Class 9 NCERT Solutions Maths Chapter 8: Quadrilaterals
Ex. 8.1
Ex. 8.2
Quadrilaterals Exercise Ex. 8.1
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)
In


AB = DC (opposite sides of a parallelogram are equal)
BC = BC (common)
AC = DB (given)




We know that sum of measures of angles on the same side of transversal is 180º.






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)
ABC
DCB (by SAS congruency)
AC = DB (by CPCT)
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

Now, in


AB = DC (sides of square are equal to each other)


BC = BC (common side)


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)
Now in






AB = CD (sides of square are always equal)


Hence, the diagonals of a square bisect each other
Now in


Now as we had proved that diagonals bisect each other
So, AO = CO
AB = CB (sides of square are equal)
BO = BO (common)




But,


2


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)


But it is given that AC bisects



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






Hence, AC bisects

(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)
Hence, ABCD is rhombus
Solution 4
(i)
Given:


and
DCA =
BCA ... (2)


Now, AD || BC and AC is a transversal






In ΔABC, we have



But BC = AD and AB = CD ... Since ABCD is a rectangle.

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.


Solution 5
(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)




AD = CB (opposite sides of parallelogram ABCD)
DP = BQ (given)


(ii) As we had observed that


(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)




AB = CD (opposite sides of parallelogram ABCD)
BQ = DP (given)


(iv) As we had observed that


(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)




AB = CD (opposite sides of parallelogram ABCD)



(ii) By using the result obtained as above


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.
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 relation
CEB =
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 AD = BC (given)
So, BC = CE


Now consider parallel lines AD and CE. AE is transversal line for them






But


From equations (1) and (2), we have


(ii) AB || CD


Also






But




(iii) In ABC and BAD
AB = BA (common side)
BC = AD (given)




(iv)


Quadrilaterals Exercise Ex. 8.2
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

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

(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.
PQ || AC and PQ =

Now using equations (1) and (2), we have
PQ || SR and PQ = SR ... (3)

(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)


In

R and S are the mid points of CD and AD respectively

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 (
QN || OM (
So, OMQN is parallelogram




But,


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)
In

P and Q are the mid-points of AB and BC respectively

Similarly in

SR || AC and SR =

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.
Now, in


But diagonals of a rectangle are equal
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

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
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)


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.

Now, in

P is the mid-point of DQ

Similarly, in

Q is the mid-point of PB

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

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,





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


AD = CD (D is the midpoint of side AC)


DM = DM (common)


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

So, CM = MA =
