite to equal sides of a triangle are equal

∵ ∆ ABP ≅ ∆ ACD | Proved in (i) above

∴ ∠ BAP = ∠ CAP …(6) | c.p.c.t.

From (4), (5) and (6) we get

∆ ABP ≅ ∆ACP | ASA Rule 

(iii) ∵ ∆ ABP ≅ ∆ACP | Proved in (ii) above c.p.c.t.

∴ ∠ BAP = ∠CAP

⇒ AP bisects ∠ A 

In ∆BDP and ∆ CDP, 

BD = CD    …(7) | From (2)

DP = DP   …(8) | Common

∵ ∆ ABP ≅ ∆ACP  | Proved in (ii) above

∴ BP = CP  …(9) | c.p.c.t.

From (7), (8) and (9) we get 

∆ BDP ≅ ∆ CDP  | SSS Rule

∴ ∠BDP = ∠ CDP | c.p.c.t.

⇒ DP bisects ∠ D

⇒ AP bisects ∠ D

(iv) ∵ ∆ BDP ≅ ∆ CDP  | Proved in (iii) above

∴ BP = CP  | c.p.c.t.

∠ BDP = ∠ CDP | c.p.c.t. 

But  ∠ BPD+∠CPD = 180⁰ | Linear Pair Axiom 

∴ ∠ BPD = ∠ CPD = 90⁰ ….(11) 

From (10) and (11) we get 

AP is the perpendicular bisector of BC

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:

(i) AD bisects BC 

(ii) AD bisects ∠ A

Ans: Given: AD-is an altitude of an isosceles triangle ABC in which AB = AC

To Prove: (i) AD bisects BC

(ii) AD bisects ∠ A

Proof: (i) In right ∆ADB and right ∆ADC,

Hyp. AB = Hyp. AC | Given 

Side AD = Side AD | Common

∴ ∆ ADB ≅ ∆ ADC  | RHS Rule

∴ BD = CD | c.p.c.t.

⇒ AD bisects BC 

(ii) ∆ ADB ≅ ∆ ADC | Proved in (i) above 

∴ ∠ BAD = ∠ CAD | c.p.c.t

⇒ AD bisects ∠ A

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 triangle PQR. Show that:

(i) ∆ ABM ≅ ∆ PQN 

(i) ∆ ABC ≅ ∆PQR

Ans: Given: Two sides AB and BC and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆ PQR

To Prove: (i) ∆ ABM ≅ ∆ PQN

(ii) ∆ ABC ≅ ∆ PQR

Proof: (i) In ∆ ABM and ∆ PQN 

AB = PQ …(1) 

AM = PN …(2) | Given 

BC = QR | Given 

⇒ 2BM = 2QN

∵ M and N are the mid-points of BC and QR respectively

⇒ BM = QN …(3)

In view of (1), (2) and (3), 

∆ ABM ≅ ∆ PQN | SSS Rule

(ii) ∵ ∆ ABM = ∆ PQN | Proved in (1) above

∴ ∠ABM = ∠PQN | c.p.c.t. 

⇒ ∠ABC = ∠PQR   …(4)

In ∆ ABC and ∆PQR, 

AB = PQ | Given 

BC = QR | Given

∠ABC = ∠PQR | From (4)

∴ Δ ABC ≅ ∆ PQR  | SAS Rule

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

Ans: Given: BE and CF are two equal altitudes of a triangle ABC.

Το Prove: ∆ ABC is isosceles. 

Proof: In right ∆ BEC and right ∆ CFB,

Side BY = Side CF | Given

Hyp. BC = Hyp. СВ | Common

∴ ∆ BEC ≅ ∆ CFB | RHS Rule

∴ ∠ BEC = ∠ CBF  | c.p.c.t.

∴ AB = AC | Sides opposite to equal angles of a triangle are equal 

∴ ∆ ABC is isosceles.

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

Ans: Given: ABC is an isosceles triangle with AB = AC 

To Prove: ∠B = ∠C

Construction: Draw AP⊥BC

Proof: In the right triangle APB and right triangle APC, 

Hyp. AB = Hyp. AC | Given 

Side AP = Side AP | Common 

∴ Δ ΑΡΒ ≅ ΔΑΡΕ | RHS Rule

∴ ∠ABP = ∠ APC | c.p.c.t.

⇒ ∠B = ∠C

Exercise 7.4

1. Show that in a right angled triangle, the hypotenuse is long. east side.

Ans: Let ABC be a right angled triangle in which ∠ B = 90⁰

Then, ∠ A+ ∠ C = 90⁰ | ∵ Sum of all the angles of a triangle is 180⁰

∴ ∠B = ∠ A+∠C 

∴ ∠B 〉∠ A and ∠ B 〉∠ C

∴ AC〉BC and AC〉AB | ∵ Side opposite to greater angle is longer

∴ AC is the longest side, i.e. hypotenuse is the longest side.

2. In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC ∠QCB. Show that AC 〉AB. 

Ans: Given: Sides AB and AC of ∆ABC are extended to points P and Q respectively. Also ∠ PBC〈 ∠ QCB 

To Prove: AC〉AB 

Proof: ∠ PBC〈 ∠QCB | Given 

⇒ – ∠ PBC -〉∠ QCB

⇒ 180⁰ – ∠ PBC 〉180⁰-∠QCB

⇒ ∠ABC 〉∠ ACB

∴ AC〉AB | ∵ Side opposite to the greater angle is longer 

3. In figure, ∠ B〈 ∠A and ∠C〈∠D. Show that AD〈 BC

Ans: Given: In figure, ∠B〈 ∠ A and ∠ C〈 ∠ D 

Prove: AD〈BC | Given 

∴ ∠ B〈 ∠A

∴ OB〈 OA …(1)  | Side opposite to greater angle is longer

∠C〈 ∠D|Given

∴ ∠D〉∠C 

∴ OC〈OD …(2)  | Side opposite to greater angle is longer

From (1) and (2), we get 

OB+OC 〉OA+OD ⇒ BC 〉AD ⇒ AD〈 BC 

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that 

∠A〉∠C and ∠B〉∠D

Ans: Given: AB and CD are respec-tively the smallest and longest sides of a quadrilateral ABCD

To Prove: ∠A〉∠C and ∠B〉∠D

Construction: Join AC

Proof: In ∆ ABC,

AB〈 BC | ∵ AB is the smallest side of quadrilateral ABCD 

⇒ BC 〉AB

∴ ∠BAC〉∠ BCA …(1) | Angle opposite to longer side is greater

In ∆ACD, 

CD〉AD  | ∵ CD is the longest side of quadrilateral ABCD

∴ ∠CAD〉∠ ACD  | Angle opposite to longer side is greater ..(2)

From (1) and (2), we obtain

∠BAC + ∠CAD〉∠ BCA + ∠ ACD 

⇒ ∠A 〉∠C 

Similarly, joining B to D, we can prove that ∠B 〉∠D

5. In figure, PR 〉PQ and PS bisects ∠QPR. Prove that ∠PSR〉∠PSQ

Ans: Given: In figure, PR〉PQ and PS bisects ∠QPR

To Prove: ∠PSR〉∠ PSQ

Proof: In ∆ PQR

PR〉PQ | Given

∴ ∠PQR 〉∠PRQ …(1)  | Angle opposite to longer side is greater 

∵ PS is the bisector of ∠QPR

∠QPS = RPS 

In ∆ PQS, …(2) 

∠PQR + ∠QPS + ∠PSQ = 180⁰…(3)  | ∵ The sum of the three angles of a ∆ is 180⁰ 

In A PRS, 

∠PRS + ∠SPR + ∠PSR = 180⁰ …(4)  | ∵ The sum of the three angles of a ∆ is 180⁰ 

From (3) and (4), 

∠PQR + ∠QPS + ∠ PSQ = ∠ PRS +∠SPR + ∠ PSR 

⇒ ∠PQR + ∠PSQ = ∠ PRS + ∠ PSR 

⇒ ∠ PRS + ∠PSR = ∠PQR + ∠PSQ 

⇒ ∠ PRS + ∠PSR 〉∠PRQ + ∠PSQ  | From (1)

⇒ ∠PRQ + ∠PSR〉∠PRS + ∠PSQ  | ∵ ∠PRQ = ∠PRS

⇒ ∠PSR〉∠PSQ

6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Ans: Given: I is a line and P is a point not lying on l. PM⊥ l N is any point on I other than M.

To Prove: PM〈 PN

Proof: In ∆ PMN

∠M = 90⁰

∴ ∠N is an acute angle.

| Angle sum property of a triangle

∴ ∠M〉∠N 

∴ PN〉PM | Side opposite to greater angle is greater

⇒ PM〈PN