**NCERT Solutions for Class 9
Maths Chapter 7 Triangles Ex 7.3**

NCERT Solutions for Class
9 Maths Chapter 7 Triangles Ex 7.3 are the part of NCERT Solutions for Class 9 Maths.
Here you can find the NCERT Solutions for Chapter 7 Triangles Ex 7.3.

**NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5**

**Ex 7.3 Class 9 Maths Question 1.****
**Î”ABC and Î”DBC are two isosceles
triangles on the same base BC and vertices A and D are on the same side of BC
(see figure). If AD is extended to intersect BC at P, show that:

**(i)**Î”ABD ≅ Î”ACD

**(ii)**Î”ABP ≅ Î”ACP

**(iii)**AP bisects ∠A as well as ∠D

**(iv)**AP is the perpendicular bisector of BC.

**Solution:
(i)** In ∆ABD and ∆ACD, we have

AB = AC [Given]

AD = DA [Common]

BD = CD [Given]

∴ ∆ABD ≅ ∆ACD [By SSS congruency rule]

∠BAD = ∠CAD [By CPCT] …(1)

**(ii)**
In ∆ABP and ∆ACP, we have

AB = AC [Given]

∠BAP = ∠CAP [From (1)]

∴ AP = PA [Common]

∴ ∆ABP ≅ ∆ACP [By SAS
congruency rule]

**(iii)**
Since, ∆ABP ≅ ∆ACP

⇒ ∠BAP = ∠CAP [By CPCT]

∴ AP is the bisector of ∠A.

Again, in ∆BDP and ∆CDP, we have

BD = CD [Given]

DP = PD [Common]

BP = CP [ ∵ ∆ABP ≅ ∆ACP]

⇒ ∆BDP = ∆CDP [By SSS congruency rule]

∴ ∠BDP = ∠CDP [By CPCT]

⇒ DP (or AP) is the bisector of ∠BDC.

∴ AP is the bisector of ∠A as well as ∠D.

**(iv)**
Since ∆ABP ≅ ∆ACP

⇒ ∠APB = ∠APC and BP = CP [By
CPCT]

But ∠APB + ∠APC = 180° [Linear
Pair]

∴ ∠APB = ∠APC = 90°

⇒ AP ⊥ BC, also BP = CP

Hence, AP is the perpendicular bisector of BC.

**Ex 7.3 Class 9 Maths Question 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:**

**(i)** In right-angled ∆ABD and ∆ACD, we have

AB = AC [Given]

∠ADB = ∠ADC [Each 90°]

AD = DA [Common]

∴ ∆ABD ≅ ∆ACD [By RHS congruency
rule]

So, BD = CD [By CPCT]

⇒ D is the mid-point of BC or AD bisects BC.

**(ii)**
Since ∆ABD ≅ ∆ACD

⇒ ∠BAD = ∠CAD [By CPCT]

So, AD bisects ∠A.

**Ex 7.3 Class 9 Maths Question 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 ∆PQR (see
figure). Show that

**(i)**∆ABM ≅ ∆PQN

**(ii)**∆ABC ≅ ∆PQR

**Solution:
**In ∆ABC, AM is the median.

∴ BM = ½ BC ……(1)

In ∆PQR, PN is the median.

∴ QN = ½ QR ……(2)

And BC = QR [Given]

⇒ ½ BC = ½ QR

⇒ BM = QN ……(3) [From eqs. (1) and (2)]

**(i)**
In ∆ABM and ∆PQN, we have

AB = PQ [Given]

AM = PN [Given]

BM = QN [From eq. (3)]

∴ ∆ABM ≅ ∆PQN [By SSS
congruency rule]

**(ii)**
Since ∆ABM ≅ ∆PQN

⇒ ∠B = ∠Q ……(4) [By CPCT]

Now, in ∆ABC and ∆PQR, we have

∠B = ∠Q [From
eq. (4)]

AB = PQ [Given]

BC = QR [Given]

∴ ∆ABC ≅ ∆PQR [By SAS congruency
rule]

**Ex 7.3 Class 9 Maths Question 4.
**BE and CF are two equal altitudes of a triangle ABC.
Using RHS congruence rule, prove that the triangle ABC is isosceles.

**Solution:**

Since BE ⊥ AC [Given]

∴ BEC is a right-angled triangle such that ∠BEC = 90°

Similarly, ∠CFB = 90°

Now, in right-angled ∆BEC and ∆CFB, we have

BE = CF [Given]

BC = CB [Common
hypotenuse]

∠BEC = ∠CFB [Each 90°]

∴ ∆BEC ≅ ∆CFB [By RHS
congruency rule]

So, ∠BCE = ∠CBF [By CPCT]

or ∠BCA = ∠CBA

Now, in ∆ABC, ∠BCA = ∠CBA

⇒ AB = AC [Sides
opposite to equal angles of a ∆ are equal]

∴ ∆ABC is an isosceles triangle.

**Ex 7.3 Class 9 Maths Question 5.**

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

**Solution:**

We have, AP ⊥ BC [Given]

∠APB = 90° and ∠APC = 90°

In ∆ABP and ∆ACP, we have

∠APB = ∠APC [Each 90°]

AB = AC [Given]

AP = AP [Common]

∴ ∆ABP ≅ ∆ACP [By RHS congruency rule]

So, ∠B = ∠C [By CPCT]

**NCERT Solutions for Maths Class 10**