give me the solutions of the exercise 8.1

12.

Extend AB. Draw a line through C, which is parallel to AD, intersecting AE at point E. 
Now, AECD is a parallelogram.

(i)   AD = CE                  (opposite sides of parallelogram AECD)  
      But AD = BC             (given)
      So, BC = CE 
     CEB = CBE         (angle opposite to equal sides are also equal) 
     Now consider parallel lines AD and CE. AE is transversal line for them 
    A + CEB = 180         (angles on the same side of transversal)
    A+ CBE = 180         (using the relationCEB = CBE)    ... (1)  
     But B + CBE = 180     (linear pair angles)            ... (2) 
     From equations (1) and (2), we have
     A = B  
(ii)  AB || CD
      A + D = 180            (angles on the same side of transversal) 
      Also C + B = 180          (angles on the same side of transversal)  
       A + D = C + B

      But A = B             [using the result obtained proved in (i)]
       C = D 

(iii)  In ABC and BAD
       AB = BA                 (common side) 
       BC = AD                 (given)
       B = A                 (proved before) 
       ABC  BAD            (SAS congruence rule) 

(iv)   ABCBAD  
       AC = BD                 (by CPCT) 

 

11.

 

(i)   Here AB = DE and AB || DE.     
      Now, if two opposite sides of a quadrilateral are equal and parallel to each other, it will be

      a parallelogram. 
      Therefore, quadrilateral ABED is a parallelogram. 

(ii)  Again BC = EF and BC || EF. 
      Therefore, quadrilateral BEFC is a parallelogram.

 

(iii)  Here ABED and BEFC are parallelograms. 
       AD = BE, and AD || BE          
       (Opposite sides of parallelogram are equal and parallel) 
       And BE = CF, and BE || CF         
       (Opposite sides of parallelogram are equal and parallel)
         AD = CF, and AD || CF 

(iv)   Here one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and

        parallel to each other, 
        so, it is a parallelogram. 
(v)    As ACFD is a parallelogram, so, pair of opposite sides will be equal and parallel to each

        other. 
         AC || DF and AC = DF 

(vi)   ABC and DEF.
        AB = DE                                (given)
        BC = EF                                (given) 
        AC = DF                                (ACFD is a parallelogram)  
        ABCDEF                (by SSS congruence rule)

 

10.

(i)  In APB and CQD
     APB = CQD         (each 90^o) 
     AB = CD             (opposite sides of parallelogram ABCD) 
     ABP = CDQ         (alternate interior angles for AB || CD) 
      APB CQD        (by AAS congruency) 

(ii) By using the result obtained as above 
     APB CQD, we have 
     AP = CQ             (by CPCT) 
9.

 

(i)  In APD and CQB
     ADP = CBQ                   (alternate interior angles for BC || AD) 
     AD = CB                       (opposite sides of parallelogram ABCD) 
     DP = BQ                       (given) 
     APD CQB         (using SAS congruence rule) 

(ii)  As we had observed that APD  CQB 
       AP = CQ                     (CPCT)

 

(iii)  In AQB and CPD
       ABQ = CDP          (alternate interior angles for AB || CD)
       AB = CD                     (opposite sides of parallelogram ABCD) 
       BQ = DP                     (given) 
       AQB  CPD               (using SAS congruence rule) 

(iv)  As we had observed that AQB CPD 
        AQ = CP             (CPCT)

(v)   From the result obtained in (ii) and (iv), we have
              AQ = CP and  AP = CQ  
       Since opposite sides in quadrilateral APCQ are equal to each other. So, APCQ is a

       parallelogram.

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