**NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1****NCERT Solutions for Class 9 Maths Chapter 8****Quadrilaterals**Ex 8.2

**NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1**

**Ex 8.1 Class 9 Maths**** ****Question 1.****
**If the diagonals of a parallelogram
are equal, then show that it is a rectangle.

**Solution:****
**Let ABCD is a parallelogram such
that AC = BD.

AC = DB [Given]

AB = DC [Opposite sides of a parallelogram]

BC = CB [Common]

∴ ∆ABC ≅ ∆DCB [By SSS congruency rule]

⇒ ∠ABC = ∠DCB [By CPCT] …(1)

Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]

From (1) and (2), we get

∠ABC = ∠DCB = 90°

i.e., ABCD is a parallelogram having an angle equal to 90°.

∴ ABCD is a rectangle.

**Ex 8.1 Class 9 Maths**** ****Question 2.****
**Show that the diagonals of a square
are equal and bisect each other at right angles.

**Solution:****
**Let ABCD be a square such that its
diagonals AC and BD intersect at O.

(i) To prove that the diagonals of a square are
equal, we need to prove AC = BD.

In ∆ABC and ∆BAD, we have

AB = BA
[Common]

BC = AD [Sides of a
square ABCD]

∠ABC = ∠BAD [Each angle is 90°]

∴ ∆ABC ≅ ∆BAD [By SAS congruency rule]

AC = BD [By CPCT] …..…(1)

(ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram]

∴ ∠DAC = ∠ACB
[Alternate interior angles are equal]

Similarly, ∠ADB = ∠DBC

Now, in ∆OAD and ∆OCB, we have

AD = CB [Sides
of a square ABCD]

∠DAO = ∠OCB
[Proved above]

∠ADO = ∠OBC
[Proved above]

∴ ∆OAD ≅ ∆OCB [By
ASA congruency rule]

⇒ OA = OC and OD = OB [By CPCT]

i.e., the diagonals AC and BD bisect each other at O. …….(2)

(iii)
In ∆OBA and ∆ODA, we have

OB = OD [Proved above]

BA = DA [Sides of
a square ABCD]

OA = OA [Common]

∴ ∆OBA ≅ ∆ODA [By SSS congruency rule]

⇒ ∠AOB = ∠AOD [By CPCT] ….…(3)

∵ ∠AOB and ∠AOD form a linear pair.

∴ ∠AOB + ∠AOD = 180°

∴ ∠AOB = ∠AOD = 90° [By(3)]

⇒ AC ⊥ BD ……(4)

From (1), (2) and (4), we get diagonals of a
square are equal and bisect each other at right angles.

**Ex 8.1 Class 9 Maths**** ****Question 3.**

Diagonal AC of a parallelogram ABCD
bisects ∠A (see figure). Show that

**(i)**
it bisects ∠C also,

**(ii)**
ABCD is a rhombus.

**Solution:**

We have a parallelogram ABCD in which diagonal AC bisects ∠A.

⇒ ∠DAC = ∠BAC

**(i)**Since, ABCD is a parallelogram.

∴ AB || DC and AC is a transversal.

∴ ∠1 = ∠3 …(1) [ ∵ Alternate interior angles are equal]

Also, BC || AD and AC is a transversal.

∴ ∠2 = ∠4 …(2) [ ∵ Alternate interior angles are equal]

Also, ∠1 = ∠2 …(3) [ ∵ AC bisects ∠A]

From (1), (2) and (3), we have

∠3 = ∠4

⇒ AC bisects ∠C.

**(ii) **In
∆ABC, we have

∠1 = ∠4 [From (2) and (3)]

⇒ BC = AB …(4) [ ∵ Sides opposite to equal angles of a ∆ are equal]

Similarly, AD = DC ……..(5)

But, ABCD is a parallelogram. [Given]

∴ AB = DC …..…(6)

From (4), (5) and (6), we have

AB = BC = CD = DA

Hence, ABCD is a rhombus.

**Ex 8.1 Class 9 Maths**** ****Question 4.
**ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that

**(i)**ABCD is a square

**(ii)**diagonal BD bisects ∠B as well as ∠D.

**Solution:
**We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.

i.e., ∠1 = ∠4 and ∠2 = ∠3 ……..(1)

**(i)**
Since every rectangle is a parallelogram.

∴ ABCD is a parallelogram.

⇒ AB || DC and AC is a transversal.

∴ ∠2 = ∠4 …..…(2) [ ∵ Alternate interior
angles are equal]

From (1) and (2), we have

∠3 = ∠4

In ∆ABC, ∠3 = ∠4

⇒ AB = BC [ ∵ Sides opposite to
equal angles of a A are equal]

Similarly, CD = DA

So, ABCD is a rectangle having adjacent sides equal.

⇒ ABCD is a square.

**(ii)**
Since ABCD is a square and diagonals of a square bisect the opposite angles.

So, BD bisects ∠B as well as ∠D.

**Ex 8.1 Class 9 Maths**** ****Question 5.
**In parallelogram ABCD, two points P and Q are taken
on diagonal BD such that DP = BQ (see figure). Show that

**(i)**
Î”APD ≅ Î”CQB

**(ii)**
AP = CQ

**(iii)**
Î”AQB ≅ Î”CPD

**(iv)** AQ = CP

**(v) **APCQ is a parallelogram.

**Solution:**

We have a parallelogram ABCD, BD is the diagonal and points P and Q are such that PD = QB

**(i)**
Since AD || BC and BD is a transversal.

∴ ∠ADB = ∠CBD [ ∵ Alternate interior
angles are equal]

⇒ ∠ADP = ∠CBQ

Now, in ∆APD and ∆CQB, we have

AD = CB [Opposite
sides of a parallelogram ABCD are equal]

PD = QB [Given]

∠ADP = ∠CBQ [Proved
above]

∴ ∆APD ≅ ∆CQB [By
SAS congruency rule]

**(ii)**
Since ∆APD ≅ ∆CQB [Proved above]

⇒ AP = CQ [By CPCT]

**(iii)**
Since AB || CD and BD is a transversal.

∴ ∠ABD = ∠CDB

⇒ ∠ABQ = ∠CDP

Now, in ∆AQB and ∆CPD, we have

QB = PD [Given]

∠ABQ = ∠CDP [Proved above]

AB = CD [Opposite
sides of a parallelogram ABCD are equal]

∴ ∆AQB ≅ ∆CPD [By
SAS congruency rule]

**(iv) **Since
∆AQB ≅ ∆CPD [Proved above]

⇒ AQ = CP
[By CPCT]

**(v)**
In a quadrilateral APCQ,

Opposite sides are equal. [Proved
above]

∴ APCQ is a parallelogram.

**Ex 8.1 Class 9 Maths**** ****Question 6.
**ABCD is a parallelogram and AP and CQ are
perpendiculars from vertices A and C on diagonal BD (see figure). Show that

**(i)**
∆APB ≅ ∆CQD

**(ii)**
AP = CQ

**Solution:
(i)** In ∆APB and ∆CQD, we have

∠APB = ∠CQD [Each 90°]

AB = CD [ ∵ Opposite sides of a parallelogram ABCD are equal]

∠ABP = ∠CDQ

[ ∵ Alternate angles are equal as AB || CD and BD is a transversal]

∴ ∆APB ≅ ∆CQD [By AAS congruency rule]

**(ii)**
Since ∆APB ≅ ∆CQD [Proved
above]

⇒ AP = CQ [By CPCT]

**Ex 8.1 Class 9 Maths**** ****Question 7.
**ABCD is a trapezium in which AB || CD and AD = BC
(see figure). Show that

**(i)**∠A = ∠B

**(ii)**∠C = ∠D

**(iii)**∆ABC ≅ ∆BAD

**(iv)**
diagonal AC = diagonal BD

[**Hint:** Extend AB and draw a line through C parallel to DA intersecting
AB produced at E].

**Solution:**

We have given a trapezium ABCD in which AB || CD and
AD = BC.

**(i)**
Produce AB to E and draw CE || AD

⇒ AE || DC Also AD || CE

∴ AECD is a parallelogram.

⇒ AD = CE …..…(1)

[ ∵ Opposite sides of the parallelogram are equal]

But AD = BC …..…(2) [Given]

From (1) and (2), we get BC = CE

Now, in ∆BCE, we have BC = CE

⇒ ∠CEB = ∠CBE ….…(3)

[∵ Angles opposite to equal sides of a triangle are equal]

Also, ∠ABC + ∠CBE = 180° …..…(4) [Linear pair]

and ∠A + ∠CEB = 180° …..…(5) [Co-interior angles of a
parallelogram ADCE]

From (4) and (5), we get

∠ABC + ∠CBE = ∠A + ∠CEB

⇒ ∠ABC = ∠A [From
(3)]

⇒ ∠B = ∠A …..…(6)

**(ii)**
AB || CD and AD is a transversal.

∴ ∠A + ∠D = 180° ….…(7)
[Co-interior angles]

Similarly, ∠B + ∠C = 180° ….…(8)

From (7) and (8), we get

∠A + ∠D = ∠B + ∠C

⇒ ∠C = ∠D [From (6)]

**(iii)**
In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Given]

∠ABC = ∠BAD [Proved
above]

∴ ∆ABC ≅ ∆BAD [By
SAS congruency rule]

**(iv)**
Since ∆ABC ≅ ∆BAD [Proved above]

⇒ AC = BD [By CPCT]

**Related Links:**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11**