NCERT Solutions For Class 7 Maths Chapter 7 Congruence of Triangles are given here in a simple and detailed way. While solving the exercise questions from the 7th class maths book, students often face difficulty and eventually pile up their doubts. These class 7 ncert solutions can be extremely helpful for the students to clear all their doubts, easily and understand the basics of this chapter in a comprehensive manner.

The class 7 congruence of triangles ncert solutions given here, are easily understandable in such a way that students do not face any difficulties while solving problems related to the topic.

## Class 7 Maths NCERT Solutions – Congruence of Triangles

The congruence of triangles is among the most important topics in the entire CBSE class 7 maths book. These concepts are also used in further classes, therefore students are suggested to learn this chapter thoroughly to comprehend the higher level classes as well.

These NCERT solutions explains the topics covered in Congruence of Triangles for each question and are designed by experts as per CBSE syllabus (2018-2019). Major topics covered in this chapter 7 of class 7 are;

- Congruence Of Triangles
- Congruence Of Plane Figures
- Congruence Among Line Segments
- Congruence Of Angles
- Criteria For Congruence Of Triangles
- Congruence Among Right-Angled Triangles

These can also help the students to have an in-depth understanding of the concepts and develop a deeper interest in the concepts. The NCERT solutions for 7th class maths Congruence of Triangles PDF is also available which students can download and study offline as well. Students appearing for state level and national level Maths competitive exams in 2019, can also refer to these solutions to clear their concepts.

### NCERT Class 7 Maths Solutions For Congruence of Triangles 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

#### Exercise 7.1

**1. In quadrilateral ACBD, AC = AD and AB bisects \(\angle\)A . Show that \(\Delta\)ABC \(\cong\) \(\Delta\)ABD. What can you say about BC and BD?**

**Sol. **

Given that,

AC = AD and AB bisects \(\angle\)A

To prove that,

\(\Delta\)ABC \(\cong\) \(\Delta\)ABD

Proof,

In \(\Delta\)ABC and \(\Delta\)ABD,

AB = AB

AC = AD

\(\angle\)CAB =\(\angle\)DAB (AB is bisector)

Therefore, \(\Delta\)ABC \(\cong\) \(\Delta\)ABD by SAS congruence condition.

BC and BD are of equal length.

**2. ABCD is a quadrilateral in which AD = BC and \(\angle\)DAB = \(\angle\)CBA . Prove that**

**(i) \(\Delta\)ABD \(\cong\) \(\Delta\)BAC
(ii) BD = AC
(iii) \(\angle\)ABD = \(\angle\)BAC.**

**Sol.**

Given that ,

AD = BC and \(\angle\) DAB = \(\angle\)CBA

(i) In \(\Delta\)ABD and \(\Delta\)BAC,

AB = BA (Common)

\(\angle\) DAB = \(\angle\)CBA (Given)

AD = BC (Given)

Therefore, \(\Delta\)ABD \(\cong\) \(\Delta\)BAC by SAS congruence condition.

(ii) Since, \(\Delta\)ABD \(\cong\) \(\Delta\)BAC

Therefore BD = AC by CPCT

(iii) Since, \(\Delta\)ABD \(\cong\) \(\Delta\)BAC

Therefore \(\angle\) ABD \(\cong\) \(\angle\)BAC by CPCT

**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 \(\Delta\)AOD and \(\Delta\)BOC,

\(\angle\)A = \(\angle\)B (Perpendicular)

\(\angle\)AOD = \(\angle\)BOC (Vertically opposite angles)

AD = BC (Given)

Therefore, \(\Delta\)AOD \(\cong\) \(\Delta\)BOC by AAS congruence condition.

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 \(\Delta\)ABC \(\cong\) \(\Delta\)CDA.**

**Sol. **

Given that,

l \(\parallel\) m and p \(\parallel\) q

To prove,

\(\Delta\)ABC ≅ \(\Delta\)CDA

Then Proof that,

In \(\Delta\)ABC and \(\Delta\)CDA,

\(\angle\)BCA =\(\angle\)DAC (Alternate interior angles)

AC = CA (Common)

\(\angle\)BAC = \(\angle\)DCA (Alternate interior angles)

Therefore, \(\Delta\)ABC \(\cong\) \(\Delta\)CDA by ASA congruence condition.

**5. Line l is the bisector of an angle \(\angle\)A and B is any point on l. BP and BQ are perpendiculars from B to the arms of \(\angle\)A . Show that:**

**(i) \(\Delta\)APB \(\cong\) \(\Delta\)AQB**

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

**Sol.**

Given,

l is the bisector of an angle \(\angle\)A.

BP and BQ are perpendiculars.

(i) In \(\Delta\)APB and \(\Delta\)AQB,

\(\angle\)P = \(\angle\)Q (Right angles)

\(\angle\)BAP =\(\angle\)BAQ (l is bisector)

AB = AB (Common)

Therefore, \(\Delta\)APB \(\cong\) \(\Delta\)AQB by AAS congruence condition.

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

**6. AC = AE, AB = AD and \(\angle\)BAD = ****\(\angle\)****EAC. Show that BC = DE.**** **

** **

**Sol.**

Given that,

AC = AE, AB = AD and \(\angle\)BAD = \(\angle\)EAC

To show,

BC = DE

then Proof that,

\(\angle\)BAD = \(\angle\)EAC (Adding \(\angle\)DAC both sides)

\(\angle\)BAD + \(\angle\)DAC = \(\angle\)EAC + \(\angle\)DAC

\(\Rightarrow\) \(\angle\)BAC = \(\angle\)EAD

In \(\Delta\)ABC and \(\Delta\)ADE,

AC = AE (Given)

\(\angle\)BAC = \(\angle\)EAD

AB = AD (Given)

Therefore, \(\Delta\)ABC \(\cong\) \(\cong\)\(\Delta\)ADE by SAS congruence condition.

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 \(\angle\)BAD = \(\angle\)ABE and \(\angle\)EPA = \(\angle\)DPB . Show that**** (i) \(\Delta\)DAP \(\cong\) \(\Delta\)EBP**** (ii) AD = BE**

**Sol.**

Given,

P is mid-point of AB.

\(\angle\)BAD = \(\angle\)ABE and \(\angle\)EPA = \(\angle\)DPB

(i) \(\angle\)EPA =\(\angle\)DPB (Adding \(\angle\)DPE both sides)

\(\angle\)EPA + \(\angle\)DPE = \(\angle\)DPB + \(\angle\)DPE

\(\Rightarrow\) \(\angle\)DPA = \(\angle\)EPB

In \(\Delta\)DAP \(\cong\) \(\Delta\)EBP,

\(\angle\)DPA = \(\angle\)EPB

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

\(\angle\)BAD =\(\angle\)ABE (Given)

Therefore, \(\Delta\)DAP \(\cong\) \(\Delta\)EBP by ASA congruence condition.

(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) \(\Delta\)AMC \(\cong\) \(\Delta\)BMD**

**(ii) \(\angle\)DBC is a right angle.**

**(iii) \(\Delta\)DBC \(\cong\) \(\Delta\)ACB**

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

**Sol.**

Given,

\(\angle\)C = 90°, M is the mid-point of AB and DM = CM

(i) In \(\Delta\)AMC and \(\Delta\)BMD,

AM = BM (M is the mid-point)

\(\angle\)CMA = \(\angle\)DMB (Vertically opposite angles)

CM = DM (Given)

Therefore, \(\Delta\)AMC \(\cong\) \(\Delta\)BMD by SAS congruence condition.

(ii) \(\angle\)ACM = \(\angle\)BDM (by CPCT)

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

Now,

\(\angle\)ACB + \(\angle\)DBC = 180° (co-interiors angles)

\(\Rightarrow\) 90° + \(\angle\)B = 180°

\(\Rightarrow\) \(\angle\)DBC = 90°

(iii) In \(\Delta\)DBC and \(\Delta\)ACB,

BC = CB (Common)

\(\angle\)ACB = \(\angle\)DBC (Right angles)

DB = AC

Therefore, \(\Delta\)DBC \(\cong\) \(\Delta\)ACB by SAS congruence condition.

(iv) DC = AB (\(\Delta\)DBC \(\cong\) \(\Delta\)ACB)

\(\Rightarrow\) DM = CM = AM = BM (M is mid-point)

\(\Rightarrow\) DM + CM = AM + BM

\(\Rightarrow\) CM + CM = AB

\(\Rightarrow\) CM = 1/2AB

#### Exercise 7.2

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

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

**Sol.**

Sol. Given that,

AB = AC, the bisectors of \(\angle\)B and \(\angle\)C intersect each other at O

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

∴ \(\angle\)B = \(\angle\)C

⇒ 1/2\(\angle\)B = 1/2\(\angle\)C

⇒ \(\angle\)OBC = \(\angle\)OCB (Angle bisectors.)

⇒ 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.

\(\angle\)BAO = \(\angle\)CAO (by CPCT)

Thus, AO bisects \(\angle\)A.

**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.

Students can use NCERT solutions for Class 7, as a worksheet for Chapter 7 and prepare for their final exams. They are also suggested to solve previous year question papers and sample papers to get an idea of the type of questions asked from this section. You can also find out NCERT Question Papers For Class 7 here with us in BYJU’S.

Apart from these solutions for chapter 6, congruence of triangles, students can also get the NCERT Solutions For Class 7 Maths for all the chapters, exercise-wise, only at BYJU’S. And to have a better understanding of the concepts of congruence of triangles and to get a new learning experience, download BYJU’S – The learning App, where you can learn from educational videos.