# NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Written by Team Trustudies
Updated at 2021-02-14

## NCERT solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1

Q1 ) The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: The ratios of the angles of quadrilateral are 3: 5: 9: 13.

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.

We know that, sum of angles of a quadrilateral = ${360}^{?}$
$?$ 3x + 5x + 9x + 13x = ${360}^{?}$
$?$ 30x = ${360}^{?}$
$?$ x = ${12}^{?}$

Thus, we get,
=> 3x = 3 × ${12}^{?}$ = ${36}^{?}$
=>5x = 5 × ${12}^{?}$ = ${60}^{?}$
=> 9x = 9 × ${12}^{?}$ = ${108}^{?}$
=> 13x = 13 × ${12}^{?}$ = ${156}^{?}$

Q2 ) If the diagonals of a parallelogram are equal, then show that it is a rectangle.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: Parallelogram is ABCD whose diagonals AC and BD are equal.

$?$ AC = BD

To prove: ABCD is a rectangle.

Proof:
In $\mathrm{?}ABC$ and $\mathrm{?}DCB$, we have,
AC = BD ...(Given)
AB = CD
(Opposite sides of parallelogram)
BC = CB ...(Common sides)
$?$ $\mathrm{?}ABC$$?$$\mathrm{?}DCB$
(By SSS rule)
$?$ $\mathrm{?}ABC$ = $\mathrm{?}DCB$ ...(i)(By CPCT)

But from figure, DC || AB and transversal CB intersects them

$\mathrm{?}ABC$ + $\mathrm{?}DCB$ = ${180}^{?}$
(interior angles on same side of transversal)
$\mathrm{?}ABC$ + $\mathrm{?}ABC$ = ${180}^{?}$
(from(i))
$?$ 2 $\mathrm{?}ABC$ = ${180}^{?}$
$?$ $\mathrm{?}ABC$ = ${90}^{?}$
Also, $\mathrm{?}DCB$ = ${90}^{?}$

Thus, we can say that, ABCD is a parallelogram and some angles are ${90}^{?}$.

Hence it is proved that, ABCD is a rectangle.

Q3 ) Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.

$?$ OA = OC and OB = OD
And also,
$\mathrm{?}AOD=\mathrm{?}AOB=\mathrm{?}COD=\mathrm{?}BOC={90}^{?}$.

To prove: ABCD is a rhombus.

Proof:
In $\mathrm{?}OAB$ and $\mathrm{?}ODC$, we have,

OA = OC and OB = OD ...(given)
$\mathrm{?}AOB$ = $\mathrm{?}COD$
(vertically opposite angles)
$?$ $\mathrm{?}OAB$ $?$ $\mathrm{?}OCD$
(By SAS rule)
$?$ AB = CD ...(i)(By CPCT)

Similarly, in $\mathrm{?}OAD$ and $\mathrm{?}OBC$, we have,
OA = OC and OD = OB ...(given)
$\mathrm{?}AOD$ = $\mathrm{?}BOC$
(vertically opposite angles)
$?$ $\mathrm{?}OAD$ $?$ $\mathrm{?}OCB$ ...(By SAS rule)
$?$ AD = BC ...(ii)(By CPCT)
Similarly, we can prove that,
AB = CD and
CD = BC ...(iii)

Hence, from (i), (ii) and (iii), we get,
AB = BC = AD = CD
Hence, it is proved that ABCD is a rhombus.

Q4 ) Show that the diagonals of a square are equal and bisect each other at right angles.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: A square ABCD whose diagonals AC and BD intersect at O.

To prove: Diagonals are equal and bisect each other at right angles.

i.e, AC = BD, OD = OB, OA = OC and $AC?BD$

Proof: In $\mathrm{?}ABC$ and $\mathrm{?}BAD$, we have,
AB = BA ...(Common side)
$\mathrm{?}ABC$ = $\mathrm{?}BAD$ = ${90}^{?}$
$?$ $\mathrm{?}ABC$ $?$ $\mathrm{?}BAD$ ...(By SAS rule)
$?$ AC = BD ...(By CPCT)

Similarly, in $\mathrm{?}OAB$ and $\mathrm{?}OCD$, we have,
AB = DC ...(given)
$\mathrm{?}OAB$ = $\mathrm{?}DCO$
($?$ AB || CD and transversal AC intersect)
$\mathrm{?}OBA$ = $\mathrm{?}BDC$
($?$ AB || CD and transversal BD intersect)
$?$ $\mathrm{?}OAB$ $?$ $\mathrm{?}OCD$
(By SAS rule)
$?$ OA = OC and OB = OD
(By CPCT)

Now, in $\mathrm{?}AOB$ and $\mathrm{?}AOD$, we have,
OA = OD ...(proved earlier)
(sides of square)
AO = OA
(Commom side)
$?$ $\mathrm{?}AOB$ $?$ $\mathrm{?}AOD$
(By SSS rule)
$?$ $\mathrm{?}AOB$ = $\mathrm{?}AOD$ ...(By CPCT)

$\mathrm{?}AOB$ + $\mathrm{?}AOD$ = ${180}^{?}$
(linear pair axiom)
$\mathrm{?}AOB$ = $\mathrm{?}AOD$ = ${90}^{?}$
$?$ $AO?BD$
$?$ $AC?BD$.
Also, AC = BD, OA = OC, OB = OD and $AC?BD$

Hence, it is proved that diagonals are equal and bisect each other at right angles.

Q5 ) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: A quadrilateral ABCD in which AC = BD and AC $?BD$ such that OA = OC and OB = OD. So, ABCD is a parallelogram.

To prove: ABCD is a square.

Proof: Let AC and BD intersect at a point O.
In $\mathrm{?}ABO$ and $\mathrm{?}ADO$, we have,
BO = OD ...(given)
AO = OA ...(Common side)
$\mathrm{?}AOB$ = $\mathrm{?}AOD$ = ${90}^{?}$ ...(given)
$?$ $\mathrm{?}ABO$ $?$ $\mathrm{?}ADO$
(By SAS rule)
$?$ AB = AD ...(By CPCT)
Also, AB = DC and AD = BC
(opposite sides of parallelogram)
$?$ AB = BC = DC = AD ...(i)

Similarly, in $\mathrm{?}ABC$ and $\mathrm{?}BAD$, we have,
AC = BD ...(given)
AB = BA ...(Common side)
$?$ $\mathrm{?}ABC$ $?$ $\mathrm{?}BAD$
(By SSS rule)
$?$ $\mathrm{?}ABC$ = $\mathrm{?}BAD$ ...(ii)(By CPCT)

But $\mathrm{?}ABC$ + $\mathrm{?}BAD$ = ${180}^{?}$ >br> (linear pair axiom)
$\mathrm{?}ABC$ = $\mathrm{?}BAD$ = ${90}^{?}$ ...(from (ii))
$?$ AB = BC = CD = DA, and $\mathrm{?}A$ = ${90}^{?}$

Hence, it is proved that, ABCD is a square.

Q6 ) Diagonal AC of a parallelogram ABCD bisects $\mathrm{?}A$ (see figure). Show that
(i) It bisects $\mathrm{?}C$ also,
(ii) ABCD is a rhombus.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given, diagonal AC of a parallelogram ABCD bisects $\mathrm{?}A$.

$?$ $\mathrm{?}DAC$ = $\mathrm{?}BAC$ = $\frac{1}{2}$ $\mathrm{?}BAD$ ...(i)
Here, AB || CD and AC is the transversal.
$\mathrm{?}DCA$ = $\mathrm{?}CAB$ ...(ii)(alternate angles)
$\mathrm{?}BCA$ = $\mathrm{?}DAC$ ...(iii)(alternate angles)

From eq. (i), (ii) and (iii), we get,
$\mathrm{?}DAC$ = $\mathrm{?}BAC$ = $\mathrm{?}DCA$ = $\mathrm{?}BCA$
Now, $\mathrm{?}BCD$ = $\mathrm{?}BCA$ + $\mathrm{?}DCA$
$?$ $\mathrm{?}BCD$= $\mathrm{?}DAC$ + $\mathrm{?}CAB$
$?$ $\mathrm{?}BCD$ = $\mathrm{?}BAD$
Thus, we can say that, diagonal AC bisects $\mathrm{?}C$.

Now, in $\mathrm{?}OAD$ and $\mathrm{?}OCD$, we have,
OA = OC
(since, diagonals bisects each other)
DO = OD ...(Common side)
$\mathrm{?}AOD$ = $\mathrm{?}COD$ = ${90}^{?}$
$?$ $\mathrm{?}OAD$ $?$ $\mathrm{?}OCD$
(By SAS rule)
$?$ AD = CD ...(By CPCT)

Now, AB = CD and AD = BC
(sides of parallelogram)
$?$ AB = CD = AD = BC

Hence it is proved that, ABCD is a rhombus.

Q7 ) ABCD is a rhombus. Show that diagonal AC bisects $\mathrm{?}A$ as well as $\mathrm{?}C$ and diagonal BD bisects $\mathrm{?}B$ as well as $\mathrm{?}D$.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: ABCD is a rhombus.
i.e., AD = AB = BC = CD ... (i)

To prove: (i) Diagonal AC bisect $\mathrm{?}A$ as well as $\mathrm{?}C$.
(ii) Diagonal BD bisects $\mathrm{?}B$ as well as $\mathrm{?}D$.

Proof: Let AC and BD are the diagonals of rhombus ABCD.
In $\mathrm{?}ABC$ and $\mathrm{?}ADC$, we have,
AC = CA ...(Common side)
CD = BC ...(From eq. (i))
$?$ $\mathrm{?}ABC$ $?$ $\mathrm{?}ADC$ ...(By SSS rule)
Thus, $\mathrm{?}DAC$ = $\mathrm{?}BAC$ ...(By CPCT)
Also, $\mathrm{?}DCA$ = $\mathrm{?}BCA$
Also, $\mathrm{?}DAC$ = $\mathrm{?}DCA$
And $\mathrm{?}BAC$ = $\mathrm{?}BCA$
This shows that, Diagonal AC bisect $\mathrm{?}A$ as well as $\mathrm{?}C$.

Now, in $\mathrm{?}BDC$ and $\mathrm{?}BDA$, we have,
AB = BC ...(Given)
BD = BD ...(Common side)
$?$ $\mathrm{?}BDC$ $?$ $\mathrm{?}BDA$ ...(By SSS rule)
Thus, $\mathrm{?}BDA$ = $\mathrm{?}BDC$ ...(By CPCT)
Also, $\mathrm{?}DBA$ = $\mathrm{?}DBC$
Also, $\mathrm{?}BDA$ = $\mathrm{?}DBA$
And $\mathrm{?}BDC$ = $\mathrm{?}DBC$
This shows that, Diagonal BD bisects $\mathrm{?}B$ as well as $\mathrm{?}D$.
Hence, proved.

Q8 ) ABCD is a rectangle in which diagonal AC bisects $\mathrm{?}A$ as well as $\mathrm{?}C$. Show that:
(i) ABCD is a square
(ii) Diagonal BD bisects $\mathrm{?}B$ as well as $\mathrm{?}D$.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: ABCD is a rectangle. AB = CD and BC = AD ...(i)

To prove: (i) ABCD is a square. i.e., AB = BC = CD = DA
(ii) Diagonal BD bisects $\mathrm{?}B$ as well as $\mathrm{?}D$.

Proof: In $\mathrm{?}ADC$ and $\mathrm{?}ABC$, we have,
$\mathrm{?}DAC$ = $\mathrm{?}BAC$
(Since, AB || DC and AC is transversal that intersects)
Similarly, $\mathrm{?}DCA$ = $\mathrm{?}BCA$
AC = CA ...(Common side)
$?$ $\mathrm{?}ADC$ $?$ $\mathrm{?}ABC$ ...(By ASA rule)
$?$AD = AB ...(By CPCT)
Also, CD = BC ...(ii)
Thus, from eq. (i) and(ii), we get,
AB = BC = AD = CD

Hence, it is proved that ABCD is a square.

Now, in $\mathrm{?}AOB$ and $\mathrm{?}COB$, we have,
AB = BC ...(Given)
BO = OB ...(Common side)
OA = OC
(Since,diagonal bisectS each other)
$?$ $\mathrm{?}AOB$ $?$ $\mathrm{?}COB$ ...(By SSS rule)
$?$ $\mathrm{?}OBA$ = $\mathrm{?}OBC$ ...(By CPCT)
This shows that, Diagonal BD bisects $\mathrm{?}B$.

Similarly, Now, in $\mathrm{?}AOD$ and $\mathrm{?}COD$, we have,
OD = DO ...(Common side)
OA = OC
(Since,diagonal bisectS each other)
$?$ $\mathrm{?}AOD$ $?$ $\mathrm{?}COD$ ...(By SSS rule)
$?$ $\mathrm{?}ADO$ = $\mathrm{?}CDO$ ...(By CPCT)

Hence, it is proved that, Diagonal BD bisects $\mathrm{?}D$, too.

Q9 ) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that
i) $\mathrm{?}APD$ $?$ $\mathrm{?}CQB$
ii) AP = CQ
iii) $\mathrm{?}AQB$ $?$ $\mathrm{?}CPD$
iv) AQ = CP
v) APCQ is a parallelogram.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given, ABCD is a parallelogram and P and Q are lie on BD such that DP = BQ ...(i)

i) Now,in $\mathrm{?}APD$ and $\mathrm{?}CQB$, we have,
DP = BQ ...(Given)
(Opposite sides are equal in parallelogram)
$\mathrm{?}ADP$ = $\mathrm{?}QBC$
(Since, AD || BC and BD is a transversal)
$?$ $\mathrm{?}APD$ $?$ $\mathrm{?}CQB$ ...(By SAS rule)

ii)
$?$ $\mathrm{?}APD$ $?$ $\mathrm{?}CQB$
$?$ AP = CQ ...(By CPCT)

iii) Here, Now, in $\mathrm{?}AQB$ and $\mathrm{?}CPD$, we have,
DP = BQ ...(Given)
AB = CD
(Opposite sides are equal in parallelogram)
$\mathrm{?}ABQ$ = $\mathrm{?}CDP$
(Since, AB || CD and BD is a transversal)
$?$ $\mathrm{?}AQB$ $?$ $\mathrm{?}CPD$ ...(By SAS rule)

iv)
$?$ $\mathrm{?}AQB$ $?$ $\mathrm{?}CPD$
$?$ AQ = CP...(By CPCT)

v) Now,in $\mathrm{?}APQ$ and $\mathrm{?}PCQ$, we have,
AQ = CP (From part iv))
AP = CQ (From part ii))
PQ = QP ...(Common side)
$?$ $\mathrm{?}APQ$ $?$ $\mathrm{?}PCQ$ ...(By SSS rule)
$?$ $\mathrm{?}APQ$ = $\mathrm{?}PQC$
And $\mathrm{?}AQP$ = $\mathrm{?}CPQ$
(Vertically Opposite angles)

Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.

$?$ AP || CQ and AQ || CP

Now, both pairs of opposite sides of quadrilateral APCQ are parallel.

Hence, it is proved that APCQ is a parallelogram.

Q10 ) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that
(i) $\mathrm{?}APB$ $?$ $\mathrm{?}CQD$
(ii) AP = CQ

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD.

$?$ AB || CD and BD is a transversal, we get,

$\mathrm{?}CDB$ = $\mathrm{?}DBA$ ...(i)
Now, in $\mathrm{?}APB$ and $\mathrm{?}CQD$, we have,
CD = AB ...(Sides of parallelogram)
$\mathrm{?}CQD$ = $\mathrm{?}APB$ = ${90}^{?}$ ...(Given)
$\mathrm{?}CDQ$ = $\mathrm{?}ABP$ ...(From Eq. (i))
$?$ $\mathrm{?}APB$ $?$ $\mathrm{?}CQD$ ...(By ASA rule)
$?$ AP = CQ ...(By CPCT)
Hence, it is proved.

Q11 ) In $\mathrm{?}ABC$ and $\mathrm{?}DEF$, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure).

Show that
(i) Quadrilateral ABED is a parallelogram
(ii) Quadrilateral BEFC is a parallelogram
(iv) Quadrilateral ACFD is a Parallelogram
(v) AC = DF
(vi) $\mathrm{?}ABC$ $?$ $\mathrm{?}DEF$

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: In $\mathrm{?}ABC$ and $\mathrm{?}DEF$, AB = DE, AB || DE, BC = EF and BC || EF

AB = DE and AB || DE ...(Given)
Since, a pair of opposite sides is equal and parallel
Therefore, ABED is a parallelogram.

BC = EF and BC || EF ...(Given)
Since, a pair of opposite sides is equal and parallel
Therefore, BEFC is a parallelogram.

(iii)Since, ABED is a parallelogram,
Also, BEFC in a parallelogram,
CF || BE and CF = BE ...(ii)
Thus, from Eq. (i) and (ii), we get,
Therefore, ACFD is a parallelogram.

(v)Since, ACFD is a parallelogram.
we get, AC = DF and AC || DF

(vi)Now, in $\mathrm{?}ABC$ and $\mathrm{?}DEF$,
AB = DE ...(Given)
BC = EF ...(Given)
and AC = DF ...(From part (v))
Therefore, $\mathrm{?}ABC$ $?$ $\mathrm{?}DEF$ ...(By SSS test)

Q12 ) ABCD is a trapezium in which AB || CD and AD = BC (see figure).

Show that
i) $\mathrm{?}A$ = $\mathrm{?}B$
ii) $\mathrm{?}C$ = $\mathrm{?}D$
iii) $\mathrm{?}ABC$ $?$ $\mathrm{?}BAD$
iv) Diagonal AC = Diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: ABCD is trapezium.
AB || CD and AD = BC
Now, extend AB and draw a line through C parallel to DA intersecting AB produced at E.

$?$ AD || CE and AD = CE
$?$ AD = BC = CE

i) We know that, $\mathrm{?}A$ + $\mathrm{?}E$ = ${180}^{?}$
(Since, interior angles on the same side of the transversal )
$?$ $\mathrm{?}E$ = ${180}^{?}$ - $\mathrm{?}A$
$?$ $\mathrm{?}E$ = ${180}^{?}$ - $\mathrm{?}A$
(Since, BC = EC)

Also, $\mathrm{?}ABC$ = ${180}^{?}$ - $\mathrm{?}CBE$
(Since, ABE is straight line)
$\mathrm{?}ABC$ = ${180}^{?}$ - ${180}^{?}$ + $\mathrm{?}A$
$\mathrm{?}B$ = $\mathrm{?}A$
Hence, it is proved.

ii)Now, $\mathrm{?}A$ + $\mathrm{?}D$ = ${180}^{?}$
(Since, interior angles on the same side of the transversal)
$?$ $\mathrm{?}D$ = ${180}^{?}$ - $\mathrm{?}A$
$?$ $\mathrm{?}D$ = ${180}^{?}$ - $\mathrm{?}B$
(from eq.(i)) ...(ii)

Also, $\mathrm{?}C$ + $\mathrm{?}B$ = ${180}^{?}$ ...180 (Since, interior angles on the same side of the transversal BC)
$?$ $\mathrm{?}C$ = ${180}^{?}$ - $\mathrm{?}B$ ...(iii)
from Eq. (ii) and (iii), we get,
$\mathrm{?}C$ = $\mathrm{?}D$
Hence, proved.

iii)Now, in $\mathrm{?}ABC$ and $\mathrm{?}BAD$, we have,
$\mathrm{?}A$ = $\mathrm{?}B$ ...(From Eq.(i))
AB = BA ...(Common side)
$?$ $\mathrm{?}ABC$ $?$ $\mathrm{?}BAD$ ...(By SAS rule)
Hence, proved.

iv)
$?$ $\mathrm{?}ABC$ $?$ $\mathrm{?}BAD$
$?$ AC = BD
Hence, it is proved.

## NCERT solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2

Q1 ) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that
i) SR || AC and SR = (1/2) AC
ii) PQ = SR
iii) PQRS is a parallelogram.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: P, Q, R, and S are mid-points of the sides.
Therefore, AP = PB, BQ = CQ CR = DR and AS = DS

i) Now, in $\mathrm{?}ADC$, we have,
S is the midpoint of AD and R is the mid point of CD.
As, we know that,
By midpoint theorem, the line segment joining the mid points of two sides of a triangle is parallel to the third side.
Thus, we can say that, SR || AC ...(i)
and SR = ($\frac{1}{2}$ ) AC ...(ii)

ii) Similarly, now, in $\mathrm{?}ABC$, we have,
PQ || AC ...(iii)
and PQ = ($\frac{1}{2}$ ) AC ...(iv)
Now, from (ii) and (iv), we get,
SR = PQ = ($\frac{1}{2}$ ) AC ...(v)

Now, from (i), (iii) and (v), we get,
PQ || SR and PQ = SR.
$?$ PQRS is a parallelogram
(Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel)
Hence, proved.

Q2 ) ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA.

By midpoint theorm, Now, in $\mathrm{?}ADC$, we have,
Thus, SR || AC ...(i)
and SR = ($\frac{1}{2}$ ) AC ...(ii)
Similarly, in $\mathrm{?}ABC$, we have,
PQ || AC ...(iii)
and PQ = ($\frac{1}{2}$ ) AC ...(iv)
from (i), (ii), (iii) and (iv), we get,
SR = PQ = ($\frac{1}{2}$ ) AC and PQ || SR
$?$ Quadrilateral PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.
$?$ $\mathrm{?}EOF$ = ${90}^{?}$.

Now, By midpoint theorem, we have,
RQ || BD
Thus, RE || OF
As SR || AC ...(from (i))
Thus, FR || OE

$?$ OERF is a parallelogram
So, $\mathrm{?}ERF$ = $\mathrm{?}EOF=90°$
(since, Opposite angle of a quadrilateral is equal)

Thus, PQRS is a parallelogram with $\mathrm{?}R=90°$
Hence, it is proved that PQRS is a rectangle.

Q3 ) ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.
Show that the quadrilateral PQRS is a rhombus.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

$?$ $\mathrm{?}A$ = $\mathrm{?}B$ = $\mathrm{?}C$ = $\mathrm{?}D$ = ${90}^{?}$ and
PQ || BD and PQ = $\frac{1}{2}$ BD and
SR || AC and SR = $\frac{1}{2}$ AC