Our ncert solutions for Solutions 7. Triangles - Exercise 7.1 | Class 9 Mathematics - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Solutions 7. Triangles - Exercise 7.1 | Class 9 Mathematics - Toppers Study

Topper Study classes prepares ncert solutions on practical base problems and comes out with the best result that helps the students and teachers as well as tutors and so many ecademic coaching classes that they need in practical life. Our ncert solutions for Solutions 7. Triangles - Exercise 7.1 | Class 9 Mathematics - Toppers Study is the best material for English Medium students cbse board and other state boards students.

## 7. Triangles

### | Exercise 7.1 |

## Solutions 7. Triangles - Exercise 7.1 | Class 9 Mathematics - Toppers Study

**Chapter 7. Triangles**

**Exercise 7.1 **

**Q1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that Δ ABC ≅ Δ ABD.**

**Solution:**

**Given:** AC = AD and AB bisects ∠A

**To prove:** Δ ABC ≅ Δ ABD.

**Proof:** In Δ ABC and Δ ABD.

AC = AD [given]

∠CAB = ∠BAD [AB bisect ∠A]

AB = AB [Common]

By SAS Congruence Criterion Rule

Δ ABC ≅ Δ ABD

BC = BD [By CPCT] Proved

**Q2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig.7.17). Prove that**

**(i) Δ ABD ≅ Δ BAC**

**(ii) BD = AC**

**(iii) ∠ ABD = ∠ BAC **

**Solution: **

**Given: **ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA

**To prove**:

(i) Δ ABD ≅ Δ BAC

(ii) BD = AC

(iii) ∠ ABD = ∠ BAC

**Proof**: (i) In Δ ABD and Δ BAC

AD = BC [given]

∠ DAB = ∠ CBA [given]

AB = AB [Common]

By SAS Congruency Criterion Rule

Δ ABD ≅ Δ BAC

(ii) BD = AC [CPCT]

(iii) ∠ ABD = ∠ BAC [CPCT]

**Q3. AD and BC are equal perpendiculars to a line segment AB (see the given figure). Show that CD bisects AB.**

**Solution:**

**Given: **AD and BC are equal perpendiculars to a line segment AB.

**To prove: **CD bisects AB.

**Proof:**

In ∆BOC and ∆AOD

∠ BOC = ∠AOD (Vertically opposite angles)

∠CBO = ∠DAO (Each 90º)

BC = AD (Given)

By AAS Congruence Criterion Rule

∆BOC ≅ ∆AOD

BO = AO (By CPCT)

Hence, CD bisects AB.

**Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q (See the given figure). Show that ∆ABC ≅ ∆CDA**

**Solution:**

**Given:** l and m are two parallel lines intersected by another pair of parallel lines p and q.

**To prove: **∆ABC ≅ ∆CDA

**Proof: **

In ∆ABC and ∆CDA,

∠ BAC = ∠DCA (Alternate interior angles, as p || q)

AC = CA (Common)

∠ BCA = ∠DAC (Alternate interior angles, as l || m)

By AAS Congruence Criterion Rule

∆ABC ≅ ∆CDA

**Q5. 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 (see the given figure). Show that: (i) ∆APB ≅ ∆AQB (ii) BP = BQ or B is equidistant from the arms of ∠A.**

**Solution:**

**Given:** 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.

**To prove: **

(i) ∆APB ≅ ∆AQB

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

**Proof: **

In ∆APB and ∆AQB,

∠ APB = ∠AQB (Each 90º)

∠ PAB = ∠QAB (l is the angle bisector of A)

AB = AB (Common)

By AAS Congruence Criterion Rule

∆APB ≅ ∆AQB

BP = BQ [CPCT]

it can be said that B is equidistant from the A.

**Q6. In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.**

**Solution:**

**Given: **AC = AE, AB = AD and ∠BAD = ∠EAC.

**To prove: **BC = DE.

**Proof: **∠BAD = ∠EAC

BAD + DAC = EAC + DAC

BAC = DAE

In ∆BAC and ∆DAE

AC = AE (Given)

AB = AD (Given)

∠BAC = ∠DAE (proved above)

By SAS Congruence Criterion Rule

∆BAC ≅ ∆DAE

BC = DE (CPCT)

**Q7. 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 (See the given figure).**

**Show that: **

**(i) ∆DAP ≅ ∆EBP**

**(ii) AD = BE**

**Solution:**

**Given: **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.

**To prove: **

(i) ∆DAP ≅ ∆EBP

(ii) AD = BE

**Proof: **In ∆ DPA and ∆ EPB

∠EPA = ∠DPB

EPA + DPE = DPB + DPE

∠ DPA = ∠EPB

∠BAD =∠ ABE (Given)

∠EPA = ∠DPB (Given)

AP =BP (P is the midpoint of AB)

By AAS Congruence Criterion Rule

∆DAP ≅ ∆EBP

AD = BE (CPCT)

##### Other Pages of this Chapter: 7. Triangles

### Select Your CBSE Classes

Important Study materials for classes 06, 07, 08,09,10, 11 and 12. Like CBSE Notes, Notes for Science, Notes for maths, Notes for Social Science, Notes for Accountancy, Notes for Economics, Notes for political Science, Noes for History, Notes For Bussiness Study, Physical Educations, Sample Papers, Test Papers, Mock Test Papers, Support Materials and Books.

*Mathematics Class - 11th*

NCERT Maths book for CBSE Students.

books

## Study Materials List:

##### Solutions ⇒ Class 9th ⇒ Mathematics

## Topper's Study

New Books