NCERT Class 9 Maths Solutions Chapter - 7 Triangles, Ex - 7.3

Ex - 7.3

Question 1.  ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). 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 the perpendicular bisector of BC.

Solution

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

      

(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 = 180^o             (linear pair angles)

         BPD + BPD = 180^o 

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

        BPD = 90^o                                                                    ...(5)

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

Question 2.  AD is an altitude of an isosceles triangles ABC in which AB = AC.  Show that 

(i) AD bisects BC    (ii) AD bisects A.

Solution

(i)   In BAD and CAD
        ADB = ADC                                      (each 90^o as AD is an altitude)
        AB = AC                                                  (given)
        AD = AD                                                  (common)

  

(ii)     Also by CPCT,
         BAD = CAD
          Hence, AD bisects A.

Question 3.  Two side AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR (see the given figure). Show that: 

(i)    ABM  PQN
(ii)    ABC  PQR

Solution

(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)

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

In BEC and CFB
BEC = CFB                                              (each 90^o )
BC = CB                                                         (common)
BE = CF                                                         (given)

                                                 (Sides opposite to equal angles of a triangle are equal)

Hence, ABC is isosceles.

Question 5.  ABC is an isosceles triangle with AB = AC. Drawn AP  BC to show that B = C.

Solution

In APB and APC
APB = APC                                             (each 90^o)
AB =AC                                                          (given)
AP = AP                                                   

     (common)

 B = C                                                (by using CPCT)

Ex-7.2

Ex-7.4