NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals
Learn to provide Mathematical proofs easily with TopperLearning’s NCERT Solutions for CBSE Class 9 Mathematics Chapter 8 Quadrilaterals. Can a parallelogram be a rhombus? How? Learn by practising with our solutions by Maths experts. Understand how to express your understanding of quadrilaterals, such as a parallelogram and rhombus, and provide clear explanations in your answers.
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Chapter 8 - Quadrilaterals Exercise Ex. 8.1
Given that the diagonals of ABCD are equal and bisect each other at right angles.




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


AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)




And


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
From equations (1) and (2), we have
ABCD is a parallelogram
Now, in


AO = CO (Diagonals bisect each other)


OD = OD (common)



But, AD = BC and AB = CD (opposite sides of parallelogram ABCD)
So, all the sides quadrilateral ABCD are equal to each other
Now, in


AD = BC (Already proved)
AC = BD (given)
DC = CD (Common)










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.
Now, AECD is a parallelogram.
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)


Since the sum of all interior angles of a quadrilateral is 360.
30x = 360
x = 12
Hence, the angles are
3x = 3

5x = 5

9x = 9

13x = 13

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.
i.e. OA = OC, OB = OD and




To prove ABCD a rhombus, we need to prove ABCD is a parallelogram and all sides of ABCD are equal.
Now, in


OA = OC (Diagonal bisects each other)


OD = OD (common)


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.
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)


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.




But it is given that AC bisects









Hence, AC bisects

(ii)


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

BC = AB (side of a rhombus are equal to each other)


But




So, AC bisects

Also,




So, AC bisects

Similarly, we can prove that BD bisects B and D as well.


Or


CD = DA (sides opposite to equal angles are also equal)
But DA = BC and AB = CD (opposite sides of rectangle are equal)
Hence, ABCD is a square
In

BC = CD (side of a square are equal to each other)


But






Also








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


(ii) As we had observed that






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




AB = CD (opposite sides of parallelogram ABCD)



(ii) By using the result obtained as above


AP = CQ (by CPCT)
Now, if two opposite sides of a quadrilateral are equal and parallel to each other, it will be
Therefore, quadrilateral ABED is a parallelogram.
(ii) Again BC = EF and BC || EF.
Therefore, quadrilateral BEFC is a parallelogram.
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)
(iv) Here one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and
so, it is a parallelogram.
(v) As ACFD is a parallelogram, so, pair of opposite sides will be equal and parallel to each
(vi)


AB = DE (given)
BC = EF (given)
AC = DF (ACFD is a parallelogram)


Chapter 8 - Quadrilaterals Exercise Ex. 8.2


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.

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,







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


DM = DM (common)


But AM =

So, CM = MA =


In a triangle the line segment joining the mid points of any two sides of the triangle is

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

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.
Join PQ, QR, RS, SP and BD.
In

So, By using mid-point theorem, we can say that
SP || BD and SP =

Similarly in

QR || BD and QR =

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.
Hence, PR and QS bisect each other.
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