NCERT Solutions For Class 7 Maths Chapter 7 Congruence of Triangles are given here in a simple and detailed way. These NCERT Solutions for class 7 maths chapter 7 can be extremely helpful for the students to clear all their doubts easily and understand the basics of this chapter in a better and detailed way.

NCERT class 7 maths chapter 7 Congruence of Triangles solutions given here are very easily understandable so that students does not face any difficulties regarding any of the solutions. The NCERT solutions for class 7 maths chapter 7 Congruence of Triangles PDF is also available here that the students can download and study.

### NCERT Solutions For Class 7 Maths Chapter 7 Exercises

- NCERT Solutions For Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.1
- NCERT Solutions For Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.2

**Excerise 7.1**

**1. In quadrilateral ACBD, AC = AD and AB bisects ∠A . Show that ΔABC ≅ ΔABD. What can you say about BC and BD?**

**Sol. **

Given that,

AC = AD and AB bisects

To prove that,

Proof,

In

AB = AB

AC = AD

Therefore,

BC and BD are of equal length.

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

**(i) ΔABD ≅ ΔBAC**

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

**Sol.**

Given that ,

AD = BC and

(i) In

AB = BA (Common)

AD = BC (Given)

Therefore,

(ii) Since,

Therefore BD = AC by CPCT

(iii) Since,

Therefore

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

** **

**Sol. **

Given that,

AD and BC are equal perpendiculars to AB.

To prove,

CD bisects AB

Proof,

In

AD = BC (Given)

Therefore,

Now,

AO = OB (CPCT). CD bisects AB.

**4. l and m are two parallel lines intersected by another pair of parallel lines p and q . Show that ΔABC ≅ ΔCDA.**

**Sol. **

Given that,

l

To prove,

Then Proof that,

In

AC = CA (Common)

Therefore,

**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 . Show that:**

**(i) ΔAPB ≅ ΔAQB**

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

**Sol.**

Given,

l is the bisector of an angle

BP and BQ are perpendiculars.

(i) In

AB = AB (Common)

Therefore,

(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of

**6. AC = AE, AB = AD and ∠BAD = **

∠

**EAC. Show that BC = DE.**

** **

**Sol.**

Given that,

AC = AE, AB = AD and

To show,

BC = DE

then Proof that,

In

AC = AE (Given)

AB = AD (Given)

Therefore,

BC = DE by CPCT.

**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)**Δ DAP ≅ Δ EBP

**(ii) AD = BE**

**Sol.**

Given,

P is mid-point of AB.

(i)

In

AP = BP (P is mid-point of AB)

Therefore,

(ii) AD = BE by CPCT.

**8. A 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. Show that:**

**(i) ΔAMC ≅ ΔBMD**

**(ii) ∠DBC is a right angle.**

**(iii) ΔDBC ≅ ΔACB**

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

**Sol.**

Given,

(i) In

AM = BM (M is the mid-point)

CM = DM (Given)

Therefore,

(ii)

Therefore, AC || BD as alternate interior angles are equal.

Now,

(iii) In

BC = CB (Common)

DB = AC

Therefore,

(iv) DC = AB (

__Exercise 7.2__

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

**(a) OB = OC (b) AO bisects ****A**

**Sol.**

Sol. Given that,

AB = AC, the bisectors of

(i) Since ABC is an isosceles with AB = AC,

∴

⇒ 1/2

⇒

⇒ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given)

AO = AO (Common)

OB = OC (Proved above)

Therefore, ΔAOB ≅ ΔAOC by SSS congruence condition.

Thus, AO bisects

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

**Sol.**

Given,

AD is the perpendicular bisector of BC

To show,

AB = AC

Proof,

In ΔADB and ΔADC,

AD = AD (Common)

∠ADB = ∠ADC

BD = CD (AD is the perpendicular bisector)

Therefore, ΔADB ≅ ΔADC by SAS congruence condition.

AB = AC

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

**Sol.**

Given that,

BE and CF are altitudes.

AC = AB

To show,

BE = CF

Proof,

In ΔAEB and ΔAFC,

∠A = ∠A (Common)

∠AEB = ∠AFC (Right angles)

AB = AC (Given)

Therefore, ΔAEB ≅ ΔAFC by AAS congruence condition.

Thus, BE = CF by CPCT.

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

** (i) ΔABE ≅ ΔACF**

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

**Sol.**

Given,

BE = CF

(i) In ΔABE and ΔACF,

∠A = ∠A (Common)

∠AEB = ∠AFC (Right angles)

BE = CF (Given)

Therefore, ΔABE ≅ ΔACF by AAS congruence condition.

(ii) Thus, AB = AC by CPCT and therefore ABC is an isosceles triangle.

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

**Sol.**

Sol. Given that,

ABC and DBC are two isosceles triangles.

To show,

∠ABD = ∠ACD

Proof,

In ΔABD and ΔACD,

AD = AD (Common)

AB = AC (ABC is an isosceles triangle.)

BD = CD (BCD is an isosceles triangle.)

Therefore, ΔABD ≅ ΔACD by SSS congruence condition. Thus, ∠ABD = ∠ACD by CPCT.

**6. ΔABC 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.**

**Sol.**

Sol. Given that,

AB = AC and AD = AB

To show,

∠BCD is a right angle.

Proof,

In ΔABC,

AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)

In ΔACD,

AD = AB

⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)

Now,

In ΔABC,

∠CAB + ∠ACB + ∠ABC = 180°

⇒ ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — (i)

Similarly in ΔADC,

∠CAD = 180° – 2∠ACD — (ii)

also,

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding (i) and (ii)

∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB – 2∠ACD

⇒ 2(∠ACB + ∠ACD) = 180°

⇒ ∠BCD = 90°

**7.ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.**

**Sol.**

Given,

∠A = 90° and AB = AC

A/q,

AB = AC

⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)

Now,

∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)

⇒ 90° + 2∠B = 180°

⇒ 2∠B = 90°

⇒ ∠B = 45°

Thus, ∠B = ∠C = 45°

**8. Show that the angles of an equilateral triangle are 60° each.**

**Sol.**

Let ABC be an equilateral triangle.

BC = AC = AB (Length of all sides is same)

⇒ ∠A = ∠B = ∠C (Sides opposite to the equal angles are equal.)

Also,

∠A + ∠B + ∠C = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

Therefore, ∠A = ∠B = ∠C = 60°

Thus, the angles of an equilateral triangle are 60° each.