NCERT Class 9 Maths Solutions Chapter - 7 Triangles, Ex - 7.2

Ex - 7.2

Question 1.  In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersect each other at O. Join A to O. Show that:
(i) OB = OC        (ii) AO bisects A

Solution

(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.)

Question 2.  In ABC, AD is the perpendicular bisector of BC (see the given figure).  Show that ABC is an isosceles triangle in which AB = AC.

Solution

In ADC and ADB
    AD = AD                                         (Common)
    ADC =ADB                              (each 90^o)
    CD = BD                                        (AD is the perpendicular bisector of BC)

Question 3.  ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.

Solution

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

Question 4.  ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure) show that 

    (i)  ABE  ACF 
    (ii) AB = AC, i.e., ABC is an isosceles triangle.

Solution

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

     

(ii) We have already proved
    AEB  AFC
     AB = AC                                      (by CPCT)

Question 5.  ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that

ABD = ACD.

Solution

Let us join AD
In ABD and ACD
AB = AC                                    (Given)
BD = CD                                    (Given)
AD = AD                                    (Common side)

Question 6.  ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure).  Show that BCD is a right angle.

Solution

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

^{Question 7.}  ABC is a right angled triangle in which A = 90^o and AB = AC.  Find B and C.

Solution

Given that

AB = AC

 C = B                      (angles opposite to equal sides are also equal)

In ABC,

A + B + C = 180^o     (angle sum property of a triangle)

 90^o + B + C = 180^o

 90^o + B + B = 180^o
 2 B = 90^o
 B = 45�

Question 8.  Show that the angles of an equilateral triangle are 60^o each.

Solution

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 = 180^o
 A + A + A = 180^o
 3A = 180^o
 A = 60^o
 A = B = C = 60^o
Hence, in an equilateral triangle all interior angles are of 60^o.

Ex-7.1

Ex-7.3