NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

# 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. In this post, you will find the NCERT Solutions for Chapter 7 Triangles Ex 7.3.

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

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]
BD = CD                     [Given]
∆ABD ∆ACD      [By SSS congruency rule]

(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
A

Solution:

(i) In right-angled ∆ABD and ∆ACD, we have
AB = AC                       [Given]
∆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
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

NCERT Solutions for Maths Class 11

NCERT Solutions for Maths Class 12