The third exercise of Chapter 7 Maths Class 9 NCERT book covers a few more criteria for congruence of triangles. This exercise will mainly focus on SSS and RHS congruence rule. This exercise contains only 5 questions which are based on proof. To help students in their studies, here we have provided the solutions to all the questions in PDF format with detailed explanations. Students can download the NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3 PDF, through the link provided further below in this page. Also, these solutions are provided by the best teachers.

To begin with, students should try to solve the Exercise problem 7.3 by their own selves. After that they should go through the solution pdf. This will assist them in getting to know the best way of writing the answers that will help them in scoring more marks in the annual Maths paper. Also, their doubts related to any particular question will get resolved.

### Access Answers to NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.3

**1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that**

(i) ΔABD ΔACD

(ii) ΔABP ΔACP

(iii) AP bisects A as well as D.

(iv) AP is the perpendicular bisector of BC.

**Solution:**

In the above question, it is given that ΔABC and ΔDBC are two isosceles triangles.

(i) ΔABD and ΔACD are similar by SSS congruency because:

AD = AD (It is the common arm)

AB = AC (Since ΔABC is isosceles)

BD = CD (Since ΔDBC is isosceles)

∴ ΔABD ΔACD.

(ii) ΔABP and ΔACP are similar as:

AP = AP (It is the common side)

PAB = PAC (by CPCT since ΔABD ΔACD)

AB = AC (Since ΔABC is isosceles)

So, ΔABP ΔACP by SAS congruency condition.

(iii) PAB = PAC by CPCT as ΔABD ΔACD.

AP bisects A. — (i)

Also, ΔBPD and ΔCPD are similar by SSS congruency as

PD = PD (It is the common side)

BD = CD (Since ΔDBC is isosceles.)

BP = CP (by CPCT as ΔABP ΔACP)

So, ΔBPD ΔCPD.

Thus, BDP = CDP by CPCT. — (ii)

Now by comparing (i) and (ii) it can be said that AP bisects A as well as D.

(iv) BPD = CPD (by CPCT as ΔBPD ΔCPD)

and BP = CP — (i)

also,

BPD +CPD = 180° (Since BC is a straight line.)

⇒ 2BPD = 180°

⇒ BPD = 90° —(ii)

Now, from equations (i) and (ii), it can be said that

AP is the perpendicular bisector of BC.

**2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that**

(i) AD bisects BC (ii) AD bisects A.

**Solution:**

It is given that AD is an altitude and AB = AC. The diagram is as follows:

(i) In ΔABD and ΔACD,

ADB = ADC = 90°

AB = AC (It is given in the question)

AD = AD (Common arm)

∴ ΔABD ΔACD by RHS congruence condition.

Now, by the rule of CPCT,

BD = CD.

So, AD bisects BC

(ii) Again, by the rule of CPCT, BAD = CAD

Hence, AD bisects A.

**3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that:**

(i) ΔABM ΔPQN

(ii) ΔABC ΔPQR

**Solution:**

Given parameters are:

AB = PQ,

BC = QR and

AM = PN

(i) ½ BC = BM and ½ QR = QN (Since AM and PN are medians)

Also, BC = QR

So, ½ BC = ½ QR

⇒ BM = QN

In ΔABM and ΔPQN,

AM = PN and AB = PQ (As given in the question)

BM = QN (Already proved)

∴ ΔABM ΔPQN by SSS congruency.

(ii) In ΔABC and ΔPQR,

AB = PQ and BC = QR (As given in the question)

ABC = PQR (by CPCT)

So, ΔABC ΔPQR by SAS congruency.

**4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.**

.

**Solution:**

It is known that BE and CF are two equal altitudes.

Now, in ΔBEC and ΔCFB,

BEC = CFB = 90° (Same Altitudes)

BC = CB (Common side)

BE = CF (Common side)

So, ΔBEC ΔCFB by RHS congruence criterion.

Also, C = B (by CPCT)

Therefore, AB = AC as sides opposite to the equal angles is always equal.

**5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that B = C.**

**Solution:**

In the question, it is given that AB = AC

Now, ΔABP and ΔACP are similar by RHS congruency as

APB = APC = 90° (AP is altitude)

AB = AC (Given in the question)

AP = AP (Common side)

So, ΔABP ΔACP.

∴ B = C (by CPCT)

Students have studied the AAA congruence and other congruence rules related to angle of a triangle in the previous two exercises. Now, here students will learn some more new Congruence rules, which will be related to the sides of a triangle. These congruence rules are described in two theorems as mentioned below:

**Theorem 7.4 (SSS congruence rule):**If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.**Theorem 7.5 (RHS congruence rule):**If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

After this, the examples are given in the NCERT book to illustrate them in a better way for students’ understanding. The question in the exercise are fully based on SSS and RHS congruence rules. So, the students should first go through the examples and then solve the exercise problems.

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