NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

Let the common ratio between the angles is x. So, the angles will be 3x, 5x, 9x and 13x respectively.
Since the sum of all interior angles of a quadrilateral is 360.
 3x + 5x + 9x + 13x = 360

30x = 360
    x = 12
Hence, the angles are
3x = 3  12 = 36
5x = 5  12 = 60
9x = 9  12 = 108
13x = 13  12 = 156^o

                                            

 

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 quadrilateral, whose diagonals AC and BD bisect each other at right angle
 i.e. OA = OC, OB = OD and AOB = BOC = COD = AOD = 90
To prove ABCD a rhombus, we need to prove ABCD is a parallelogram and all sides of ABCD are equal.
Now, in AOD and COD
OA = OC                 (Diagonal bisects each other)
AOD = COD            (given)
OD = OD                 (common)

AOD COD             (by SAS congruence rule)

AD = CD                 (1)

Similarly we can prove that
AD = AB and CD = BC        (2)
From equations (1) and (2), we can say that
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, so, we can say that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, so, we can say that ABCD is a rhombus.

                                         

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.  

                                           

 

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O.
Given that the diagonals of ABCD are equal and bisect each other at right angles.

So, AC = BD, OA = OC, OB = OD     and

    AOB = BOC = COD = AOD = 90.
To prove ABCD a square, we need to prove ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angle is 90.

Now, in AOB and COD
AO = CO                              (Diagonals bisect each other)
OB = OD                              (Diagonals bisect each other)
AOB = COD                   (Vertically opposite angles)
AOB  COD           (SAS congruence rule)
 AB = CD                           (by CPCT)        ... (1)

And OAB = OCD             (by CPCT)
But these are alternate interior angles for line AB and CD and alternate interior angle are equal to each other only when the two lines are parallel
 AB || CD                              ... (2)

From equations (1) and (2), we have
ABCD is a parallelogram
Now, in AOD and COD
AO = CO                               (Diagonals bisect each other)
AOD = COD                    (Given that each is 90)
OD = OD                               (common)
AOD  COD                         (SAS congruence rule)
 AD = DC                            ... (3)

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

So, all the sides quadrilateral ABCD are equal to each other

Now, in ADC and BCD
AD = BC                                 (Already proved)
AC = BD                                 (given)
DC = CD                                (Common)
ADC BCD              (SSS Congruence rule)
ADC = BCD                 (by CPCT)

But ADC + BCD = 180^o     (co-interior angles)
ADC + ADC = 180^o
ADC = 180^o
ADC = 90^o
One of interior angle of ABCD quadrilateral is a right angle

Now, we have ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angle is 90. Therefore, ABCD is a square.

 

(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

                                          

 

Let us join AC
In ABC
BC = AB        (side of a rhombus are equal to each other)
1 = 2         (angles opposite to equal sides of a triangle are equal)

But 1 = 3         (alternate interior angles for parallel lines AB and CD)
2 = 3
So, AC bisects C.
Also, 2 = 4    (alternate interior angles for || lines BC and DA)
1 = 4
So, AC bisects A
Similarly, we can prove that BD bisects B and D as well.

                            

 

(i) Given that AC is bisector of A and C.
    Or DAC = DCA         
    CD = DA                             (sides opposite to equal angles are also equal)
    But DA = BC and AB = CD     (opposite sides of rectangle are equal)
     AB = BC = CD = DA

 

     ABCD is a rectangle and all of its sides are equal.
     Hence, ABCD is a square

 

(ii) Let us join BD
     In BCD
     BC = CD                               (side of a square are equal to each other)
     CDB = CBD                    (angles opposite to equal sides are equal)
     But CDB = ABD              (alternate interior angles for AB || CD)
      CBD = ABD

      BD bisects  B.
      Also CBD = ADB        (alternate interior angles for BC || AD)
      CDB = ADB
        BD bisects D.

 

(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.

(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)

 

(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)

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.

                                          

Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.
Join PQ, QR, RS, SP and BD.
In ABD, S and P are mid points of AD and AB respectively.
So, By using mid-point theorem, we can say that
SP || BD and SP =   BD             ... (1)  
Similarly in BCD
QR || BD and QR =  BD               ... (2)
From equations (1) and (2), we have
SP || QR and SP = QR  
As in quadrilateral SPQR one pair of opposite sides are equal and parallel to
each other.
So, SPQR is a parallelogram.

Since, diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

                                            

 

(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