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Quadrilaterals
We know that quadrilateral is a closed
figure with four sides. But by definition
suppose P, Q, R and S are four points
in a plane such that
i.
3 of these points are not collinear
ii. If sides PQ, QR, RS, SP are such that any of the
two such segments have a common point, it is an
endpoint only.
iii. If we draw a line containing any one of these
segments the remaining two points lie on the
same side of this line and
iv. If seg PR and seg QS intersect in the points other
than P, Q, R, S then the union of 4 segments
PQ, QR, RS, SP is called a quadrilateral.
is a symbol of ‘quadrilateral’.
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(And Interior of the quadrilateral is a convex set
but quadrilateral is not a convex set)
Each quadrilateral has 4 sides, 4 angles, 4
vertices, 2 diagonals, 4 pairs of adjacent angles,
2 pairs of opposite sides, 4 pairs of adjacent or
consecutive sides and 2 pairs of opposite angles.
There are main 8 types of quadrilaterals
(as shown in the figures they are
i.
Parallelogram
ii. Rectangle
iii. Rhombus
iv. Square
v. Trapezium (non parallel sides are not
congruent)
vi. Isosceles trapezium
vii. Kite
viii. Kite (with all side, congruent)
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Table
S. No. Quadrilateral
Type
1.
Parallelogram
2.
Rectangle
3.
Rhombus
4.
Square
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5.
Trapezium
6.
Isosceles Trapezium
7.
Kite I
8.
Kite II
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Property of a quadrilateral
Theorem 1
[Angle sum property of a quadrilateral]
The sum of the measures of all angles of a
quadrilateral is 360º
Given
:
To prove
:
Construction :
ABCD is a given quadrilateral
∠A + ∠B + ∠C + ∠D = 360º
Join the points A and C
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Proof
:
In
ABC,
∠BAC + ∠ ABC + ∠BCA = 180º
.....sum of angles of
ABC .... (1)
and, In
ADC,
∠DAC + ∠ADC + ∠DCA = 180º
.....sum of angles of
ADC .... (2)
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Adding (1) and (2) we get
∠BAC + ∠ABC + ∠BCA + ∠DAC
+ ∠ADC + ∠DCA = 180º + 180º
∴ ∠BAC + ∠DAC + ∠ABC + ∠BCA
+ ∠DCA + ∠ADC = 360º
......(3)
But ∠BAC + ∠DAC = ∠BAD
....Angle addition property....(4)
and ∠BCA + ∠DCA = ∠BCD
....Angle addition property....(5)
∴ from (3), (4) and (5) we get
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∠BAD + ∠ABC + ∠BCD + ∠ADC = 360º
i.e. ∠A + ∠B + ∠C + ∠D = 360º
Hence, sum of the measures of all
angles of a quadrilateral is 360º
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Theorem 2
The opposite sides of a parallelogram are congruent.
Given
ABCD is a parallelogram in which
:
side AB || side CD and side BC || side
DA
:
Construction :
Proof
:
side AB
side CD and
side BC
To prove
side DA
Draw diagonal AC.
ABCD is a parallelogram in
which side AB || side CD and
seg AC is a transversals.
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∴∠BAC
∠DCA ...(alternate angle)....(1)
Similarly,
side BC || side DA and seg AC is a
transversal.
∴ ∠BCA
∠DAC ......(alternate
angle) ...(2)
In
ABC and
∠BAC
seg AC
∠BCA
∴
CDA
∠DCA
......(from 1)
seg CA .......(common side)
∠ DAC
ABC
CDA
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........(from 2)
.......(ASA test)

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∴ side AB
side CD
and side BC
......(c.s.c.t)
side DA ......(c.s.c.t.)
Hence, the opposite sides of a
parallelogram are congruent.
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Theorem 3
If opposite sides of a quadrilateral are congruent then
the quadrilateral is parallelogram.
Given
:
In
ABCD
side AB
side DC
and side AD
To prove
:
Construction :
side BC
ABCD is a parallelogram
Draw a diagonal BD
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Proof
:
In
ABD and
CDB
side AB
side DC
side AD
side BC
side BD
side DB
......(given)
....(common side)
∴
ABD
CDB
∴ ∠ABD
∠CDB
.....(SSS test)
......(c.a.c.t.)
∴ side AB || side CD .......(1)
...(alternate angles test)
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Similarly, we can prove that
side AD || side BC
......(2)
....(alternate angels test)
∴ from (1) and (2) we have
ABCD is a parallelogram
Hence, if opposite sides of a
quadrilateral are congruent then it is a
parallelogram.
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Theorem 4
The opposite angles of a parallelogram are congruent
Given
PQRS is a parallelogram
:
To prove :
∠SPQ
∠QRS
and ∠PSR
Construction :
Proof
:
∠RQP
Draw a diagonal SQ
PQRS is a parallelogram
∴ side PS || side QR and seg SQ
is a transversal
∴ ∠PSQ
∠RQS
..(Alternate angles) ...(1)
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Also, side PQ || side SR and
seg SQ is a transversal.
∴ ∠PQS
∠RSQ
...(Alternate angles)..(2)
In
PQS and
∠PSQ
RSQ
∠RQS ......from (1)
side SQ
side QS
....(common side)
∠PQS
∴
∠RSQ
PQS
......from (2)
RSQ
......(ASA test)
∴ ∠SPQ
17
∠QRS
......(c.a.c.t.)

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Similarly, we can prove by drawing
diagonal PR.
∠PSR
∠ RQP
Hence, the opposite angles of a
parallelogram are congruent.
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Theorem 5
A quadrilateral is a parallelogram if its opposite angles
are congruent.
Given
:
PQRS is a parallelogram in which
∠SPQ
∠PQR
To prove :
Proof
:
∠QRS and
∠RSP
PQRS is a parallelogram
Let ∠SPQ = ∠QRS = xº
And ∠PQR = ∠RSP = yº ......Opposite
angle of a quadrilateral.
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∠SPQ + ∠PQR + ∠QRS + ∠RSP= 360º.
…(Angle sum property of a
quadrilateral)
∴ x + y + x + y = 360º
∴ 2x + 2y = 360º
∴ x + y = 180º
.....(dividing by 2)
∴ ∠SPQ + ∠RSP = 180º
∴ side PQ || side SR ... (interior angles
test) ....(1)
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Similarly, we can prove that
side PS || side QR .......(2)
∴
PQRS is a parallelogram
...... from (1) and (2)
Hence, if opposite angles of a
quadrilateral are congruent, then it
is a parallelogram.
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Theorem 6
The diagonals of a parallelogram bisect each other.
Given
:
ABCD is a parallelogram in which
the diagonals AC and BD intersect in M.
To prove :
seg AM
seg CM
and seg BM
Proof
:
since
seg DM
ABCD is a parallelogram
side AB || side CD and AC is a
transversal.
∴ ∠BAC
∠DCA ...(Alternate angles)
i.e. ∠BAM
∠DCM ...(A – M – C)..(1)
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Also, side AB || side DC and seg DB is
a transversal.
∴ ∠ABD
∠CDB ...(Alternate angles)
i.e. ∠ABM
∠CDM ...(B – M – D) ..(2)
Now, In
ABM and
∠BAM
∠DCM ......(from 1)
side AB
∠ABM
∴
CDM
side DC .....(opposite sides)
∠CDM ......(from 2)
ABM
CDM ......(ASA test)
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∴ seg AM
seg CM
and seg BM
.....(c.s.c.t.)
seg DM
Hence, diagonals of a parallelogram
bisect each other.
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Theorem 7
If the diagonals of a quadrilateral bisect each other,
then the quadrilateral is a parallelogram.
Given
:
PQRS is a quadrilateral in which
diagonals PR and QS intersect in M.
seg PM
seg QM
To prove :
Proof
:
seg RM and
seg SM
PQRS is a parallelogram
In
PMQ and
seg PM
RMS
seg RM
25
......(given)

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∠PMQ
∠RMS
.....(verticallyopposite angles)
seg QM
∴
seg SM .....(given)
PMQ
RMS
∴ ∠PQM
∠RSM
i.e. ∠PQS
......(SAS test)
......(c.a.c.t)
∠RSQ ......(S – M – Q)
∴ side PQ || side SR ....(alternate
angles test) ...(1)
Similarly, we can prove that
side PS || side QR
26
.....(2)

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PQRS is a parallelogram
.........from (1) and (2)
Hence, if the diagonals of a
quadrilateral bisect each other then it
is a parallelogram.
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Theorem 8
A quadrilateral is a parallelogram if a pair of opposite
sides is parallel and congruent.
Given
LMNK is a given quadrilateral in
:
which
side LM || side NK and
side LM
To prove
:
Construction :
side NK
LMNK is a parallelogram
Draw diagonal MK
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Proof
:
since
LMNK is a quadrilateral in
which side LM || side NK
and seg MK is a transversal.
∴∠LMK
∠NKM
..(Alternate angles) ...(1)
Now, In
KLM and
MNK
seg LM
seg NK
.....(given)
∠LMK
∠NKM
seg KM
.......(from 1)
seg MK
.......(common side)
∴
KLM
MNK
......(SAS test)
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∴ ∠LKM
∠NMK
.......(c.a.c.t)
∴ side LK || side MN
.....(alternate
angles test) ....(2)
and side LM || side NK
...(given) .....(3)
∴ from (2) and (3) we have
LKMN is
a parallelogram
Hence, if a pair of opposite sides of a
quadrilateral is parallel and congruent
then the quadrilateral is a parallelogram.
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Sums
1. The angles of a quadrilateral are in the ratio
2 : 3 : 5 : 8. Find all the angles of the quadrilateral.
Sol.
The angles of the quadrilateral are in ratio
2 : 3 : 5 : 8.
Let the measures of the four angles are 2xº, 3xº,
5xº and 8xº.
According to the angle sum property of a
quadrilateral, we have
2xº + 3xº + 5xº + 8xº = 360º
∴ 18xº = 360º
3600
∴x =
18
0
∴ x 0 = 200
2xº = 2 × 20º = 40º, 3xº = 3 × 20º = 60º,
5xº = 5 × 20º = 100º and 8xº = 8 × 20º = 160º
∴
The measures of the angles of the
quadrilateral are 40º, 60º, 100º and 160º.
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2. In the figure, PQRS is a quadrilateral. The
bisectors of ∠P and ∠Q meet at the point A.
∠S = 50º, ∠R = 110º. Find the measure of∠PAQ.
Sol.
Refer above fig.
According to the angle sum property of a
quadrilateral,
∠SPQ + ∠PQR + ∠R + ∠S = 360º
∴ ∠SPQ + ∠PQR + 110º + 50º = 360º
… (substituting the given values)
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∴ ∠SPQ + ∠PQR + 160º = 360º
∴ ∠SPQ + ∠PQR = 360º - 160º
∴ ∠SPQ + ∠PQR = 200º
… (1)
Now, it is given that Ray PA and ray QA are the
bisectors of ∠SPQ and ∠RQP respectively.
∴ ∠APQ = ½ ∠SPQ and ∠AQP = ½ ∠RQP
∴ ∠APQ + ∠AQP = ½ (∠SPQ + ∠RQP) … (2)
From (2) and (1), ∠APQ + ∠AQP = ½ (200º)
∴ ∠APQ + ∠AQP = 100º
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… (3)

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Also, In APQ, ∠PAQ + ∠APQ + ∠AQP = 180º
… (Angles of a triangle)
∴ ∠PAQ + 100º = 180º
… [From (3)]
∴ ∠PAQ = 180º - 100º
∴ ∠PAQ = 80º
∴The measure of ∠PAQ is 80º.
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35.
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3. The sides BA and DC of quadrilateral ABCD are
extended as shown in the figure. Prove that
m + n = p + q.
Proof :
Consider ∠DAB b = a and ∠BCD = c.
According to the angle sum property of
a quadrilateral,
In quadrilateral ABCD,
a + q + c + p = 360º
35
… (1)

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From (2), m + c + n + a = 360º … (3)
From (1) and (3),
a+q+c+p=m+c+n+a
∴q+p=m+n
i.e. m + n = p + q.
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4. In a parallelogram ABCD, ∠A = xº,
∠B = (3x + 20)º. Find x, ∠C and ∠D.
Sol.
We know that the adjacent angles of a
parallelogram are supplementary.
∴ ∠A + ∠B = 180º
Put the given values
∴ xº + 3xº + 20º = 180º
∴ 4xº = 180º - 20º
∴ 4xº = 160º
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∴ xº = 40º
3xº + 20º = 3 × 40º + 20º = 120º + 20º = 140º
∴ ∠A = 40º and ∠B = 140º
Also, the opposite angles of a parallelogram are
congruent.
∴ ∠C = ∠ A = 40º and ∠ D = ∠ B = 140º
∴ x = 40. ∠ C = 40º, ∠ D = 140º.
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39.
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5. The perimeter of a parallelogram is 150 cm. One
of its sides is greater than the other by 25 cm. Find
the lengths of all the sides of the parallelogram.
Sol:
Consider that the smaller of the two adjacent sides
be x cm.
Then, it is given that the other side is (x + 25) cm
Now, the perimeter of a parallelogram is equal to
the sum of the lengths of four sides.
∴
x + (x + 25) + x + (x + 25) = 150
∴
4x + 50 = 150
∴
x = 25 and x + 25 = 25 + 25 = 50
∴
The lengths of the sides of the parallelogram
∴ 4x = 150 – 50 ∴ 4x = 100
are 25 cm, 50 cm, 25 cm and 50 cm.
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40.
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6. In the figure,
WXYZ is a parallelogram. From the
information given in the figure, find the values of x
and y.
Sol.
As XY & ZW are opposite sides of a parallelogram
side XY || side ZW
seg XZ is a transversal.
∴ ∠ ZXY = ∠ XZW
∴ 4y = 28º
… (alternate angles)
… (given)
∴ y = 7º
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Similarly, side XW || side YZ and seg XZ is the
transversal.
∴ ∠ ZXW = ∠ XZY.
∴ 10x = 60º
… (given)
∴ x = 6º.
∴The value of x is 6º and that of y is 7º.
41

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7. Prove that two opposite vertices of a parallelogram
are equidistant from the diagonal not containing
these vertices.
Given :
1.
ABCD is a parallelogram.
2. Seg AM ⊥ diagonal BD and
Seg CN ⊥ diagonal BD.
We have to prove seg AM ≅ seg CN.
First, draw diagonal AC intersecting diagonal BD
in the point P.
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Proof :
P is the point of intersection of the diagonals of
parallelogram ABCD.
∴
seg AP ≅ seg CP
… (1)
In APM and CPN,
seg AP ≅ seg CP
… [From (1)]
∠ APM ≅ ∠ CPN
… (vertically opposite angles)
∠ AMP ≅ ∠CNP
… (each a right angle)
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∴ APM ≅ CPN
… (SAA test)
∴ seg AM ≅ seg CN
… (c.s.c.t.)
Therefore, the opposite vertices of a parallelogram
are at equal distance from the diagonal not
containing these vertices.
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45.
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8. The ratio of two sides of a parallelogram is 3 : 4. If
its perimeter is 112 cm, find the lengths of the
sides of the parallelogram.
Sol:
It is given that the ratio of the lengths of two sides
of the parallelogram is 3:4.
Let the lengths of those two sides be 3x cm and 4x
cm respectively.
We know that the opposite sides of a
parallelogram are congruent.
∴ The sides of the parallelogram are 3x cm, 4x cm,
3x cm and 4x cm.
Now, the perimeter of a parallelogram = the sum
of the lengths of four sides
∴
112 = 3x + 4x + 3x + 4x
… (given perimeter 112 cm)
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∴
14x = 112
∴
x=8
3x = 3 × 8 = 24 and 4x = 4 × 8 = 32
∴
The lengths of the sides of the given
parallelogram are 24 cm, 32 cm, 24 cm and 32
cm.
46

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Definitions
Parallelogram
A quadrilateral is a parallelogram if its opposite sides
are parallel.
Rectangle
A parallelogram (or a quadrilateral) in which each
angle is a right angle, is called a rectangle.
Rhombus
A quadrilateral having all sides congruent is called a
rhombus.
Square
A quadrilateral is called a square if all its sides and
angles are congruent.
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Trapezium
If only one pair of opposite sides is parallel, the
quadrilateral is said to be a trapezium.
Isosceles trapezium
A trapezium in which nonparallel sides are congruent,
it is called as isosceles trapezium.
Kite
If in
ABCD, AB
AD and CB
CD and diagonal
AC is perpendicular bisector of diagonal BD, then
ABCD is called as a kite.
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Theorem 9
Diagonals of a rectangle are congruent.
Theorem 10
If diagonals of a parallelogram are congruent then it is
a rectangle.
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Theorem 11
Diagonals of a rhombus are perpendicular bisectors of
each other.
Theorem 12
If the diagonals of a quadrilateral bisect each other at
right angle then the quadrilateral is a rhombus.
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Theorem 13
Diagonals of a square are congruent and
perpendicular bisectors of each other.
Theorem 14
If diagonals of quadrilateral are congruent and
perpendicular bisectors of each other then it is square.
51

52.
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Sums
1. Adjacent sides of a rectangle are of lengths 7 cm
and 24 cm. Find the lengths of its diagonals.
Sol.
Consider
ABCD as a rectangle in which
AB = 24 cm and BC = 7 cm.
In right-angled ABC,
By Pythagoras’ theorem,
AC² = AB² + BC²
= (24)² + (7)² = 576 + 49 = 625
∴ AC = 25
52

53.
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But, the diagonals of a rectangle are congruent.
∴ BD = AC = 25
∴ The length of diagonal AC = 25 cm and that of
BD = 25 cm.
53

54.
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2.
ABCD is a trapezium in which AB || DC. M and
N are the midpoints of side AD and side BC
respectively. If AB = 12 cm and MN = 14 cm, find
CD.
Sol.
The length of the segment joining the midpoints of
non - parallel sides of a trapezium is half the sum
of the lengths of its parallel sides.
∴ MN = ½ (AB + CD)
∴ 14 = ½ (12 + CD)
… (substituting the given values)
54

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∴
28 = 12 + CD
… (Multiplying both the sides by 2)
∴ CD = 28 – 12
∴ CD = 16.
55

56.
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3. Prove that the base angles of an isosceles
trapezium are congruent.
Given
:
□ ABCD is an isosceles
trapezium in which
side AB || side CD and
side AD
To prove
:
Construction :
∠ADC
side BC.
∠BCD.
Draw seg BE || side AD and
D-E-C.
56

57.
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Proof
:
IN □ABED, side AB || side DE.
… (given)
seg BE || side AD
… (construction )
∴ □ABED is a parallelogram.
The opposite sides and the
opposite angles of a
parallelogram are congruent.
∴ side AD
and ∠D
57
side BE … (1)
∠ABE
… (2)

58.
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side AD ≅ side BC … (given)
… (3)
From (1) and (3),
side BE ≅ side BC
∴ ∠ BEC ≅ ∠C …(Isosceles
triangle theorem) … (4)
side AB || side DC
… (given)
side BE is the transversal.
∴ ∠ABE ≅ ∠BEC
… (alternate angles) … (5)
From (2), (4) and (5),
∠D ≅ ∠C.
i.e. ∠ADC ≅ ∠BCD.
[Similarly, we can prove that
∠DAB
58
∠CBA.]

59.
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4. The lengths of the diagonals PR and QS of a
rhombus are 20 cm and 48 cm respectively. Find
the length of each side of the rhombus.
Sol.
We know that the diagonals of a rhombus bisect
each other at right angles.
∴ MQ = ½ QS = ½ × 48 cm
∴ MQ = 24 cm
MP = ½ PR = ½ × 20 cm
∴ MP = 10 cm.
∠ PMQ = 90º
59

60.
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In right-angled PMQ,
By Pythagoras’ theorem
PQ² = PM² + QM²
= (10)² + (24)²
= 100 + 576
= 676
∴ PQ = 26.
But all the sides of a rhombus are congruent.
∴ The length of each side of the given rhombus
is 26 cm.
60

61.
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5.
ABCD is a kite in which diagonal AC and
diagonal BD intersect at point O.
∠ OBC = 20º and ∠OCD = 40º.
Find : i. ∠ABC
ii. ∠ADC
iii. ∠BAD.
Sol.
Refer above fig,
The diagonal joining the unequal sides of a kite
bisects the other diagonal at right angles.
∴ Each angle at O is a right angle.
Also, side BA
side BC.
61

62.
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Side DA
side DC and seg AO
seg CO
… (perpendicular bisector theorem)
… (1)
In BOC, ∠BOC = 90º, ∠OBC = 20º
… (given)
∴ ∠OCB = 180º - (90º + 20º)
= 180º - 110º
∴ ∠OCB = 70º
… (2)
62

63.
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In BAC, BA = BC
… [From (1)]
∴ ∠BCA = ∠BAC
… (angles opposite to
equal sides)
i.e. ∠OCB = ∠OAB
∴ ∠OAB = 70º
… [From (2)]
In DAC, DA = DC
∴ ∠DCA = ∠DAC
… [From (1)]
… (angles opposite to
equal sides)
i.e. ∠DCO = ∠DAO
63
… (3)

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∴ ∠DAO = 40º
… [Given : ∠DCO = 40º]
… (4)
From (3) and (4),
∠OAB + ∠DAO = 70º + 40º = 110º … (5)
∠BAO + ∠DAO = ∠BAD
… (Angle addition postulate)
… (6)
From (5) and (6),
∠BAD = 110º.
In DAC, ∠ADC + ∠DAC + ∠DCA = 180º
… (angles of a triangle)
∴ ∠ ADC + 40º + 40º = 180º
… [From (4) and given]
64

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∴ ∠ADC = 180º - 80º
∴ ∠ADC = 100º
… (7)
In BAC, ∠ABC + ∠BAC + ∠BCA = 180º
… (angles of a triangle)
∴ ∠ABC + 70º + 70º = 180º
… [From (2) and 3)]
∴ ∠ABC = 180º - 140º
∴ ∠ABC = 40º
Ans. : i. ∠ABC = 40º
ii. ∠ADC = 100º
iii. ∠BAD = 110º.
65

66.
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Theorem 15
Diagonals of an isosceles trapezium are congruent.
Given
ABCD is an isosceles trapezium in
:
which seg AC and seg BD are
diagonals side AD
side BC and side
AB || side DC
To Prove
:
Construction :
seg AC
seg BD
Draw seg AM ⊥ seg DC,
seg BN ⊥ seg DC
Proof
:
In ∆AMD and ∆BNC
side AD
side BC
66
….(given)

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∠AMD
seg AM
∠BNC
…..(90º each)
seg BN …..(perpendicular
distance between
two parallel lines)
∴ ∆AMD
∆BNC ……(hypotenuse
side test)
∴ ∠ADM
∠BCN …..(c. a. c. t.)
i.e. ∠ADC
∠BCD .(D - M - C)…(1)
Now, In ∆ADC and ∆BCD
side AD
∠ADC
side BC
∠BCD
67
….(given)
…..(from (1))

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side DC
∴ ∆ADC
side CD …..(common side)
∆BCD
∴ seg AC
…..(SAS test)
seg BD ……(c.s.c.t.)
Hence, diagonals of an isosceles
trapezium are congruent.
68

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Theorem 16
(Intercepts made, by three parallel lines)
If three parallel lines make congruent intercepts on a
transversal then they make congruent intercepts on
any other transversal.
Given
:
line
|| line m || line n
and lines t1 and t2 are transversals.
seg AB
Prove
:
Construction :
seg BC
seg DE
seg EF
Draw a line GI parallel to the line t1
through E.
69

70.
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Proof
:
seg AG || seg BE ……(given)
seg AB || seg GE ……. (construction)
∴
ABEG is a parallelogram.
∴ seg AB
seg GE…(opposite sides
of a parallelogram …(1)
Similarly,
∴ seg BC
BCIE is a parallelogram.
seg EI …..(opposite sides
of a parallelogram….(2)
But seg AB
seg BC ….. (given)...(3)
Hence seg GE
seg EI ….(4) from
(1), (2) and (3)
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71.
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Now, In ∆GED and ∆IEF
∠DGE
∠FIE …(alternate angles)
seg GE
∠GED
seg EI …..(from (4))
∠IEF
……(vertically
opposite angles)
∴ ∆GED
∴ seg DE
∆IEF ……(ASA test)
seg EF ……(c.s.c.t.)
Hence, if three parallel lines make
congruent intercepts on a transversal
then they make congruent intercepts
on any other transversal.
71

72.
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Theorem 17
Mid - Point theorem
The line segment joining the midpoints of any two
sides of a triangle is parallel to the third side and is
half of it.
Given
:
In ∆ABC, P and Q are midpoints of
sides AB and AC respectively.
To prove :
i.
seg PQ || side BC
ii. seg PQ =
1
side BC
2
Construction : Take the point R on ray PQ such that
seg PQ
seg QR. Join C and R
72

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Proof
:
In ∆AQP and ∆CQR
seg AQ
∠AQP
seg CQ ……(given)
∠CQR ….(vertically opposite
angles)
seg PQ
seg RQ ……(construction)
∴ ∆AQP
∆CQR
…… (SAS test)
seg AP
seg CR ……(c.s.c.t.)
seg BP
seg CR ...(
and ∠PAQ
∠RCQ
∴ seg AP || seg CR …
AP = PB)…(1)
……(c.a.c.t)
(alternate
angles test)
∴ seg BP || seg CR …(AP = PB)....(2)
73

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PBCR is a parallelogram (from (1)
and (2))
∴ side PR || side BC …..(opposite
sides of a parallelogram)
seg PQ || side BC
……(P - Q - P)
1
Also, seg PQ = side PR
2
…(construction)
∴ seg PQ =
1
side BC …(
2
PR = BC)
Hence, In a triangle, line segment
joining mid - points of any two sides is
parallel to the third side and is half of
it.
74

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Theorem 18
Converse of mid - point theorem
If a line drawn through the mid - point of one side of a
triangle is parallel to second side then it bisects the
third side.
In ∆ABC, A - D - B if AD = DB and line DE || side BC
then, line DE bisects side AC. i.e. AE = CE
Try to write the proof.
75

76.
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Sums
14. In the figure, □PQRS and □LMNR are rectangles,
where M is the midpoint of PR. Prove that:
i. SL = LR
ii. LN = ½ SQ.
Proof
:
In above fig.
□LMNR is a rectangle.
∴ LM || RN
… (Opposite
sides of a rectangle)
i.e. LM || RQ
… (R-N-Q)
… (1)
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77.
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RQ || SP
… (opposite sides
of a rectangle)
…(2)
From (1) and (2), we have
LM || RQ || SP
… (3)
In PSR, it is given that M is the
midpoint of PR and LM || RQ
… [From (1)]
∴ By the converse of midpoint
theorem, point L is the midpoint of
side SR.
∴ SL = LR
77
… (4)
…i

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Again, LR || MN and LR || PQ
… (opposite sides of
rectangle LMNR and
rectangle PQRS
respectively)
∴ MN || PQ
… (property for
parallel sides)
∴ In RPQ,
M is the midpoint of PR.
78

79.
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∴ By the converse of midpoint
theorem,
point N is the midpoint of RQ.
… (5)
From (4) and (5), L and N are the
midpoints of SR and RQ
respectively.,
∴ In RSQ, by midpoint theorem,
LN = ½ SQ
Hence proved.
79
… ii

80.
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15. In the figure, seg PD is the median of PQR. Point
PM 1
G is the midpoint of seg PD. Show that
= .
PR 3
Proof
:
Draw seg DN || seg QM.
We are given that, In PDN, G is
the midpoint of seg PD.
Seg QM (i.e. seg GM) || seg DN
… (by construction)
80

81.
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∴ By the converse of midpoint
theorem, M is the midpoint of seg
PN.
∴ PM = MN
… (1)
In QRM, As PD is median, D is the
midpoint of side QR
And seg QM || seg DN
… (construction)
∴ By the converse of midpoint
theorem, N is the midpoint of
seg MR.
∴ MN = NR
81
… (2)

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From (1) and (2), PM = MN = NR
… (3)
Now, PR = PM + MN + NR
… (4)
∴ PR = 3PM
… [From (3) and (4)]
i.e. 3PM = PR
82
PM 1
∴
= .
PR 3

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16. Prove that a diagonal of a rhombus bisects two
opposite angles.
Proof :
Refer above fig.
ABCD is a rhombus & AC is its
diagonal.
In ABC and ADC,
side AB
side BC
side AD and
side DC
… (sides of a rhombus)
side AC
side AC
… (common side)
∴ ABC
ADC
83
… (SSS test)

84.
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∴ ∠BAC
and ∠BCA
∠DAC
… (1)
∠DCA
… (2)
… (c.a.c.t.)
seg AC bisects ∠BAD.
…[from(1)]
seg AC bisects ∠BCD.
…[From(2)]
Therefore diagonal of a rhombus
bisects two opposite angels.
84

85.
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17. In the figure,
BHIR is a kite. If ∠HIR = 50º, find a.
∠HRI and b. ∠BHI.
Sol.
Consider abive fig. □BHIR is a kite.
a. In IRH, side IR = side IH
∴ ∠HRI = ∠RHI
… (given)
… (angles opposite to
equal sides)
… (1)
∠I + ∠HRI + ∠RHI = 180º
… (angles of a triangle)
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86.
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∴ 50º + ∠HRI + ∠HRI = 180º
… [Given and from (1)]
∴ 2∠HRI = 180º - 50º
∴ 2∠HRI = 130º
∴ ∠HRI = 65º
… (2)
∴ ∠RHI = 65º … [From (1) and (2)] … (3)
86

87.
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b.
BHR is an equilateral triangle.
… (given)
∴ ∠BHR = 60º
… (angle of an
equilateral triangle)
… (4)
∠BHI = ∠BHR + ∠RHI
… (angle addition postulate)
= 60º + 65º … [From (4) and (3)]
∴ ∠BHI = 125º
87

88.
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18. In CBS, seg BC
seg SC. Ray CE bisects
exterior ∠DCS. Ray SE || ray BC. Prove that
CBSE is a parallelogram.
Proof :
We are given that
seg CB
∴ ∠CBS
seg CS
∠CSB
… (angles opposite to
congruent sides) … (1)
∠DCS is the exterior angle of CBS.
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89.
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∴ ∠DCS = ∠CBS + ∠CSB
… (Remote interior angle
theorem)
… (2)
From (1) and (2), .
∠CBS = ∠CSB = ½∠DCS
… (3)
Now, it is given that Ray CE bisects
∠DCS.
∴ ∠DCE = ∠ECS = ½∠DCS
… (4)
89

90.
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From (3) and (4), ∠CSB = ∠ECS
∴ seg BS || seg CE
… (Alternate angles test
for parallel lines)
seg SE || seg BC … (given)
∴ Both the pairs of opposite sides
of
∴
CBSE are parallel.
CBSE is a parallelogram.
90

91.
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19. In the figure,
LAXM is a parallelogram. Point I is
the midpoint of diagonal LX. PQ is a line passing
through I. P and Q are the points of intersection
with the sides LA and MX respectively.
Prove that seg PI
Proof
:
seg IQ.
In LPI and XQI,
∠LIP & ∠XIQ are vertically
opposite angles
∴∠LIP
∠XIQ
seg LI
seg XI
… (As I is the midpoint of LX)
91

92.
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∠PLI
∠QXI
… (Alternate angles)
∴ LPI
XQI
∴ seg PI
seg QI
Hence proved.
92
… (ASA test)
… (c.s.c.t.)

93.
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20.
MNOP is a rhombus. Q is a point in the interior
of the rhombus such that QM = QO. Prove that Q
lies on diagonal NP.
Proof :
In MPN and OPN,
seg PM
seg PO and
seg MN
seg ON
… (sides of a rhombus)
seg PN
sg PN … (common side)
∴By SSS test
∴ MPN
OPN
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94.
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∴ ∠MPN
∠OPN
… (c.a.c.t.)
i.e. diagonal PN is the bisector of
∠MPO … (1)
In MPQ and OPQ,
seg MP
seg OP
… (sides of a rhombus)
seg QM
sg QO
… (given)
seg PQ
seg PQ … (common
side)
∴ By SSS test
∴ MPQ
OPQ
∴∠MPQ
∠OPQ
94
… (c.a.c.t.)

95.
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i.e. seg PQ is the bisector of ∠MPO
… (2)
Point Q is in the interior of
MNOP.
… (3)
The bisector of an angle is unique.
∴From (1), (2) and (3) bisectors PN
and PQ of ∠MPO are one and the
same.
∴ Point Q lies on the diagonal NP.
Hence proved.
95

96.
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21.
ABCD is a square. P and Q are the points such
that seg AQ
seg DP.
Prove that seg AQ ⊥ seg DP.
Proof :
In DAP and ABQ,
∠DAP
∠ABQ
… (Each is a right angle)
It is given that
hypotenuse DP
hypotenuse AQ
side DA ≅ side AB
… (side of a square)
∴By hypotenuse-side theorem,
96

97.
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∴ DAP
ABQ
∴ ∠DPA
∠AQB
… (c.a.c.t.)
and ∠ADP
… (1)
∠BAQ
… (c.a.c.t.)
… (2)
From (1), Assume
∠DPA = ∠AQB = α
… (3)
From (2), Assume
∠ADP = ∠BAQ = β
97
… (4)

98.
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Then, In DAP,
α+β+
DAP = 180º
… (Angles of a triangle)
∴ α + β + 90º = 180º
… (∠DAP = 90º)
∴ α +β = 90º
… (5)
Now, In APT,
α + β + ∠ATP = 180º
… (Angles of a triangle)
∴ 90º + ∠ATP = 180º
… [From (5)]
∴ ∠ATP = 90º
98

99.
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Point T is the point of
intersection of seg AQ and seg
PD.
∴ seg AQ ⊥ seg DP.
Hence proved!!
99

100.
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22.
ABCD is a kite. AB = AD and CB = CD. Prove
that
i.
diagonal AC ⊥ diagonal BD.
ii. diagonal AC bisects diagonal BD.
Proof :
We are given, AB = AD
∴ Point A is equidistant from points
B and D of seg BD.
Also, CB = CD
… (1)
… (given)
∴ Point C is equidistant from points B
and D of seg BD.
100
… (2)

101.
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From (1) and (2), points A and C are
equidistant from points B and D of
seg BD.
Therefore by perpendicular bisector
theorem, AC is the perpendicular
bisector of BD.
i.e.
i. diagonal AC ⊥ diagonal BD and
ii. diagonal AC bisects diagonal BD.
Hence proved!!
101

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23. Let points A and B be on one side of line ℓ. Draw
seg AD ⊥ line ℓ and seg BE ⊥ line ℓ. Let point C
be the midpoint of seg AB. Prove that
seg CD
seg CE.
Construction :
Draw seg CM ⊥ lin l.
Proof
Refer above fig.
:
We can have,
Seg AD, seg BE and seg CM
are perpendiculars to line l
∴ seg AD || seg BE || seg CM.
102

103.
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Seg CA
seg CB
… (As C is the
midpoint of seg AB)
Now, Seg CA and seg CB
are the intercepts made by
three parallel lines AD, BE
and CM.
As DM & ME are intercepts
made by the same parallel
lines on other transversal,
we have
∴ intercept DM
103
intercept ME

104.
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i.e. seg DM
seg ME
… (1)
In CDM and CEM,
seg DM
seg EM
… [From (1)]
∠CMD
∠CME
… (Each a right
angle : construction)
seg CM
seg CM
… (common side)
∴ CDM
CEM
… (SAS test)
∴ seg CD
104
seg CE … (c.s.c.t.)

105.
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24. □ABCD is a parallelogram. P, Q, R and S are the
points on sides AB, BC, CD and DA respectively
such that seg AP
Prove that
Proof :
seg BQ
seg CR
seg DS.
PQRS is a parallelogram.
AB = AP + PB
… (A-P-B)
… (1)
CD = CR + RD
… (C-R-D)
… (2)
side AB
side CD
… (opposite sides of a parallelogram)
∴ AB = CD
… (3)
105

106.
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From (1), (2) and (3),
AP + PB = CR + RD
… (4)
seg AP
… (given)
seg CR
∴ AP = CR
… (5)
From (4) and (5), PB = RD
∴ seg PB
seg RD
… (6)
In PBQ and RDS,
seg PB
seg RD
… [From (6)]
Now, ∠B &∠D are opposite angles of a
parallelogram
∴∠B
∠D
106

107.
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seg BQ
seg DS
… (given)
∴By SAS test
∴ PBQ
RDS
∴ seg PQ ≅ seg RS … (c.s.c.t.)
Similarly, seg PS
proved.
seg QR can be
… (8)
From (7) and (8), opposite sides of
PQRS are congruent.
∴
PQRS is a parallelogram.
Hence proved!!
107

108.
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108