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# Class 9 NCERT Solutions Maths Chapter 8 - Quadrilaterals

### Solution 1

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.

### Solution 2

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.

### Solution 3

(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

### Solution 4

(i)
Given:
DAC = CAB   ... (1)
and DCA = BCA   ... (2)
Now, AD || BC and AC is a transversal

DAC = BCA
CAB = BCA
In ΔABC, we have

CAB = BCA
BC = AB  ... (Sides opposite to equal angles are equal)
But BC = AD and AB = CD ... Since ABCD is a rectangle.

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.

### Solution 5

(i)  In APD and CQB
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.

### Solution 6

(i)  In APB and CQD
APB = CQD         (each 90o)
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)

### Solution 7

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)
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

AB = BA                 (common side)
B = A                 (proved before)

AC = BD                 (by CPCT)

### Solution 1

(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.

### Solution 2

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

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.
MQ || ON                (  PQ || AC)
QN || OM                 ( QR || BD)
So, OMQN is parallelogram
MQN = NOM
PQR = NOM

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

PQR = 90o

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

### Solution 3

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)

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.

### Solution 4

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.

### Solution 5

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.

### Solution 6

(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)