NCERT Solutions Class 9 – Chapter 8 Quadrilaterals – Exercise 8.1

Question 1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:

As mentioned parallelogram, so let PQRS be a parallelogram

where, given 

PR = QS

In ∆PQR and ∆QRS,

PR = QS      …………………[Given]

PQ = RS      …………………[Opposite sides of a parallelogram]

QR = RQ     …………………[Common side]

∴ ∆PQR ≅ ∆QRS           [By SSS congruency]

so, ∠PQR = ∠QRS [By C.P.C.T.] ………………………………………….(1)

Now, PQ || RS and QR is a transversal. …………………….[PQRS is a parallelogram]

∴ ∠PQR + ∠QRS = 180°  [Co-interior angles of parallelogram]…………………………………… (2)

From (1) and (2), we have

∠PQR = ∠QRS = 90°

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

Hence, PQRS is a rectangle. (having all angles equal to 90° and opposite sides are equal)

Question 2. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:

As mentioned quadrilateral, so let PQRS be a quadrilateral

where, given

PO = RO

SO = QO

In ∆POQ and ∆POS, we have

PO = PO [Common]

OQ = OS [O is the mid-point of QS]

∠POQ= ∠POS [Each 90°]

∴ ∆POQ ≅ ∆POS [By, SAS congruency]

∴ PQ = PS [By C.P.C.T.] ……..    (1)

Similarly, PQ = QR …………………..(2)

QR = RS ……………………………………..(3)

RS = SP ………………………………………(4)

∴ From (1), (2), (3) and (4), we have

PQ = QR = RS = SP

Thus, the quadrilateral PQRS is a rhombus.

Alternative Solution:

So as it is given that diagonals of a quadrilateral PQRS bisect each other

According to Theorem 8.7 NCERT it is a parallelogram

PQRS can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.

 In ∆POQ and ∆POS, we have

PO = PO [Common]

OQ = OS [O is the mid-point of QS]

∠POQ= ∠POS [Each 90°]

∴ ∆POQ ≅ ∆POS [By,SAS congruency]

∴ PQ = PS [By C.P.C.T.] ……..    (1)

Similarly, PQ = QR …………………..(2)

QR = RS ……………………………………..(3)

RS = SP ………………………………………(4)

From (1), (2), (3) and (4), we have

PQ = QR = RS = SP

Hence, as a parallelogram has all sides equal then it is called a rhombus. 

Question 3. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Solution:

(i) As, ABCD is a parallelogram.

∠BAC = ∠DCA ……………………………(1) [Alternate interior angles are equal]

∠CAD = ∠BCA ………………………….(2) [Alternate interior angles are equal]

Also, ∠CAD = ∠CAB …………………….(3) [ (Given) as AC bisects ∠A]

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

∠DCA = ∠BCA

Hence, AC bisects ∠C.

(ii) In ∆ABC, 

∠BAC = ∠DCA ………………………….. [Alternate interior angles are equal]

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) (opposite sides of parallelogram)

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

AB = BC = CD = DA

As, ABCD is a parallelogram having all sides equal then it is a rhombus.

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:

There is rectangle ABCD such that AC bisects ∠A as well as ∠C, so

∠BAC = ∠DAC and,

∠DCA = ∠BCA ………………………………(1)

(i) As we know that every rectangle is a parallelogram.

 ABCD is a parallelogram.

∠BCA = ∠DAC …………………….(2) [ Alternate interior angles are equal]

From (1) and (2), we have

∠DCA= ∠DAC……………………….(3)

In ∆ABC, ∠DCA= ∠DAC then, 

 CD = DA [Sides opposite to equal angles of a ∆ are equal]

Similarly, AB = BC

So, ABCD is a rectangle having adjacent sides equal.

ABCD is a square.

(ii) Since, ABCD is a square 

AB = BC = CD = DA

so, In ∆ABD, as AB = AD 

 ∠ABD = ∠ADB [Angles opposite to equal sides of a ∆ are equal]……………………..(1)

Similarly, ∠CBD = ∠CDB…………………..(2)

∠CBD = ∠ADB               [Alternate interior angles are equal]………………(3)

From (1) and (3)

∠CBD = ∠ABD

From (2) and (3)

∠ADB = ∠CBD

So, BD bisects ∠B as well as ∠D.

Question 5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that :

(i) ∆APD ≅ ∆CQB

(ii) AP = CQ

(iii) ∆AQB ≅ ∆CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Solution:

ABCD is a parallelogram

DP = BQ

(i) As ABCD is a parallelogram

∠ADB = ∠CBD [Alternate interior angles are equal]……………….(1)

∠ABD = ∠CDB   [Alternate interior angles are equal]…………………(2)

Now, in ∆APD and ∆CQB, we have

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

PD = QB [Given]

∠ADP = ∠CBQ [Proved]

Hence, ∆APD ≅ ∆CQB [By SAS congruency]

(ii) As, ∆APD ≅ ∆CQB [Proved]

AP = CQ [By C.P.C.T.]…………………(3)

(iii) Now, in ∆AQB and ∆CPD, we have

QB = PD [Given]

∠ABQ = ∠CDP [Proved]

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

Hence, ∆AQB ≅ ∆CPD [By SAS congruency]

(iv) As, ∆AQB ≅ ∆CPD [Proved]

 AQ = CP [By C.P.C.T.] …………………………..(4)

(v) In a quadrilateral APCQ,

Opposite sides are equal. [From (3) and (4)]

Hence, APCQ is a parallelogram. (NCERT Theorem 8.3)

Question 6. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that

(i) ∆APB ≅ ∆CQD

(ii) AP = CQ

Solution:

ABCD is a parallelogram

DP = BQ

(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]

Hence, ∆APB  ≅  ∆CQD [By AAS congruency]

(ii) As, ∆APB ≅ ∆CQD [Proved]

AP = CQ [By C.P.C.T.]

Question 7. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). 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:

so basically here is a trapezium ABCD in which AB || CD and AD = BC.

Extended AB and draw a line through C parallel to DA intersecting AB produced at E

(i) AB || DC , AE || DC Also AD || CE

then, AECD is a parallelogram.

AD = CE …………………………(1) [Opposite sides of the parallelogram are equal]

But AD = BC …(2) [Given]

From (1) and (2), 

BC = CE

Now, in ∆BCF, we have BC = CF

∠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 in parallelogram]

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]

Hence, ∆ABC ≅ ∆BAD [By SAS congruency]

(iv) Since, ∆ABC ≅ ∆BAD [Proved]

AC = BD [By C.P.C.T.]