 # NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1

Exercise 7.1 of Chapter 7 of NCERT Class 9 Maths deals with the “Congruence of Triangles”. You must be wondering why this topic has been included in the syllabus although we rarely see its applications. Well, in real life two triangles are rarely congruent. However, studying this topic becomes crucial because it is used in the construction of large buildings or architectural designing. The mathematician found the Triangle as the most stable shape and congruence is needed to create even surfaces. Also, you must have noticed congruent triangles in the geometric art, carpeting designs, stepping stone designs, architectural designs, etc. Thus, this chapter plays a major role in daily life.

The NCERT Class 9 Maths Exercise is full of proofs. Therefore, it’s important that you first know theorems related to it and then go for the exercise questions. Here we have provided the step by step NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1. Students can refer to it whenever they find difficulty in solving the questions. These solutions are created by the subject experts and are solved in simpler way. These solutions will also help you to learn the answer writing skills so that you can score high marks in the annual exams. You can download the NCERT Solution PDF, from the link below.

### Download PDF of NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.1      ### Access other Exercise Solutions of Chapter 7 – Triangles

You should practise all the questions of Exercise 7.1. To get the detailed solutions of all the exercises of NCERT Class 9 Chapter 7, visit the links provided below:

Exercise 7.2 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Question)

Exercise 7.3 Solution 5 Questions (3 Short Answer Questions, 2 Long Answer Question)

Exercise 7.4 Solution 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

Exercise 7.5 (Optional) Solution 4 Questions

### Access Answers to NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.1

1. In quadrilateral ACBD, AC = AD and AB bisect A (see Fig. 7.16). Show that ΔABC ΔABD. What can you say about BC and BD? Solution:

It is given that AC and AD are equal i.e. AC = AD and the line segment AB bisects A.

We will have to now prove that the two triangles ABC and ABD are similar i.e. ΔABC ΔABD

Proof:

Consider the triangles ΔABC and ΔABD,

(i) AC = AD (It is given in the question)

(ii) AB = AB (Common)

(iii) CAB = DAB (Since AB is the bisector of angle A)

So, by SAS congruency criterion, ΔABC ΔABD.

For the 2nd part of the question, BC and BD are of equal lengths by the rule of C.P.C.T.

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

(i) ΔABD ΔBAC

(ii) BD = AC

(iii) ABD = BAC. Solution:

The given parameters from the questions are DAB = CBA and AD = BC.

(i) ΔABD and ΔBAC are similar by SAS congruency as

AB = BA (It is the common arm)

DAB = CBA and AD = BC (These are given in the question)

So, triangles ABD and BAC are similar i.e. ΔABD ΔBAC. (Hence proved).

(ii) It is now known that ΔABD ΔBAC so,

BD = AC (by the rule of CPCT).

(iii) Since ΔABD ΔBAC so,

Angles ABD = BAC (by the rule of CPCT).

3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB. Solution:

It is given that AD and BC are two equal perpendiculars to AB.

We will have to prove that CD is the bisector of AB

Now,

Triangles ΔAOD and ΔBOC are similar by AAS congruency since:

(i) A = B (They are perpendiculars)

(ii) AD = BC (As given in the question)

(iii) AOD = BOC (They are vertically opposite angles)

∴ ΔAOD ΔBOC.

So, AO = OB (by the rule of CPCT).

Thus, CD bisects AB (Hence proved).

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ΔCDA. Solution:

It is given that p q and l m

To prove:

Triangles ABC and CDA are similar i.e. ΔABC ΔCDA

Proof:

Consider the ΔABC and ΔCDA,

(i) BCA = DAC and BAC = DCA Since they are alternate interior angles

(ii) AC = CA as it is the common arm

So, by ASA congruency criterion, ΔABC ΔCDA.

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 Fig. 7.20). Show that:

(i) ΔAPB ΔAQB

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

It is given that the line “l” is the bisector of angle A and the line segments BP and BQ are perpendiculars drawn from l.

(i) ΔAPB and ΔAQB are similar by AAS congruency because:

P = Q (They are the two right angles)

AB = AB (It is the common arm)

BAP = BAQ (As line l is the bisector of angle A)

So, ΔAPB ΔAQB.

(ii) By the rule of CPCT, BP = BQ. So, it can be said the point B is equidistant from the arms of A.

6. In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. Solution:

It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC

To prove:

The line segment BC and DE are similar i.e. BC = DE

Proof:

We know that BAD = EAC

Now, by adding DAC on both sides we get,

BAD + DAC = EAC +DAC

Now, ΔABC and ΔADE are similar by SAS congruency since:

(i) AC = AE (As given in the question)

(iii) AB = AD (It is also given in the question)

So, by the rule of CPCT, it can be said that BC = DE.

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 Fig. 7.22). Show that

(i) ΔDAP ΔEBP Solutions:

In the question, it is given that P is the mid-point of line segment AB. Also, BAD = ABE and EPA = DPB

(i) It is given that EPA = DPB

Now, add DPE on both sides,

EPA +DPE = DPB+DPE

This implies that angles DPA and EPB are equal i.e. DPA = EPB

Now, consider the triangles DAP and EBP.

DPA = EPB

AP = BP (Since P is the mid-point of the line segment AB)

BAD = ABE (As given in the question)

So, by ASA congruency, ΔDAP ΔEBP.

(ii) By the rule of CPCT, AD = BE.

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 Fig. 7.23). Show that:

(i) ΔAMC ΔBMD

(ii) DBC is a right angle.

(iii) ΔDBC ΔACB

(iv) CM = ½ AB Solution:

It is given that M is the mid-point of the line segment AB, C = 90°, and DM = CM

(i) Consider the triangles ΔAMC and ΔBMD:

AM = BM (Since M is the mid-point)

CM = DM (Given in the question)

CMA = DMB (They are vertically opposite angles)

So, by SAS congruency criterion, ΔAMC ΔBMD.

(ii) ACM = BDM (by CPCT)

∴ AC BD as alternate interior angles are equal.

Now, ACB +DBC = 180° (Since they are co-interiors angles)

⇒ 90° +B = 180°

∴ DBC = 90°

(iii) In ΔDBC and ΔACB,

BC = CB (Common side)

ACB = DBC (They are right angles)

DB = AC (by CPCT)

So, ΔDBC ΔACB by SAS congruency.

(iv) DC = AB (Since ΔDBC ΔACB)

⇒ DM = CM = AM = BM (Since M the is mid-point)

So, DM + CM = BM+AM

Hence, CM + CM = AB

⇒ CM = (½) AB

The students of Class 9 will learn the concepts of Congruence of Triangles, Criteria for Congruence and theorem related to it in Exercise 7.1. In the exercise, the questions are related to proof. So, while solving these questions it’s very important that students should mention the congruence rule. Also, they should not miss any steps while solving it. To know how to write answers in a better way to score good marks, download the NCERT solution PDF.

We hope this information on “NCERT Solution Class 9 Maths Chapter 7 Triangles Exercise 7.1” is useful for students.

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