**NCERT Class 9 Mathematics Chapter 7 Triangles** **Solutions,** **NCERT Solutions For Class 9 Maths, NCERT Class 9 Mathematics Chapter 7 Triangles Notes** to each chapter is provided in the list so that you can easily browse throughout different chapter **NCERT Class 9 Mathematics Chapter 7 Triangles****Notes** and select needs one.

**NCERT Class 9 Mathematics Chapter 7 Triangles**

**NCERT Class 9 Mathematics Chapter 7 Triangles**Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. **NCERT****Class 9 Mathematics Textual Question Answer. **These solutions are part of NCERT All Subject Solutions. Here we have given

**for All Subject, You can practice these here.**

**Solutions****NCERT Class 9 Mathematics Chapter 7 Triangles****Triangles**

**Triangles****Chapter – 7**

Exercise 7.1 |

**Q.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)

**Q.2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ BCA**

**Prove that**

**(i) ∆AB D ≅ ∆ BAC**

Ans: In ∆ ABD and ∆BAC

**(ii) BD = AC**

Ans: AD = BC (Given)

**(iii) ∠ ABD =∠BAC**

Ans: ∠ DAB = ∠CBA

(Given)

(Common)

and AB = BA

By SAS Congruence Condition

∆ABD ≅ ∆BAC

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

(iii) ∠ABD = ∠ BAC

(Again by C.P.C.T)

**Q.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.

**Q.4. L 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

**Q.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**

Ans:

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 or B is equidistant from the arms of ∠ A**

Ans: BP = BQ (By C.P.C.T.)

**Q.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 Congruence Condition

∆ BAC ≅ ∆ DAE

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

**Q.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: ∆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.)

**Q.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 Congruence 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 Congruence Condition.

∆ DBC ≅ ∠ ACB

**(iv) CM = 1/2 AB**

Ans: Since ∆ DBC ≅ ∠ ACB

DC = AB

CM = AM

Exercise 7.2 |

**Q.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 **

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

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) AO bisects ∠ A**

Ans: In ∆OAB and ∆OAC

AB = AC

OB = OC

AB = AC

| Given Proved (i) above

| 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

∴ ∠ OAB =∠OAC

∴ AO bisects ∠ A

|By SAS Rule

|c.p.c.t.

**Q.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

**Q.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 tri- angle 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.

**Q.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.

**Q.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

**Q.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. D**

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 …(3) ⇒ ⇒ ⇒

⇒ ∠ BCD is a right angle.

**Q.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 tri angle is 180⁰

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

⇒∠ B+∠C = 90⁰ …(2)

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

**Q.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 |

**Q.1. ∆ ABC and ∆BDE 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

Ans: 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) ∆ ABP≅ACP

Ans:

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) AP bisects ∠ A as well as ∠ D

Ans: ∆ 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) AP is the perpendicular bisector of BC

Ans: ∆ 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

**Q.2. AD is an altitude of an isosceles triangle ABC in which AB = AC. **

**Show that:**

**(i) AD bisects BC **

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) AD bisects ∠ A**

Ans: ∆ ADB ≅ ∆ ADC | Proved in (i) above

∴ ∠ BAD = ∠ CAD | c.p.c.t

⇒ AD bisects ∠ A

**Q.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 **

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

Ans:

To Prove: ∆ ABM ≅ ∆ PQN

∆ 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) ∆ ABC ≅ ∆PQR**

Ans: ∆ ABM = ∆ PQN | Proved in (1) above

∴ ∠ABM = ∠PQN | c.p.c.t. …(4)

⇒∠ABC = ∠PQR

In ∆ ABC and ∆PQR,

AB = PQ | Given

BC = QR | Given

∠ABC = ∠PQR | From (4)

∴ Δ ABC ≅∆ PQR | SAS Rule

**Q.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.

**Q.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 Ξ ΔΑΡΕ | RHS Rule

∴ ∠ABP = ∠ APC | c.p.c.t.

⇒ ∠B = ∠C