Quadrilaterals

Exercise (8.1)

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

Difficulty Level:
Easy

What is known/given?
The angles of quadrilateral are in the ratio 3:5:9:13.

What is unknown?
All the angles of the quadrilateral.

Reasoning:
Sum of angles in any quadrilateral is 360 degree.

Solution:
Let the common ratio between the angles be x.
Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.
As the sum of all interior angles of a quadrilateral is 3600 ,
 3x + 5 x + 9 x + 13x = 3600
30 x = 3600
x = 120
Hence, the angles are
3 x = 3×12 = 360
5 x = 5×12 = 600
9 x = 9×12 = 1080
13x = 13×12 = 1560

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

Difficulty Level:
Easy

What is known/given?
The diagonals of a parallelogram are equal.

What is unknown?
How we can show that it is a rectangle.
Reasoning:
To show that given parallelogram is a rectangle, we have to prove that one of its interior
angles is 900 and this can be done by showing two triangles congruent.

Solution:

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that
one of its interior angles is 900 .
In ∆ABC and ∆DCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
∴ ∆ABC ≅ ∆DCB (By SSS Congruence rule)
⇒ ∠ABC = ∠DCB

It is known that the sum of the measures of angles on the same side of transversal is 180 0 .
ABC + DCB = 1800 ( AB || CD )
 ABC + ABC = 1800
 2ABC = 1800
 ABC = 900
Since ABCD is a parallelogram and one of its interior angles is 900 , ABCD is a rectangle.

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

Difficulty Level:
Easy

What is known/given?
The diagonals of a quadrilateral bisect each other at right angles.

What is unknown?
How we can show that it is a rhombus.

Reasoning:
To show that given quadrilateral is a rhombus, we have to show it is a parallelogram and
all the sides are equal.
Solution:

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right
angle i.e . OA = OC, OB = OD, and AOB = BOC = COD = AOD = 900.

To prove ABCD a rhombus,

We have to prove ABCD is a parallelogram and all the sides of ABCD are equal.

In AOD and COD,

OA = OC ( Diagonals bisect each other )
AOD = COD ( Given )
OD = OD ( Common )
AOD  COD ( By SAS congruence rule )
 AD = CD ............................................................ (1)

Similarly, it can be proved that

AD = AB and CD = BC ............................................ ( 2 )
From Equations (1) and (2),
AB = BC = CD = AD

Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a
parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that
ABCD is a rhombus.

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

Difficulty Level:
Medium

What is known/given?
Given quadrilateral is a square.

What is unknown?
How we can show that diagonals of a square are equal and bisect each other at
right angles.
Reasoning:
By showing two triangles congruent consist diagonals and then we can say corresponding
parts of congruent triangles are equal.

Solution:

Let ABCD be a square.

Let the diagonals AC and BD intersect each other at a point O.
To prove that the diagonals of a square are equal and bisect each other at right angles, we
have to prove,
AC = BD, OA = OC, OB = OD, and AOB = 900
In ABC and DCB,
AB = DC ( Sides of a square are equal to each other )
ABC = DCB (All interior angles are of 90o )
BC = CB ( Common side )
ABC  DCB ( By SAS congruency )
 AC = DB ( By CPCT )

Hence, the diagonals of a square are equal in length.

In AOB and COD,
AOB = COD ( Vertically opposite angles )
ABO = CDO ( Alternate interior angles )
AB = CD ( Sides of a square are always equal )
AOB  COD ( By AAS congruence rule )
 AO = CO and OB = OD ( By CPCT )

Hence, the diagonals of a square bisect each other.

In AOB and COB,
As we had proved that diagonals bisect each other, therefore,
AO = CO
AB = CB ( Sides of a square are equal )
BO = BO ( Common )
AOB  COB ( By SSS congruency )
AOB = COB ( By CPCT )
However,
AOB + COB = 1800
2AOB = 1800
AOB = 900

Hence, the diagonals of a square 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.

Difficulty Level:
Medium

What is known/given?
The diagonals of a quadrilateral are equal and bisect each other at right angle.

What is unknown?
How we can show that it is a square.

Reasoning:
We have to show that given quadrilateral is a parallelogram in which all sides are equal
and one of its interior angles is 900 .

Solution:

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each
other at O.

It is given that the diagonals of ABCD are equal and bisect each other at right angles.
Therefore, AC = BD, OA = OC, OB = OD, and AOB = BOC = COD = AOD = 900

To prove ABCD is a square

We have to prove that ABCD is a parallelogram in which AB = BC = CD = AD, and
one of its interior angles is 900 .
In AOB and COD,
AO = CO ( Diagonals bisect each other )
OB = OD ( Diagonals bisect each other )
AOB = COD ( Vertically opposite angles )
AOB  COD ( SAS congruence rule )
 AB = CD ( By CPCT ) .......................................... (1)

And, ∠OAB = ∠OCD (By CPCT)

However, these are alternate interior angles for line AB and CD and alternate interior
angles are equal to each other only when the two lines are parallel.
AB || CD ... (2)

From Equations (1) and (2), we obtain

ABCD is a parallelogram.

In AOD and COD,

AO = CO ( Diagonals bisect each other )
AOD = COD ( Given that each is 900 )
OD = OD ( Common )
AOD  COD ( SAS congruence rule )
 AD = DC .................................................................. ( 3)

However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)

AB = BC = CD = DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.
In ADC and BCD,
AD = BC ( Already proved )
AC = BD ( Given )
DC = CD ( Common )
ADC  BCD ( SSS Congruence rule )
ADC = BCD ( By CPCT )

However, ADC + BCD = 1800 (Co-interior angles)

ADC + ADC = 1800
2ADC = 1800
ADC = 900
One of the interior angles of quadrilateral ABCD is a right angle.
Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its
interior angles is 900 .

Therefore, ABCD is a square.

Q6. Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure).
Show that
(i) It bisects ∠C also,
(ii) ABCD is a rhombus.

Difficulty Level:
Medium

What is known/given?
The diagonal AC of a parallelogram ABCD bisects ∠A.

What is unknown?
How we can show that (i) It bisects ∠C also, (ii) ABCD is a rhombus.

Reasoning:
We can use alternate interior angles property to show diagonal AC bisects angle C also,
by showing all sides equal it can be said rhombus.

Solution:
(i) ABCD is a parallelogram.
 DAC =  BCA ( Alternate interiorangles ) ................................. (1)
And,  BAC =  DCA ( Alternate interiorangles ) ................................. ( 2)

However, it is given that AC bisects ∠A.

 DAC = BAC ............................................................................................ ( 3)

From Equations (1), (2), and (3), we obtain

 DAC =  BCA =  BAC =  DCA .......................................................... ( 4)
 DCA =  BCA
Hence, AC bisects ∠C.

(ii) From Equation (4), we obtain

 DAC =  DCA
DA = DC ( Side opposite to equal angles are equal )

However, DA = BC and AB = CD (Opposite sides of a parallelogram)

AB = BC = CD = DA
Hence, ABCD is a rhombus.
Q7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and
diagonal BD bisects ∠B as well as ∠D.

Difficulty Level:
Medium

What is known/given?
ABCD is a rhombus.

What is unknown?
How we can show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects
∠B as well as ∠D.

Reasoning:
We can use alternate interior angles property to show diagonal AC bisects angles A and
C, similarly diagonal BD bisects angles B and D.

Solution:

Let us join AC.

In ABC,
BC = AB (Sides of a rhombus are equal to each other)
BAC = BCA (Angles opposite to equal sides of a triangle are equal)

However, BAC = DCA (Alternate interior angles for parallel lines AB

and CD)
BCA = DCA
Therefore, AC bisects ∠C.

Also, BCA = DAC (Alternate interior angles for || lines BC and DA)
BAC = DAC

Therefore, AC bisects ∠A as well as ∠C.

Similarly, it can be proved that BD bisects ∠B and ∠D as well.
Q8. 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.

Difficulty Level:
Medium

What is known/given?
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

What is unknown?
How we can show that (i) ABCD is a square
(ii) Diagonal BD bisects ∠B as well as ∠D.

Reasoning:
We can use angle bisector property and isosceles triangle property to show given
rectangle as square and alternate interior angles property to show BD bisects angles B
and D.

Solution:

(i) It is given that ABCD is a rectangle.

A = C
1 1
 A = C
2 2
 DAC=DCA (AC bisects A and C)
CD = DA (Sides opposite to equal angles are also equal)

However, DA = BC and AB = CD (Opposite sides of a rectangle are equal)

AB = BC = CD = DA
ABCD is a rectangle and all the sides are equal.

Hence, ABCD is a square.

(ii) Let us join BD.
In BCD,
BC = CD (Sides of a square are equal to each other)
∠CDB = ∠CBD (Angles opposite to equal sides are equal)

However, ∠CDB = ∠ABD (Alternate interior angles for AB || CD)

∠CBD = ∠ABD
BD bisects ∠B.

Also, ∠CBD = ∠ADB (Alternate interior angles for BC || AD)

∠CDB = ∠ABD

BD bisects ∠D and ∠B.

Q9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such
that DP = BQ (see the given figure). Show that:

(i) APD  CQB

( ii ) AP = CQ
( iii ) AQB  CPD
( iv ) AQ = CP
( v) APCQ is a parallelogram

Difficulty Level:
Medium

What is known/given?
ABCD is a parallelogram and DP = BQ

What is unknown?
How we can show that
( i ) APD  CQB
( ii ) AP = CQ
( iii ) AQB  CPD
( iv ) AQ = CP
( v) APCQ is a parallelogram
Reasoning:
We can use alternate interior angles and parallelogram property for triangles congruence
criterion to show triangles congruent then we can say corresponding parts of congruent
triangles will be equal.

Solution:
(i) In APD and CQB,
ADP = CBQ ( Alternate interior angles for BC || AD )
AD = CB ( Opposite sides of parallelogram ABCD )
DP = BQ ( Given )
APD  CQB ( Using SAS congruence rule )

(ii) As we had observed that APD  CQB,

∴ AP = CQ (CPCT)

(iii) In AQB and CPD,

ABQ = CDP ( Alternate interior angles for AB || CD )
AB = CD ( Opposite sides of parallelogram ABCD )
BQ = DP ( Given )
AQB  CPD ( Using SAS congruence rule )

(iv) As we had observed that AQB  CPD,

∴ AQ = CP (CPCT)

(v) From the result obtained in (ii) and (iv),

AQ = CP and
AP = CQ

Since opposite sides in quadrilateral APCQ are equal to each other, APCQ is a
parallelogram.

Q10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A

and C on diagonal BD (See the given figure). Show that
(i) APB  CQD
(ii) AP = CQ

Difficulty Level:
Medium

What is known/given?
ABCD is a parallelogram and AP ⊥ DB, CQ ⊥ DB

What is unknown?
How we can show that (i) APB  CQD
(ii) AP = CQ

Reasoning:
We can use alternate interior angles and parallelogram property for triangles congruence
criterion to show triangles congruent then we can say corresponding parts of congruent
triangles will be equal.

Solution:

(i) In APB and CQD,

∠APB = ∠CQD (Each 90°)
AB = CD (Opposite sides of parallelogram ABCD)
∠ABP = ∠CDQ (Alternate interior angles for AB || CD)
APB  CQD ( By AAS congruency )

(ii) By using the above result

APB  CQD, we obtainAP = CQ ( By CPCT )

Q11. In ABC and 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 the given
figure). Show that
(i) Quadrilateral ABED is a parallelogram
(ii) Quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) Quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ABC  DEF.

Difficulty Level:
Medium

What is known/given?
In ABC and DEF, AB = DE, AB || DE, BC = EF and BC || EF.

What is unknown?
How we can show that (i) Quadrilateral ABED is a parallelogram
(ii) Quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) Quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ABC  DEF.

Reasoning:
We can use the fact that in a quadrilateral if one pair of opposite sides are parallel and
equal to each other then it will be a parallelogram and converse is also true. Also, by
using suitable congruence criterion we can show triangles congruent then we can say
corresponding parts of congruent triangles will be equal.

Solution:
(i) It is given that AB = DE and AB || DE.
If one pair of opposite sides of a quadrilateral are equal and parallel to each other, then it
will be a parallelogram.
Therefore, quadrilateral ABED is a parallelogram.

(ii) Again, BC = EF and BC || EF

Therefore, quadrilateral BCFE is a parallelogram.

(iii) As we had observed that ABED and BEFC are parallelograms, therefore
AD = BE and AD || BE
(Opposite sides of a parallelogram are equal and parallel)
And, BE = CF and BE || CF
(Opposite sides of a parallelogram are equal and parallel)
∴ AD = CF and AD || CF

(iv) As we had observed that one pair of opposite sides (AD and CF) of quadrilateral
ACFD are equal and parallel to each other, therefore, it is a parallelogram.

(v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and
parallel to each other.
∴ AC || DF and AC = DF
(vi) ∆ABC and ∆DEF,
AB = DE (Given)
BC = EF (Given)
AC = DF (ACFD is a parallelogram)
∴ ∆ABC ≅ ∆DEF (By SSS congruence rule)

Q12. ABCD is a trapezium in which AB || CD and AD = BC (see the given 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.]

Difficulty Level:
Medium

What is known/given?
ABCD is a trapezium in which AB || CD and AD = BC.

What is unknown?
How we can show that (i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD
(iv) diagonal AC = diagonal BD

Reasoning:
Consider parallel lines AD and CE, AE is the transversal line for them then sum of
co-interior angles will be 180 degree also angles CBE and CBA are linear pairs gives sum
180, angles CEB equals to CBE because opposite to equal sides in triangle BCE. Using
these observations angle A is equal to angle B. Then AB is parallel to CD will help to
show angle C is equal to angle D. Also, by using suitable congruence criterion we can
show triangles congruent then we can say corresponding parts of congruent triangles will
be equal.
Solution:
Let us extend AB. Then, draw a line through C, which is parallel to AD intersecting AB
at point E. It is clear that AECD is a parallelogram.

(i) AD = CE (Opposite sides of parallelogram AECD)

However, AD = BC (Given)
Therefore, BC = CE
∠CEB = ∠CBE (Angle opposite to equal sides are also equal)

Consider parallel lines AD and CE. AE is the transversal line for them.
∠A + ∠CEB = 1800 (Angles on the same side of transversal)
∠A + ∠CBE = 1800 (Using the relation ∠CEB = ∠CBE) ... (1)

However, ∠B + ∠CBE = 1800 (Linear pair angles) ... (2)

From Equations (1) and (2), we obtain
∠A = ∠B

(ii) AB || CD
∠A + ∠D = 1800 (Angles on the same side of the transversal)
Also, ∠C + ∠B = 180° (Angles on the same side of the transversal)
∴ ∠A + ∠D = ∠C + ∠B
However, ∠A = ∠B [Using the result obtained in (i)]
∴ ∠C = ∠D

(iii) In ABC and BAD,

AB = BA (Common side)
BC = AD (Given)
∠B = ∠A (Proved before)
ABC  BAD ( SAS congruence rule )

(iv) We had observed that,

ABC  BAD
 AC = BD ( By CPCT )
 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 the given figure). AC is a diagonal. Show that :

1
i) SR AC and SR = AC
2
ii) PQ = SR
iii) PQRS is a parallelogram.

Difficulty Level:
Medium

What is known/given?
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD
and DA. AC is a diagonal.

What is unknown?
1
How we can show that (i) SR AC and SR = AC
2
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Reasoning:
In a triangle, the line segment joining the mid-points of any two sides of the triangle is
parallel to the third side and is half of it. Also, if one pair of opposite sides of
quadrilateral is parallel and equal to each other then it is a parallelogram.
Solution:
i) In  ADC, S and R are the mid-points of sides AD and CD respectively. In a
triangle, the line segment joining the mid-points of any two sides of the triangle
is parallelto the third side and is half of it.
1
SR AC and SR = AC ... (1)
2
ii) In ABC, P and Q are mid-points of sides AB and BC respectively. Therefore,
by using mid-point theorem,
1
PQ AC and PQ = AC ... (2)
2
Using Equations (1) and (2), we obtain PQ SR and PQ = SR ... (3)
PQ = SR
iii) From Equation (3), we obtained PQ SR and PQ = SR

Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.
Hence, PQRS is a parallelogram.

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.

Difficulty Level:
Medium

What is known/given?
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and
DA respectively.

What is unknown?
How we can show that quadrilateral PQRS is a rectangle.

Reasoning:
In a triangle, the line segment joining the mid-points of any two sides of the triangle is
parallel to the third side and is half of it. Also if one pair of opposite sides of quadrilateral
is parallel and equal to each other then it is a parallelogram. By showing one of the angle
in parallelogram as 90 degree then we can say it is rectangle.

Solution:
In ABC, P and Q are the mid-points of sides AB and BC respectively.
1
PQ AC and PQ = AC (Using mid-point theorem) ... (1)
2
In ADC,
R and S are the mid-points of CD and AD respectively.
1
RS AC and RS = AC (Using mid-point theorem) ... (2)
2
From Equations (1) and (2), we obtain
PQ RS and PQ = RS

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other,
it is a parallelogram.

Let the diagonals of rhombus ABCD intersect each other at point O.

In quadrilateral OMQN,
MQ ON ( PQ AC)
QN OM ( QR BD)

Therefore, OMQN is a parallelogram.

  MQN =  NOM
  PQR =  NOM

However,  NOM = 90° (Diagonals of a rhombus are perpendicular to each other)

  PQR = 90°

Clearly, PQRS is a parallelogram having one of its interior angles as 900 .

Hence, 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.

Difficulty Level:
Medium

What is known/given?
ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and
DA respectively.

What is unknown?
How we can show that quadrilateral PQRS is a rhombus.

Reasoning:
In a triangle, the line segment joining the mid-points of any two sides of the triangle is
parallel to the third side and is half of it. Also if one pair of opposite sides of quadrilateral
is parallel and equal to each other then it is a parallelogram. By showing all four sides
equal in parallelogram we can say it is rhombus.
Solution:

Let us join AC and BD.

In  ABC,
P and Q are the mid-points of AB and BC respectively.
1
PQ AC and PQ = AC (Mid-point theorem) ... (1)
2
Similarly, in ADC,
1
SR AC and SR = AC (Mid-point theorem) ... (2)
2
Clearly, PQ SR and PQ = SR
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other,
it is a parallelogram.
PS QR and PS = QR (Opposite sides of parallelogram) ... (3)

In  BCD, Q and R are the mid-points of side BC and CD respectively.

1
QR BD and QR = BD (Mid-point theorem) ... (4)
2
However, the diagonals of a rectangle are equal.
AC = BD ... (5)
By using Equations (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus.

Q4. ABCD is a trapezium in which AB DC, BD is a diagonal and E is the mid -

point of AD. A line is drawn through E parallel to AB intersecting BC at F
(see the given figure). Show that F is the mid-point of BC.
Difficulty Level:
Medium

What is known/given?
ABCD is a trapezium in which AB DC, BD is a diagonal and E is the mid-point of AD.
A line is drawn through E parallel to AB intersecting BC at F.

What is unknown?
How we can show that F is the mid-point of BC.

Reasoning:
By converse of mid-point theorem, we know that a line drawn through the mid-point of
any side of a triangle and parallel to another side, bisects the third side.

Solution:
Let EF intersect DB at G.

By converse of mid-point theorem, we know that a line drawn through the mid-point of
any side of a triangle and parallel to another side, bisects the third side.

In ABD,
EF AB and E is the mid-point of AD.

Therefore, G will be the mid-point of DB.

As EF AB and AB CD,
EF CD (Two lines parallel to the same line are parallel to each other)
In BCD, GF CD and G is the mid-point of line BD. Therefore, by using converse of
mid-point theorem, F is the mid-point of BC.
Q5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD
respectively (see the given figure). Show that the line segments AF and EC
trisect the diagonal BD.

Difficulty Level:
Medium

What is known/given?
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.

What is unknown?
How we can show that the line segments AF and EC trisect the diagonal BD.

Reasoning:
In a quadrilateral if one pair of opposite sides is parallel and equal to each other. Then it
is a parallelogram. Also, by converse of mid-point theorem, we know that a line drawn
through the mid-point of any side of a triangle and parallel to another side, bisects the
third side.

Solution:
ABCD is a parallelogram.
AB CD
And hence, AE FC
Again, AB = CD (Opposite sides of parallelogram ABCD)
1 1
AB = CD
2 2
AE = FC (E and F are mid-points of side AB and CD)
In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to
each other. Therefore, AECF is a parallelogram.
 AF EC (Opposite sides of a parallelogram)

In DQC, F is the mid-point of side DC and FP CQ (as AF EC). Therefore, by using

the converse of mid-point theorem, it can be said that P is the mid-point of DQ.
DP = PQ ... (1)
Similarly, in APB, E is the mid-point of side AB and EQ AP (as AF EC).
Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-
point of PB.
PQ = QB ... (2)

From Equations (1) and (2),

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.

Q6. Show that the line segments joining the mid-points of the opposite sides of a
quadrilateral bisect each other.

Difficulty Level:
Medium

What is known/given?
ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD,
and DA respectively.

What is unknown?
How we can show that the line segments joining the mid-points of the opposite sides of a
quadrilateral bisect each other.

Reasoning:
In a triangle, the line segment joining the mid-points of any two sides of the triangle is
parallel to the third side and is half of it. In a quadrilateral if one pair of opposite sides is
parallel and equal to each other then it is a parallelogram.

Solution:

Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC,
CD, and DA respectively. Join PQ, QR, RS, SP, and BD.

In ABD, S and P are the mid-points of AD and AB respectively. Therefore, by using

mid-point theorem, it can be said that
1
SP BD and SP = BD ... (1)
2
Similarly, in  BCD,
1
QR BD and QR = BD ... (2)
2
From Equations (1) and (2), we obtain
SP QR and SP = QR

In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other.
Therefore, SPQR is a parallelogram.
We know that diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Q7. ABC is a triangle right angled at C. A line through the mid-point M of

hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
1
(iii) CM = MA = AB
2
Difficulty Level:
Hard

What is known/given?
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB
and parallel to BC intersects AC at D.

What is unknown?
How we can show that (i) D is the mid-point of AC
(ii) MD ⊥ AC
1
(iii) CM = MA = AB
2
Reasoning:
By converse of mid-point theorem, we know that a line drawn through the mid-point of
any side of a triangle and parallel to another side, bisects the third side. To show MD
perpendicular to AC we know that sum of interior angles is 180. Also, to get CM=MA=
half of AB we can show two triangles congruent.

Solution:
(i) In ABC,
It is given that M is the mid-point of AB and MD BC.
Therefore, D is the mid-point of AC. (Converse of mid-point theorem)

(ii) As DM CB and AC is a transversal line for them, therefore,

 MDC +  DCB = 1800 (Co-interior angles)
 MDC + 900 = 1800
 MDC = 900
 MD ⊥ AC

(iii) Join MC.

In AMD and CMD,
AD = CD (D is the mid-point of side AC)
 ADM =  CDM (Each 900 )

DM = DM (Common)
AMD  CMD (By SAS congruence rule)

Therefore, AM = CM (By CPCT)

1
However, AM = AB (M is the mid-point of AB)
2

Therefore, it can be said that

1
CM = AM = AB
2