NCERT Solutions for Exercise 8.1 Class 9 Maths Chapter 8 - Quadrilaterals

Q1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:
Given: ABCD is a parallelogram with AC = BD.
To prove: ABCD is a rectangle.
Proof: In $\mathrm{△}$ ABC and $\mathrm{△}$ BAD,

BC = AD (Opposite sides of a parallelogram)
AC = BD (Given)
AB = AB (Common)
$\mathrm{△}$ ABC $\cong$ $\mathrm{△}$ BAD (By SSS)
$\mathrm{\angle }ABC=\mathrm{\angle }BAD$ (CPCT)
And $\mathrm{\angle }ABC+\mathrm{\angle }BAD={180}^{\circ }$ (Co - interior angles)
$2\mathrm{\angle }BAD={180}^{\circ }$
$\mathrm{\angle }BAD={90}^{\circ }$
Hence, it is a rectangle.

Q2. Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:
Given: ABCD is a square, i.e. AB = BC = CD = DA.
To prove: the diagonals of a square are equal and bisect each other at right angles i.e. AC = BD, AO = CO, BO = DO and $\mathrm{\angle }COD={90}^{\circ }$
Proof: In $\mathrm{△}$ BAD and $\mathrm{△}$ ABC,
$\mathrm{\angle }BAD=\mathrm{\angle }ABC$ (Each ${90}^{\circ }$)
AD = BC (Given)
AB = AB (Common)
$\mathrm{△}$ BAD $\cong$ $\mathrm{△}$ ABC (By SAS)
BD = AC (CPCT)
In $\mathrm{△}$ AOB and $\mathrm{△}$ COD,
$\mathrm{\angle }$ OAB= $\mathrm{\angle }$ OCD (Alternate angles)
AB = CD (Given)
$\mathrm{\angle }$ OBA= $\mathrm{\angle }$ ODC (Alternate angles)
$\mathrm{△}$ AOB $\cong$ $\mathrm{△}$ COD (By AAS)
AO = OC, BO = OD (CPCT)
In $\mathrm{△}$ AOB and $\mathrm{△}$ AOD,
OB = OD (proved above)
AB = AD (Given)
OA = OA (Common)
$\mathrm{△}$ AOB $\cong$ $\mathrm{△}$ AOD (By SSS)
$\mathrm{\angle }$ AOB = $\mathrm{\angle }$ AOD (CPCT)
$\mathrm{\angle }$ AOB+ $\mathrm{\angle }$ AOD = ${180}^{\circ }$
2$\mathrm{\angle }$ AOB = ${180}^{\circ }$
$\mathrm{\angle }$ AOB = ${90}^{\circ }$
Hence, the diagonals of a square are equal and bisect each other at right angles.

Q3 (i) Diagonal AC of a parallelogram ABCD bisects $\mathrm{\angle }A$ (see Fig. $8.19$ ). Show that it bisects $\mathrm{\angle }C$ also.

Solution:
Given: $\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BAC ..........(1)
$\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BCA.............(2) (Alternate angles)
$\mathrm{\angle }$ BAC= $\mathrm{\angle }$ ACD .............(3) (Alternate angles)
From equations (1), (2) and (3), we get
$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ BCA...............(4)
Hence, diagonal AC bisect angle C also.

Q3 (ii) Diagonal AC of a parallelogram ABCD bisects $\mathrm{\angle }A$ (see Fig. $8.19$ ). Show that ABCD is a rhombus.

Solution:
Given: $\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BAC .............(1)
$\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BCA..............(2) (Alternate angles)
$\mathrm{\angle }$ BAC= $\mathrm{\angle }$ ACD ..............(3) (Alternate angles)
From equations (1), (2), and (3), we get
$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ BCA..............(4)
From 2 and 4, we get
$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ DAC
In $\mathrm{△}$ ADC,
$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ DAC (proved above)
AD = DC (In a triangle, sides opposite to equal angles are equal)
A parallelogram whose adjacent sides are equal is a rhombus.

Thus, ABCD is a rhombus.

Q4 (i) ABCD is a rectangle in which diagonal AC bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }C$ . Show that: ABCD is a square

Solution:

Given: ABCD is a rectangle with AB=CD and BC=AD $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 2 and $\mathrm{\angle }$ 3= $\mathrm{\angle }$ 4.
To prove: ABCD is a square.
Proof : $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 4 ............(1)(Alternate angles)
$\mathrm{\angle }$ 3= $\mathrm{\angle }$ 4 ..............(2)(Given )
From 1 and 2, $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 3.......................(3)
In $\mathrm{△}$ ADC,
$\mathrm{\angle }$ 1= $\mathrm{\angle }$ 3 (From 3 )
DC = AD (In a triangle, sides opposite to equal angles are equal)
A rectangle whose adjacent sides are equal is a square.
Hence, ABCD is a square.

Q4 (ii) ABCD is a rectangle in which diagonal AC bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }C$ . Show that: diagonal BD bisects $\mathrm{\angle }B$ as well as $\mathrm{\angle }D$ .

Solution:

In $\mathrm{△}$ ADB,
AD = AB (ABCD is a square)
$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 7...............(1) (Angles opposite to equal sides are equal)
$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 8...............(2) (Alternate angles)
From (1) and (2), we have
$\mathrm{\angle }$ 7= $\mathrm{\angle }$ 8.............(3)
And $\mathrm{\angle }$ 7 = $\mathrm{\angle }$ 6.................4(Alternate angles)
From (1) and (4), we get
$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 6..............(5)
Hence, from 3 and 5, diagonal BD bisects angles B as well as angle D.

Q5 (i) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.12$ ). Show that: $\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$

Solution:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$.
To prove : $\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$
Proof:
In $\mathrm{\Delta }APDand\mathrm{\Delta }CQB,$
DP=BQ (Given)
$\mathrm{\angle }$ ADP= $\mathrm{\angle }$ CBQ (Alternate angles)
AD = BC (Opposite sides of a parallelogram)
$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$ (By SAS)

Q5 (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.12$ ). Show that: $AP=CQ$

Solution:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$.
To prove: $AP=CQ$
Proof:
In $\mathrm{\Delta }APDand\mathrm{\Delta }CQB,$
DP=BQ (Given)
$\mathrm{\angle }$ ADP= $\mathrm{\angle }$ CBQ (Alternate angles)
AD = BC (Opposite sides of a parallelogram)
$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$ (By SAS)
$AP=CQ$ (CPCT)

Q5 (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.12$ ). Show that: $\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$

Solution:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$.
To prove: $\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$
Proof:
In $\mathrm{\Delta }AQBand\mathrm{\Delta }CPD,$
DP = BQ (Given)
$\mathrm{\angle }$ ABQ= $\mathrm{\angle }$ CDP (Alternate angles)
AB = CD (Opposite sides of a parallelogram)
$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$ (By SAS)

Q5 (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.12$ ). Show that: $AQ=CP$

Solution:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$.
To prove: $AQ=CP$
Proof:
In $\mathrm{\Delta }AQBand\mathrm{\Delta }CPD,$
DP = BQ (Given)
$\mathrm{\angle }$ ABQ= $\mathrm{\angle }$ CDP (Alternate angles)
AB = CD (Opposite sides of a parallelogram)
$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$ (By SAS)
$AQ=CP$ (CPCT)

Q5 (v) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.12$ ). Show that: APCQ is a parallelogram

Solution:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$.
To prove: APCQ is a parallelogram
Proof:
In $\mathrm{\Delta }APDand\mathrm{\Delta }CQB,$
DP = BQ (Given)
$\mathrm{\angle }$ ADP= $\mathrm{\angle }$ CBQ (alternate angles)
AD = BC (Opposite sides of a parallelogram)
$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$ (By SAS)
$AP=CQ$ (CPCT)..............(1)
Also,
In $\mathrm{\Delta }AQBand\mathrm{\Delta }CPD,$
DP = BQ (Given)
$\mathrm{\angle }$ ABQ= $\mathrm{\angle }$ CDP (Alternate angles)
AB = CD (Opposite sides of a parallelogram)
$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$ (By SAS)
$AQ=CP$ (CPCT)................(2)
From equations 1 and 2, we get
$AP=CQ$
$AQ=CP$
Thus, opposite sides of quadrilateral APCQ are equal, so APCQ is a parallelogram.

Q6 (i) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. $8.13$ ). Show that $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$

Solution:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove: $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$
Proof: In $\mathrm{\Delta }APBand\mathrm{\Delta }CQD$,
$\mathrm{\angle }$ APB= $\mathrm{\angle }$ CQD (Each ${90}^{\circ }$)
$\mathrm{\angle }$ ABP= $\mathrm{\angle }$ CDQ (Alternate angles)
AB = CD (Opposite sides of a parallelogram)
Thus, $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$ (By SAS)

Q6 (ii) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. $8.13$ ). Show that $AP=CQ$

Solution:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove: $AP=CQ$
Proof: In $\mathrm{\Delta }APBand\mathrm{\Delta }CQD$,
$\mathrm{\angle }$ APB= $\mathrm{\angle }$ CQD (Each ${90}^{\circ }$)
$\mathrm{\angle }$ ABP= $\mathrm{\angle }$ CDQ (Alternate angles)
AB = CD (Opposite sides of a parallelogram )
Thus, $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$ (By SAS)
$AP=CQ$ (CPCT)

Q7 (i) ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see Fig. $8.14$ ). Show that $\mathrm{\angle }A=\mathrm{\angle }B$
[ Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Solution:
Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$
To prove: $\mathrm{\angle }A=\mathrm{\angle }B$
Proof: Let $\mathrm{\angle }$ A be $\mathrm{\angle }$ 1, $\mathrm{\angle }$ ABC be $\mathrm{\angle }$ 2, $\mathrm{\angle }$ EBC be $\mathrm{\angle }$ 3, $\mathrm{\angle }$ BEC be $\mathrm{\angle }$ 4.
In AECD,
AE || DC (Given)
AD || CE (By construction)
Hence, AECD is a parallelogram.
AD = CE.........(1) (Opposite sides of a parallelogram)
AD = BC...........(2) (Given)
From (1) and (2), we get
CE = BC
In $\mathrm{△}$ BCE,
$\mathrm{\angle }3=\mathrm{\angle }4$ ...............(3) (Opposite angles of equal sides)
$\mathrm{\angle }2+\mathrm{\angle }3={180}^{\circ }$ ..............(4) (Linear pairs)
$\mathrm{\angle }1+\mathrm{\angle }4={180}^{\circ }$ ...............(5) (Co-interior angles)
From (4) and (5), we get
$\mathrm{\angle }2+\mathrm{\angle }3=\mathrm{\angle }1+\mathrm{\angle }4$
$\therefore \mathrm{\angle }2=\mathrm{\angle }1\to \mathrm{\angle }B=\mathrm{\angle }A$ (Since, $\mathrm{\angle }3=\mathrm{\angle }4$ )

Q7 (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that $\mathrm{\angle }C=\mathrm{\angle }D$ [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Solution:
Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$
To prove: $\mathrm{\angle }C=\mathrm{\angle }D$
Proof: Let $\mathrm{\angle }$ A be $\mathrm{\angle }$ 1, $\mathrm{\angle }$ ABC be $\mathrm{\angle }$ 2, $\mathrm{\angle }$ EBC be $\mathrm{\angle }$ 3, $\mathrm{\angle }$ BEC be $\mathrm{\angle }$ 4.
$\mathrm{\angle }1+\mathrm{\angle }D={180}^{\circ }$ (Co-interior angles)
$\mathrm{\angle }2+\mathrm{\angle }C={180}^{\circ }$ (Co-interior angles)
$\therefore \mathrm{\angle }1+\mathrm{\angle }D=\mathrm{\angle }2+\mathrm{\angle }C$
Thus, $\mathrm{\angle }C=\mathrm{\angle }D$ (Since , $\mathrm{\angle }1=\mathrm{\angle }2$ )

Q7 (iii) ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see Fig. $8.13$ ). Show that $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Solution:
Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$
To prove: $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$
Proof: In $\mathrm{\Delta }ABCand\mathrm{\Delta }BAD$,
BC = AD (Given)
AB = AB (Common)
$\mathrm{\angle }ABC=\mathrm{\angle }BAD$ (Proved in (i))
Thus, $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$ (By SAS rule)

Q7 (iv) ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see Fig. $8.13$ ). Show that diagonal AC $=$ diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Solution:
Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$
To prove: diagonal AC $=$ diagonal BD
Proof: In $\mathrm{\Delta }ABCand\mathrm{\Delta }BAD$,
BC = AD (Given)
AB = AB (Common)
$\mathrm{\angle }ABC=\mathrm{\angle }BAD$ (Proved in (i))
Thus, $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$ (By SAS rule)
diagonal AC $=$ diagonal BD (CPCT)