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# Class 9 NCERT Solutions Maths Chapter 7 - Triangles

## Triangles Exercise Ex. 7.1

### Solution 1

In ABC and ABD
CAB = DAB                                 (given)
AB = AB                                             (common)

So, BC and BD are of equal length.

### Solution 2

In ABD and BAC
DAB = CBA                              (given)
AB = BA                                         (common)

And ABD = BAC                          (by CPCT)

### Solution 3

In BOC and AOD
BOC = AOD                                 (vertically opposite angles)
CBO = DAO                                 (each 90o)

### Solution 6

BAD + DAC = EAC + DAC
BAC = DAE
Now in BAC and DAE
BAC = DAE                                 (proved above)
AC = AE                                             (given)

### Solution 7

Given that EPA = DPB
EPA + DPE = DPB + DPE
DPA = EPB
Now in  DAP and  EBP
DAP = EBP                               (given)
AP = BP                                          (P is mid point of AB)
DPA = EPB                              (from above)

### Solution 8

(i)  In AMC and BMD
AM = BM                                              (M is mid point of AB)
AMC = BMD                                  (vertically opposite angles)
CM = DM                                             (given)

(ii) We have ACM = BDM
But ACM and BDM are alternate interior angles
Since alternate angles are equal.
Hence, we can say that DB || AC
DBC + ACB = 180o                   (co-interior angles) DBC + 90o = 180o
DBC + 90o = 1800
DBC          =  90o

(iii) Now in DBC and ACB
DBC = ACB                                 (each 90o )
BC = CB                                             (Common)
(iv) We have DBC  ACB

## Triangles Exercise Ex. 7.2

### Solution 1

(i)    It is given that in triangle ABC, AC = AB
ACB = ABC     (angles opposite to equal sides of a triangle are equal)
OBC = OBC
OB = OC                  (sides opposite to equal angles of a triangle are also equal)

(ii) Now in OAB and OAC
AO =AO                               (common)
AB = AC                                (given)
OB = OC                               (proved above)
So, OAB  OAC         (by SSS congruence rule)
BAO = CAO             (C.P.C.T.)

### Solution 2

CD = BD                                        (AD is the perpendicular bisector of BC)

### Solution 3

In AEB and AFC
AEB = AFC                                             (each 90o)
A = A                                                     (common angle)
AB = AC                                                        (given)

### Solution 4

(i) In AEB and AFC
AEB = AFC                          (each 90)
A = A                                      (common angle)
BE = CF                                        (given)

AEB  AFC
AB = AC                                      (by CPCT)

### Solution 5

In ABD and ACD
AB = AC                                    (Given)
BD = CD                                    (Given)

### Solution 6

In ABC
AB = AC                                                (given)
ACB = ABC                               (angles opposite to equal sides of a triangle are also equal)
Now In ACD
ADC = ACD                              (angles opposite to equal sides of a triangle are also equal)
Now, in BCD
ABC + BCD + ADC = 180o          (angle sum property of a triangle)
ACB + ACB +ACD + ACD = 180o
2(ACB + ACD) = 180o
2(BCD) = 180o
BCD = 90o

### Solution 7

Given that
AB = AC
C = B                      (angles opposite to equal sides are also equal)
In ABC,
A + B + C = 180o     (angle sum property of a triangle)
90o + B + C = 180o
90o + B + B = 180o
2 B = 90o
B = 45

### Solution 8

Let us consider that ABC is an equilateral triangle.
So, AB = BC = AC
Now, AB = AC
C = B         (angles opposite to equal sides of a triangle are equal)

We also have
AC = BC
B = A             (angles opposite to equal sides of a triangle are equal)

So, we have
A = B = C
Now, in ABC
A + B + C = 180o
A + A + A = 180o
3A = 180o
A = 60o
A = B = C = 60o
Hence, in an equilateral triangle all interior angles are of 60o.

## Triangles Exercise Ex. 7.3

### Solution 1

(i)  In ABD and ACD
AB = AC                                             (given)
BD = CD                                            (given)

(ii)  In ABP and ACP
AB = AC                                            (given).
BAP = CAP                                  [from equation (1)]
AP = AP                                             (common)

(iii)   From equation (1)
BAP = CAP
Hence, AP bisect A
Now in BDP and CDP
BD = CD                                            (given)
DP = DP                                            (common)
BP = CP                                            [from equation (2)]

(iv)   We have BDP  CDP

Now, BPD + CPD = 180o             (linear pair angles)

BPD + BPD = 180o

2BPD = 180o                                    [from equation (4)]

BPD = 90o                                                                    ...(5)

From equations (2) and (5), we can say that AP is perpendicular  bisector of BC.

### Solution 2

AB = AC                                                  (given)

(ii)     Also by CPCT,

(ii)              Also by CPCT,

### Solution 3

(i)  In ABC, AM is median to BC
BM = BC
In PQR, PN is median to QR
QN = QR
But BC = QR

BN = QN                                                     ...(i)
Now, in ABM and PQN
AB = PQ                                                       (given)
BM = QN                                                       [from equation (1)]
AM = PN                                                        (given)

(ii)  Now in ABC and PQR

AB = PQ                                                        (given)
ABC = PQR                                             [from equation (2)]
BC = QR                                                        (given)
ABC  PQR                                  (by SAS congruence rule)

### Solution 4

In BEC and CFB
BEC = CFB                                              (each 90o )
BC = CB                                                         (common)
BE = CF                                                         (given)
(Sides opposite to equal angles of a triangle are equal)

Hence, ABC is isosceles.

### Solution 5

In APB and APC
APB = APC                                             (each 90o)
AB =AC                                                          (given)
AP = AP
(common)
B = C                                                (by using CPCT)