Class 9 NCERT Solutions Maths Chapter 8 - Quadrilaterals

                                            

 

Let ABCD be a parallelogram. To show ABCD a rectangle, only we need to prove one of its interior angle is 90.
In ABC and DCB
AB = DC             (opposite sides of a parallelogram are equal)
BC = BC             (common)
AC = DB             (given)
ABC  DCB         (by SSS Congruence rule)

ABC = DCB         
We know that sum of measures of angles on the same side of transversal is 180º.
ABC + DCB = 180     (AB || CD)
ABC + ABC = 180     
ABC = 180  
ABC = 90
Since ABCD is a parallelogram and one of its interior angles is 90, therefore, ABCD is rectangle.  

                                         

Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.
 To show diagonals of a square are equal and bisect each other at right angles, we need to prove AC = BD, OA = OC, OB = OD and AOB = 90 
  Now, in ABC and DCB
  AB = DC                          (sides of square are equal to each other)
ABC = DCB                 (all interior angles are of 90 )
  BC = BC                          (common side)
 ABCDCB           (by SAS congruency)
 AC = DB                        (by CPCT)

 

Hence, the diagonals of a square are equal in length
    
Now in AOB and COD
AOB = COD                        (vertically opposite angles)
ABO = CDO                        (alternate interior angles)
AB = CD                             (sides of square are always equal)
 AOB COD          (by AAS congruence rule)
 AO = CO and OB = OD     (by CPCT)             

Hence, the diagonals of a square bisect each other
Now in AOB and COB
Now as we had proved that diagonals bisect each other
So, AO = CO                                 
AB = CB                              (sides of square are equal)
BO = BO                             (common)
AOB COB           (by SSS congruence)
 AOB = COB             (by CPCT)

But, AOB + COB = 180        (linear pair)
2AOB = 180
AOB = 90
Hence, the diagonals of a square bisect each other at right angle.  

 

(i) ABCD is a parallelogram.
     DAC = BCA        (Alternate interior angles)    ... (1)

    And BAC = DCA        (Alternate interior angles)    ... (2)
    But it is given that AC bisects A.
     DAC = BAC                        ... (3)

    From equations (1), (2) and (3), we have
    DAC = BCA = BAC = DCA                ... (4) 
    DCA = BCA
     Hence, AC bisects C.
(ii)

     From equation (4), we have
     DAC = DCA    
      DA = DC                         (side opposite to equal angles are equal)

     But DA = BC and AB = CD     (opposite sides of parallelogram)
      AB = BC = CD = DA

 

     Hence, ABCD is rhombus

(i)

Given: 

DAC = CAB   ... (1)

and DCA = BCA   ... (2)

Now, AD || BC and AC is a transversal

 

DAC = BCA

 

AB = BC = AD = CD

All the sides of this rectangle are equal.

Thus, ABCD is a square.

 

(ii)

Since, ABCD is a square and we know that the diagonals of a square bisect its angles.

  BD bisects B as well asD.

 

(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)
       AQBCPD               (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.

(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)

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)
 

(i)  In ADC, S and R are the mid points of sides AD and CD respectively.
     In a triangle the line segment joining the mid points of any two sides of the triangle is

     parallel to the third side and is half of it.
     SR || AC and SR =  AC                ... (1)

(ii)  In ABC, P and Q are mid points of sides AB and BC respectively. So, by using

      mid-point theorem, we have
      PQ || AC and PQ =  AC                ... (2)   
      Now using equations (1) and (2), we have
      PQ || SR and PQ = SR                 ... (3)
       PQ = SR
(iii)  From equations (3), we have
       PQ || SR and PQ = SR          
       Clearly one pair of opposite sides of quadrilateral PQRS is parallel and equal
       Hence, PQRS is a parallelogram.

                                         

 

In ABC, P and Q are mid points of sides AB and BC respectively.
     PQ || AC and PQ =  AC         (using mid-point theorem)    ... (1)

 
    In ADC
    R and S are the mid points of CD and AD respectively
     RS || AC and RS =  AC         (using mid-point theorem)    ... (2)

    From equations (1) and (2), we have
     PQ || RS and PQ = RS

    As in quadrilateral PQRS one pair of opposite sides are equal and parallel to each other, so, it is a parallelogram.
    Let diagonals of rhombus ABCD intersect each other at point O.
    Now in quadrilateral OMQN
    MQ || ON                (  PQ || AC)
    QN || OM                 ( QR || BD)
    So, OMQN is parallelogram
    MQN = NOM
    PQR = NOM

    But, NOM = 90^o     (diagonals of a rhombus are perpendicular to each other)

    PQR = 90^o

    Clearly PQRS is a parallelogram having one of its interior angle as 90.
    Hence, PQRS is rectangle.

                                           

 

    Let us join AC and BD
    In ABC
    P and Q are the mid-points of AB and BC respectively
    PQ || AC and PQ =  AC         (mid point theorem)        ... (1)

    Similarly in ADC
    SR || AC and SR =   AC         (mid point theorem)    ...  ... (2)
    Clearly, PQ || SR and PQ = SR
    As in quadrilateral PQRS one pair of opposite sides is equal and parallel to
    each other, so, it is a parallelogram.
     PS || QR and PS = QR        (opposite sides of parallelogram)... (3)

    Now, in BCD, Q and R are mid points of side BC and CD respectively.
     QR || BD and QR = BD         (mid point theorem)        ... (4)

    But diagonals of a rectangle are equal
     AC = BD                   ... ...  (5)

    Now, by using equation (1), (2), (3), (4), (5) we can say that
    PQ = QR = SR = PS  
    So, PQRS is a rhombus.

                                          

 

 

By converse of mid-point theorem a line drawn, through the mid point of any side of a triangle and parallel to another side bisects the third side.
Now in ABD
EF || AB and E is mid-point of AD
So, this line will intersect BD at point G and G will be the mid-point of DB.
Now as EF || AB and AB || CD
EF || CD    (Two lines parallel to a same line are parallel to each other)

Now, in BCD, GF || CD and G is the midpoint of line BD. So, by using converse of mid-point theorem, F is the mid-point of BC.

ABCD is a parallelogram
 AB || CD

So, AE || FC
Again AB = CD         (opposite sides of parallelogram ABCD)
     AB = CD
          AE = FC             (E and F are midpoints of side AB and CD)
As in quadrilateral AECF one pair of opposite sides (AE and CF) are parallel and equal to each other. So, AECF is a parallelogram.
AF || EC            (Opposite sides of a parallelogram)
Now, in DQC, F is mid point of side DC and FP || CQ (as AF || EC). So, by using converse of mid-point theorem, we can say that
P is the mid-point of DQ
DP = PQ            ... (1)
Similarly, in APB, E is mid point of side AB and EQ || AP (as AF || EC). So, by using converse of mid-point theorem, we can say that
Q is the mid-point of PB
PQ = QB            ... (2)
From equations (1) and (2), we may say that
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.

                                            

 

(i)     In ABC
        Given that M is mid point of AB and MD || BC.
        So, D is the mid-point of AC.        (Converse of mid-point     theorem)

(ii)    As DM || CB and AC is a transversal line for them.
        So, MDC + DCB = 180         (Co-interior angles)  
        MDC + 90 = 180
        MDC = 90
         MD  AC

 

(iii)   Join MC

 

                             

 

     In AMD and CMD
     AD = CD                         (D is the midpoint of side AC)
     ADM = CDM             (Each 90)
     DM = DM                         (common)
     AMDCMD        (by SAS congruence rule)

     So, AM = CM                   (by CPCT)
     But AM =  AB              (M is mid point of AB)
     So, CM = MA =  AB