Class 9 Maths Chapter 7 Exercise 7.3 Solution | Ex 7.3 Class 9 Maths NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 Triangles

In this post, you can find a complete explanation of ex 7.3 class 9 which belongs to Chapter 7 Triangles. Before you go through class 9 ex 7.3, you should read and study the theorems of chapter 7.

 

Exercise 7.3 Class 9 Solutions

Here you will find complete solutions to all questions of 9^th Maths 7.3  along with the video. You can watch videos related to class 9 ex 7.3  which will help you to understand the topic and questions of maths class 9 exercise 7.3.

If you find any difficulty please comment.

Ex.7.3 Class 9 Solutions
Question  1
Question  2
Question  3
Question  4
Question  5
Video for Ex.7.3

Class 9 Maths Chapter 7 Exercise 7.3 Question 1

Q1.∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. if AD is extended to  intersect BC at P, show that

(i).∆ABD≅∆ACD

(ii).∆ABP≅∆ACP

(iii).AP bisects ∠A as well as∠D

(iv).AP is perpendicular bisector of BC

Solution Ex 7.3 class 9

(i). In ∆s ABD and ACD,we have∶

AB=AC(given)

BD=CD(given)

AD=AD(common)

and      ∴ By SSS criterion of congruence, we have

∆ABD≅∆ACD 

(ii).  In ∆s ABP and ACp,we have:

        AB=AC

       ∠BAP=∠CAP

[∵∆ABD≅∆ACD⟹∠BAD=∠DAC⇒∠BAP=∠CAP(CPCT)]

AP=AP(common)

and        ∵By SAS criterion congruence, we have

∆ABP≅∆ACP 

(iii). Since ∆ABD≅∆ACD, therefore

        ∠BAD=∠CAD………..(i)

⇒ AD bisects ∠A ⇒AP bisects ∠A

Now, in ∆s BDP and CDP, we have:

 BD=CD(given)

BP=CP(∵∆ABP≅∆ACP⟹BP=CP)

DP=DP(common)

and

∵ By SSS criterion of congruence, we have:

∆BDP≅∆CDP 

∠BDP=∠CDP                       …………(ii)

⇒DP bisects ∠D 

⇒AP bisects ∠D 

Combining (i) and (ii) we get:

AP bisects∠A as well as∠D 

(iv). Since AP stands on BC

   ∴∠APB+∠APC=180° (Linear pari)

But ∠APB=∠APC (∵∆APB≅∆APC⟹∠APB=∠APC)

∴∠APB=∠APC=`\frac{180°}{2}=90°` 

Also BP=CP

So, AP is a perpendicular bisector of BC.

Hence proved

Class 9 Maths Chapter 7 Exercise 7.3 Question 2

Q2. AD is an altitude of an isosceles triangle ABC in which AB=AC. Show that 

(i).AD bisects BC

(ii).AD bisects ∠A

Solution Ex 7.3 class 9

AD is the altitude drawn from vertex A of an isosceles

 ∆ABC to the opposite base BC so that

 AB=AC,∠ADC=∠ADB=90°

Now, in ∆s ADB and ADC,we have:

HypAB=HypAC (given)

AD=AD(common)

And ∠ADB=∠ADC (∵each=90°)

By RHS criterion of congruence, we have:

∆ADB≅∆ADC 

BD=DC and ∠BAC=∠DAC

(∵corresponding parts of congruent triangles and equal)

Hence, AD bisects BC, which proves (i) 

AD bisects ∠A which proves (ii)

Class 9 Maths Chapter 7 Exercise 7.3 Question 3

Q3.two sides AB and  BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR.show that:

(i).∆ABM≅∆PQN

(ii).∆ABC≅∆PQR

Solution Ex 7.3 class 9

Two ∆s ABC and PQR in which AB=PQ, BC=QR, and AM=PN.

Since AM and PN are medians of ∆s ABC and PQR respectively

Now, BC=QR(given)

⟹`\frac{1}{2} BC=\frac{1}{2}` QR ⟹BM=QN ………..(i)

Now in ∆s ABM and PQN, we have:

AB=PQ(given)

BM=QN(form (i))

AM=PN(given)

∴  By SSS criterion of congruence,we have:

∆ABM≅∆PQn, which proves(i) 

So, ∠B=∠Q  …………….(ii)

(∵corresponding parts of congruent triangles(CPCT)are equal)

Now, in ∆ABC and PQR, we have;

AB=PQ(given)

∠B=∠Q (from(ii))

BC=QR(given)

∴ BY SAS criterion of congrence , we have:

∆ABC≅∆PQR which proves (ii) 

Hence proved

Class 9 Maths Chapter 7 Exercise 7.3 Question 4

Q4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution Ex 7.3 class 9

In ∆s BCF and CBE,we have:

∠BFC=∠CEB (∵each=90°)

Hyp. BC=Hyp. BC (common)

CF=BE(given)

∴ By RHS criterion of congruence, we have:

∆BCF=∆CBE 

So, ∠FBC=∠ECB 

(∵ corresponding parts of congruent triangles are equal)

Now, in ∆ABC,∠ABC=∠ACB (∵∠FBC=∠ECB)

AB=AC

(∵sides opposite to equal angles of a triangle are equal)

∴∆ABC is an isosceles triangle 

Hence proved

Class 9 Maths Chapter 7 Exercise 7.3 Question 5

Q5.ABC is an isosceles triangle with AB=AC. Draw AP⊥BC    to show that ∠B=∠C.

Solution Ex 7.3 class 9

In ∆s ABP and ACP, we have:

AB=AC(given)

AP=AP(common)

And ∠APB=∠APC (∵each=90°)

∵ By RHS criterion of congruence, we have:

∆ABP≅∆ACP 

So, ∠B=∠C

(∵Corresponding parts of congruent triangles are equal)

Hence proved

Watch Vidoe understand the Ex 7.3 Class 9

                                         👇