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

# Exercise 7.1 Class 9 maths 7. Triangles - ncert solutions - Toppers Study

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## Exercise 7.1 Class 9 maths 7. Triangles - ncert solutions - Toppers Study

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### You can Find Mathematics solution Class 9 Chapter 7. Triangles

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### Exercise 7.1 class 9 Mathematics Chapter 7. Triangles

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## 7. Triangles

### | Exercise 7.1 |

## Exercise 7.1 Class 9 maths 7. Triangles - ncert solutions - 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

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