**Download NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1 – Triangles. This Exercise contains 8 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 7 or other Chapters, you can click the link at the end of this Note.**

**1. In quadrilateral ACBD,
**

**AC = AD and AB bisects**

**∠**

**A**

**(see Figure).**

**Show that ∆ ABC**

**≅**

**∆ ABD.**

**What can you say about BC and BD?**

** **

**Solution :**

In ∆ABC and ∆ABD,

AC = AD (∵ Given)

∠CAB = ∠DAB (∵ AB bisects ∠A)

AB = AB (∵ Common)

=> ∆ABC ≅ ∆ABD (By SAS congruence rule)

=> BC = BD (By CPCT)

=> BC and BD are of equal lengths.

**Hence Proved.**

**2. ****ABCD is a quadrilateral in which AD = BC and ****∠**** DAB = ****∠**** CBA (see Figure).**

Prove that

**(i) ∆ ABD ****≅**** ∆ BAC
**

**(ii) BD = AC**

**(iii)**

**∠**

**ABD =**

**∠**

**BAC**

**.**

**Solution :**

In ∆ABD and ∆BAC,

AD = BC (∵ Given)

∠DAB = ∠CBA (∵ Given)

AB = BA (∵ Common)

=> ∆ABD ≅ ∆BAC (By SAS congruence rule)

=> BD = AC (By CPCT)

And, ∠ABD = ∠BAC (By CPCT)

**Hence Proved.**

**3. ****AD and BC are equal perpendiculars to a line segment AB (see Figure). Show that CD bisects AB.**

**Solution :**

In ∆BOC and ∆AOD,

∠BOC = ∠AOD (∵ Vertically opposite angles)

∠CBO = ∠DAO (∵ Each 90º)

BC = AD (∵ Given)

=> ∆BOC ≅ ∆AOD (AAS congruence rule)

=> BO = AO (By CPCT)

=> CD bisects AB.

**Hence Proved.**

**4. ****l and m are two parallel lines intersected by another pair of parallel lines p and q (see Figure). Show that ∆ ABC ****≅**** ∆ CDA****.**

**Solution :**

In ∆ABC and ∆CDA,

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

AC = CA (∵ Common)

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

=> ∆ABC ≅ ∆CDA (By ASA congruence rule)

**5. 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 Figure).
**

**Show that:**

**(i) ∆ APB**

**≅**

**∆ AQB**

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

**∠**

**A.**

**Solution :**

In ∆APB and ∆AQB,

∠APB = ∠AQB (∵ Each 90º)

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

AB = AB (∵ Common)

=> ∆APB ≅ ∆AQB (By AAS congruence rule)

=> BP = BQ (By CPCT)

It can be said that B is equidistant from the arms of ∠A.

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

** **

**Solution :**

Given,

∠BAD = ∠EAC

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE

In ∆BAC and ∆DAE,

AB = AD (∵ Given)

∠BAC = ∠DAE (∵ Proved above)

AC = AE (∵ Given)

=> ∆BAC ≅ ∆DAE (∵ By SAS congruence rule)

=> BC = DE (∵ By CPCT)

**Hence Proved**

**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**** (see Figure).**

**Show that
**

**(i) ∆ DAP**

**≅**

**∆ EBP**

**(ii) AD = BE**

** **

**Solution :**

Given,

∠EPA = ∠DPB

∠EPA + ∠DPE = ∠DPB + ∠DPE

=> ∠DPA = ∠EPB

In ∆DAP and ∆EBP,

∠DAP = ∠EBP (∵ Given)

AP = BP (∵ P is mid-point of AB)

∠DPA = ∠EPB (∵ From above)

=> ∆DAP ≅ ∆EBP (ASA congruence rule)

=> AD = BE (By CPCT)

**Hence Proved**

**8. ****In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Figures). Show that:
**

**(i) ∆ AMC**

**≅**

**∆ BMD**

**(ii)**

**∠**

**DBC is a right angle.**

**(iii) ∆ DBC**

**≅**

**∆ ACB**

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

**Solution :**

**(i)** In ∆AMC and ∆BMD,

AM = BM (∵ M is the mid-point of AB)

∠AMC = ∠BMD (∵ Vertically opposite angles)

CM = DM (∵ Given)

=> ∆AMC ≅ ∆BMD (By SAS congruence rule)

=> AC = BD (By CPCT)

And, ∠ACM = ∠BDM (By CPCT)

**(ii)** ∠ACM = ∠BDM

However,

∠ACM and ∠BDM are alternate interior angles.

Since alternate angles are equal,

It can be said that,

DB || AC

∠DBC + ∠ACB = 180º (∵ Co-interior angles)

∠DBC + 90º = 180º

=> ∠DBC = 90º

**(iii)** In ∆DBC and ∆ACB,

DB = AC (∵ Already proved)

∠DBC = ∠ACB (∵ Each 90 )

BC = CB (∵ Common)

=> ∆DBC ≅ ∆ACB (SAS congruence rule)

**(iv)** ∆DBC ≅ ∆ACB

AB = DC (By CPCT)

AB = 2 CM

=> CM = (1/2) AB

**Hence Proved.**

**NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1 – Triangles, has been designed by the NCERT to test the knowledge of the student on the following topics:-**

- Congruence of Triangles
- Criteria for Congruence of Triangles

– SAS congruence rule

– ASA congruence rule

**The next Exercise for** **NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 – Triangles can be accessed by clicking here.**

**Download NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1 – Triangles**

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