6.8 Parallelogram:
6.8.1
Construction of parallelogram
We have seen earlier (6.7.1) that
to construct unique quadrilaterals, we require five elements
 a suitable combination of sides, diagonals
and angles as given below
No. 
No of sides given 
Number of diagonals given 
Number of angles given 
Total Number of elements required 
1 
2 
2 
1 
5 
2 
2 
1 
2 
5 
3 
4 
1 
 
5 
4 
4 
 
1 
5 
5 
3 
 
2(Included) 
5 
6 
3 
2 
 
5 
7 
2(Adjacent) 
 
3 
5 
Since parallelogram has a special
property of opposite sides being parallel and opposite angles being equal,
we just need three elements to construct a unique parallelogram as
given below
No. 
No of sides given 
Number of diagonal given 
Number of angles given 
Total Number of elements required 
1 
2 
1 
 
3 
2 
2 
 
1 
3 
3 
 
2 
1(Intersecting ) 
3 
4 
1 
2 
 
3 
6.8.1.1 Construction of
a parallelogram given the length of 2 adjacent sides and a diagonal.
In a parallelogram, opposite sides
are equal. If lengths of 2 adjacent sides are given, we can arrive at the
lengths of all sides.
Thus, the construction is similar to that of a
quadrilateral with 4 sides and 1 diagonal given
(Refer 6.6.1).
6.8.1 Exercise 1: Construct a parallelogram whose adjacent sides
are 5cm and 3cm and diagonal is 6cm. (The sides are 5cm, 3cm, 5cm and
3cm and diagonal is 6cm) 

6.8.1.2. Construction
of a parallelogram given the length of 2 adjacent sides and an angle.
In a parallelogram, opposite sides
are equal. If lengths of 2 adjacent sides are given, we can arrive at the
lengths of all sides.
Thus, the construction is similar to that of a
quadrilateral with 4 sides and 1 angle given (Refer
6.6.1).
6.8.1 Exercise 2: Construct a parallelogram PQRS, given PQ=5cm, QR=4cm and _{}PQR = 70^{0}.^{} (The sides are PQ=5cm, QR=4cm,
RS=5cm and PS=4cm and _{}PQR = 70^{0}) 

6.8.1.3. Construction
of a parallelogram given the length of two diagonals and intersecting angles
between them.
6.8.1 Problem 1: Construct a parallelogram whose diagonals are 4cm and 5cm and the angle between them is = 70^{0}.^{}
First draw a rough diagram of ABCD
with AC and BD as diagonals.
Step 
Construction 

1 
Mark a point A and draw a line
though A 

2 
From A, draw an arc of radius
4cm to cut the above line at C
(AC=4cm) 

3 
Bisect AC at O (From A and C, draw arcs of
radius more than half the length of AC on both sides, Let they cut at X and Y.
XY is the perpendicular bisector of AC. It cuts AC at O 

4 
From O, draw a line at an angle
70^{0} to AC on both sides 

5 
From O, draw an arc of radius
2.5cm on both sides of AC to cut the above line at B and D. Join AB, BC, CD
and DA. 
ABCD
is the required parallelogram.
6.8.1.4. Construction
of a parallelogram given the length of one side and two diagonals.
With the given side as
the base, construct a triangle (with half of diagonals as other two sides) by
using the property that diagonals of a
parallelogram bisects each other.
Then extend these other two sides.
6.8.1 Exercise 3: Construct a parallelogram ABCD with BC=4.5cm and diagonals
AC=4cm and BD=5.0cm. In the figure of 6.8.1 Problem 1
use the property AO=OC and BO=OD to construct triangle BCO with BC= 4.5cm,
CO=2cm and BO=2.5cm and then extend CO and BO such that OA = 2cm and
OD=2.5cm. Join CD,DA,AB 

6.8.1.5. Construction
of a parallelogram given the length of adjacent sides and the height of
parallelogram.
6.8.1
Exercise 4: Construct a parallelogram ABCD,
with adjacent sides AB=4 cm, BC = 5cm and height corresponding to BC =3.5 cm
Step 
Construction 

1 
Mark a point B and draw a line
though B 

2 
From B, draw an arc of radius
5cm to cut the above line at C (Hence BC=5cm) 

3 
Draw 2 perpendicular lines XY
and PQ on BC at any two points U and T
on it. 

4 
From U and T cut PQ and XY at S and R by an arc of radius 3.5 cm Join RS ( Note RS is at a
distance of 3.5cm from BC) 

5 
Cut the line RS at A by an arc
of radius 4cm from B (Hence BA=4cm) 

6 
From A cut the line RS at D by
an arc of radius 5cm ( Hence AD=5cm) 
ABCD
is the required parallelogram.
6.8.2
Area of parallelogram
ABCD is a parallelogram with
ABCD, BCAD and AB=CD, AD=BC. From D and C draw perpendiculars
to AB (extend if necessary). Let them meet AB at E and F. DE and CF are the altitudes of
the parallelogram. Since ABCD, altitudes are equal (DE=CF). The triangles _{}ADE and _{}BFC are congruent triangles (_{}AD=BC, DE=CF and _{}DEA = _{}BFC =90^{0} – SAS Postulate) and hence AE=BF and hence Area
of _{}ADE = Area of _{}BFC. _{}Area of ABCD = Area of _{}ADE + Area of quadrilateral DEBC = Area of _{}BFC + Area of quadrilateral DEBC = Area of rectangle DEFC =
Length * breadth = DC * h Area of the
parallelogram = base*height 

6.8.2 Problem 1: ABCD is a parallelogram
with AB=24cm and AD=16cm. The distance between AB and DC is 10cm. Find the
distance between AD and BC.
Solution:
The distance between AB and DC
is the height (DE). Area of the parallelogram ABCD =
base*height = 24*10 = 240sq cm This is also the area of the
same parallelogram with AD as the base and AH as the height. _{} Area of ABCD = AD*AH
= 16*AH But area = 240sqcm _{} 16*AH = 240 I.e. AH = 240/16 = 15cm 

6.8.2 Problem 2: A rectangle and a
parallelogram have equal areas. The sides of the rectangle are 10m and 14m.
The base of the parallelogram is
20m. What is the altitude of the parallelogram?
Solution:
We know that the Area of
rectangle = side*side = 10*14 = 140sqm Area of parallelogram = Base*Altitude
= 20*Altitude Since both the areas are same Area of parallelogram =140sqm _{} 20*Altitude = 140 i.e. Altitude = 140/20 = 7mts 

6.8.2 Problem 3: A triangle and a
parallelogram have equal areas and equal bases. What is the ratio of their
altitudes?
Solution:
If B_{P} and H_{P}
are base and altitude of the parallelogram, then the area of parallelogram = Base
*Altitude = B_{p}*H_{p} If B_{T} and H_{T}
are base and altitude of the triangle, then the area of triangle = 1/2(Base *Altitude) = 1/2(B_{T}*H_{T}). Since the areas are same, B_{p}*H_{p}=1/2(B_{T}*H_{T}).
Since the bases are same, B_{P}=
B_{T}. We have H_{p}=1/2H_{T} i.e. 2H_{p}=H_{T} Therefore the altitude of the
triangle is twice the altitude of the parallelogram. 

6.8.2 Problem 3: Have you observed your
mother, grandmother or those who make sweets cutting .Burfis in the
shape of parallelogram than rectangles?
(Did they
study geometry?)
For
example let the sides of rectangle and parallelogram be 13 and 12 units. Area
of rectangle = base * another side = 12*13= 156 square units Area
of parallelogram = base * height =
12*12= 154 square units (
we used the pythogoras theorem to find the height of parallelogram (12^{2}= 13^{2}5^{2}) Thus
we notice that area of parallelogram
< area of rectangle 

Though the sides of rectangle and parallelogram are
same, area of rectangle is more than the area of parallelogram. Hence in the
same spread (area)
they can cut more number of
pieces in the shape of parallelograms than in the shape of rectangles.
6.8.3
Construction of Rhombus:
We have seen that, to construct a
parallelogram we need three elements (suitable combination of sides, diagonals
and angles).
Because of the special property of
rhombus, only two elements are enough to construct a rhombus uniquely.
1. Length of two
diagonals are given .
6.8.3 Problem 1: Construct a rhombus PQRS with PR
= 5cm and SQ=4cm
First draw a rough diagram
Step 
Construction 

1 
Mark a point P and draw a line
though P 

2 
Draw an arc of radius 5cm to cut
the above line at R (PR=5cm) 

3 
Bisect PR at O (From P and R, draw arcs of
radius more than half the length of PR, on both sides of PR. Let they cut at X and Y. (XY is the perpendicular
bisector of PR) 

4 
From O draw an arc of radius 2cm
to cut OX at S(OS=2cm) 

5 
From O draw an arc of radius 2cm
to cut OY at Q(OQ=2cm). Join PQ, QR, RS and SP. 
PQRS
is the required rhombus.
2. Lengths of one side
and one diagonal are given
6.8.3 Problem 2: Construct a rhombus ABCD with AB
= 3cm and AC=4cm
First draw a rough diagram
Step 
Construction 

1 
Mark a point A and draw a line
though A 

2 
Draw an arc of radius 4cm to cut
the above line at C (AC=4cm) 

3 
From A, draw arcs of radius 3cm
on both sides of AC 

4 
From C, draw arcs of radius 3cm
on both sides of AC to cut the above arcs at D and
B. Join AB, BC, CD and DA. 
ABCD
is the required rhombus
3. Lengths of one side
and one angle are given
6.8.3 Problem 3: Construct a rhombus ABCD with AB
= 3cm and _{}ABC = 120^{0}.
First draw a rough diagram
Step 
Construction 

1 
Mark a point B and draw a line
though B 

2 
Draw a line through B at an angle of 120^{0} with BC 

3 
From B, draw an arc of radius
3cm to cut the above line at A (BA=3cm and _{}ABC = 120^{0}) 

4 
From A and C, draw arcs of
radius 3cm to meet at D. Join AD and DC. 
ABCD
is the required rhombus
6.8.4
Area of Rhombus:
We have seen(in 6.6) that a
diagonal of a rhombus cuts the rhombus into 2
congruent triangles. We shall use this property to
calculate the area of rhombus, as we already know how to calculate the area of a
triangle. In the rhombus PQRS, draw the
diagonal PR and QS. As PQRS is a
rhombus, diagonals bisect each other perpendicularly (_{}SOR = _{}ROQ = 90^{0}) OS and OQ are the altitudes of _{}PRS and _{}PRQ_{} Area of _{}PRS= ½(base*height) =1/2(PR*OS) Area of _{}PRQ = ½(base*height) = 1/2(PR*OQ) _{}Area of PQRS = Area of _{}PRS + Area of _{}PRQ = 1/2(PR*OS) + 1/2(PR*OQ) =
1/2*PR* (OS+OQ) =1/2*PR*QS sq units Area of a rhombus = ½ *
product of diagonals 

Note : This formula was also
given by Bhaskracharya (Lilaviti Shloka 176)
6.8.4 Problem 1: The longer diagonal of
a rhombus is greater than the other diagonal by 10mts. The sum of diagonals is
32mts. Find its area.
Solution:
Let the shorter diagonal be x. Since the longer diagonal is
greater than the other diagonal by 10mts, its length will be x+10. Since the sum of diagonals is 32m. We have x+(x+10) = 32 i.e. 2x+10 =32 i.e. 2x=3210 =22 _{} x=11mts Hence its diagonals are 11 and 21(=11+10)
meters. We know that Area of a rhombus = 1/2
* product of diagonals = 1/2*11*21 = 231/2 sq mts 

6.8.4 Problem 2: The perimeter of a rhombus is
40cm and one of its diagonal is 16cm. Find the other diagonal and area of
rhombus.
Solution:
Let x be the side. Since perimeter
is sum of all sides and all sides are equal in a rhombus: We have 4x = 40cm _{} x=10cm Thus in As PQRS is a rhombus, diagonals
bisect each other perpendicularly ( PR=16cm(given) _{} POQ is a right angled triangle
with PQ=10 and By Pythagoras theorem PQ^{2}= PO^{2}+OQ^{2} _{} OQ^{2} =PQ^{2}PO^{2}
= 10^{2}8^{2} = 10064 =36= 6^{2}_{} OQ = 6 Since OQ=OS, we have OS=6cm and
hence QS =6+6=12cm Area = 1/2* product of diagonals
= 1/2* 16*12= 96 sq cms 

6.8.5
Construction of Trapezium:
Trapezium can be thought of as a figure
got by merging a triangle with parallelogram.
We have seen that to construct a
quadrilateral we require five elements. Since trapezium has one pair of opposite
sides parallel,
we require a maximum of four
elements to construct a trapezium uniquely.
6.8.5.1. When length of two sides and two angles are given
6.8.5 Problem 1: Construct a trapezium ABCD with
ABCD, AB=5cm, CD=3cm and (_{}DAB = 70^{0} _{}ABC = 50^{0})
First draw a rough diagram.
Step 
Construction 

1 
Mark a point A and draw a line
though A 

2 
From A, draw an arc of radius
5cm to cut the above line at B (AB=5cm) 

3 
From A, draw an arc of radius
2cm (=ABCD) to cut AB at E (AE=2cm) 

4 
Draw a line through A at an
angle of 70^{0} to AB 

5 
Draw a line through E at an
angle of 50^{0} to AB to cut
the above line at D 

6 
Draw a line through B at an
angle of 50^{0} to AB. 

7 
From D, draw an arc of radius
3cm to cut the above line at C (DC=3cm and _{}ABC = 50^{0}). Join DC. 
ABCD
is the required trapezium.
Note: In step 4, we used the
property of corresponding angles being equal when a transversal cuts two
parallel lines (
6.8.5. 2. When lengths of parallel sides and the altitude are
given.
6.8.5 Problem 2: Construct a trapezium PQRS with
PQSR, PQ=5cm, SR=6cm and PS=3cm
First draw a rough diagram.
Step 
Construction 

1 
Mark a point P and draw a line
though P 

2 
From P, draw an arc of radius
5cm to cut the above line at Q (PQ=5cm) 

3 
Draw a perpendicular from P (altitudes
form 90^{0}) 

4 
From P, draw an arc of radius
3cm to cut the above line at S (PS=3cm) 

5 
Draw a perpendicular from S (parallel
to PQ) 

6 
From S, draw an arc of radius 6cm
to cut the above line at R (SR=6cm) 
6.8.5. 3. When lengths of all sides are given.
Since trapezium can be
thought of as consisting of a triangle and Parallelogram first construct a
triangle of three sides like what is explained in 6.8.5 Problem1.
6.8.5 Exercise 1: Construct a trapezium ABCD with ABCD, AB=7cm,
DC=5cm, AD=2cm and BC= 2.5cm Hint: Draw a rough diagram as in the figure on the right
side. Mark E on AB such that AE=ABCD Construct triangle AED such that
AD =2cm, DE =BC=2.5cm From B draw a line parallel to
ED and mark C such that BC=2.5cm, Join D and C ABCD is the required trapezium 

6.8.5 Problem 3: Construct an isosceles trapezium ABCD with
CDAB, AB=7cm, AD=2cm and DC=5cm Since in an isosceles trapezium
non parallel sides are equal, we have BC=AD. Thus, we have to construct a
trapezium ABCD with four sides given (DA=2cm, AB=7cm, BC=2cm and
CD=5cm). Hint: As in 6.6.8 Problem 1,
first we need to construct a triangle AED with AE = ABDC = 75 = 3cm 1. With AD=2cm (given), AE = 3cm
(construction) and DE = CB = 2cm (parallelogram), we should be able to
construct the triangle ADE. 2. From B, draw an arc of radius
2cm and from D draw an arc of radius 5cm to cut at point C. Join DC and CB. 3. ABCD is the isosceles
trapezium. 

6.8.6
Area of Trapezium:
ABCD is a trapezium with BCAD
and AB=CD. Draw perpendiculars from A and D
to the base BC to cut BC at E and F respectively. Since BC and AD are parallel,
the altitudes AE and DF are equal (the distance between parallel lines are
always same) Note that AEFD is a rectangle.
Therefore EF =AD and hence EF = 1/2EF+1/2EF=1/2(EF+AD) Area of trapezium ABCD = Area of
_{}ABE + Area of rectangle AEFD+ Area of _{}DFC = 1/2(base*height)+ (side*side)+
1/2(base*height) = 1/2(BE*height)+ (EF*height)+
1/2(FC*height) = (BE/2 + EF + FC/2)*height = (BE/2+(EF+AD)/2+FC/2)*height (_{}EF=AD and hence EF = 1/2( EF+AD)) = 1/2(BE+EF+AD+FC)*height = 1/2(BE+EF+FC+AD)*height = 1/2(BC+AD)*height Area of trapezium =
half the product of height and sum of parallel sides 

6.8.6 Problem 1: The parallel sides of a trapezium
are in the ratio of 2:1. If the distance between the parallel sides is 6cm and
the area is 135 sq cm,
find the length of parallel sides.
Solution:
It is given that the height of
the trapezium is 6cm. Since parallel sides of the
trapezium are in the ratio of 2:1, let the sides be 2x and 1x. We know Area of trapezium = half the
product of height and sum of parallel sides = 1/2(6*(2x+x)) = 3*3x = 9x We are also given that the area
is 135sq cm _{} 9x=135 _{} x= 15cm: 2x =30cm _{} The parallel sides
of trapezium are 15cms and 30cms. 

6.8.6 Problem 2: The area of a trapezium is 204sq
cm. Its altitude is 17cm and one of the parallel sides is 16cms. Find the other
side.
Solution:
Let x be the other side Area of trapezium = half the
product of height and sum of parallel sides = 1/2(17*(16+x)) We are given that the area is
204 sq cm _{}1/2(17*(16+x)) =204 i.e. 17 (16+x) =408 i.e. 16+x =408/17 =
24 i.e. x = 2416= 8cm The other side is 8cm Verification Area of trapezium =
1/2*17*(16+8) = 17*12 = 204 sq cm which is as given in the problem. 

6.8.7 Theorems on parallelograms:
6.8.7 Theorem
1: The diagonals of a parallelogram
bisect each other.
Data: ABCD is a parallelogram. The
diagonals AC and BD meet at O.
To prove: AO=OC and BO=OD
Proof:
Steps 
Statement 
Reason 

1 
AB = CD 
Opposite sides of parallelogram 

2 
_{}BAO = _{}OCD 
Alternate angles of ABCD, AC is transversal 

3 
_{}ABO = _{}ODC 
Alternate angles of ABCD, BD is transversal 

4 
_{}ABO _{} _{}CDO 
ASA Postulate 

5 
AO=OC and BO=OD 
Corresponding sides of congruent triangles 
This proves that the diagonal of a
parallelogram bisect each other.
6.8.7 Theorem
2: Each diagonal divides a
parallelogram in to two congruent triangles.
Data: ABCD is a parallelogram, AC
is a diagonal
To prove: _{}ABC _{} _{}ACD
Proof:
Steps 
Statement

Reason 

1 
AB = CD 
Opposite sides of parallelogram 

2 
BC = AD 
Opposite sides of parallelogram 

3 
AC is common 


4 
_{}ABC _{} _{}ACD 
SSS Postulate 
This
proves that the diagonal divides the parallelogram in to two congruent
triangles.
Corollary means a result/effect of
a main event.
For example,
August 15^{th} is
celebrated as the independence day of
for schools/colleges/offices.
The main event here is
Independence Day. Corollary to that is, it is a holiday.
Similarly, based on Theorems we
derive Corollaries.
6.8.7 Corollary 1:
In a parallelogram, if one angle is a right angle then it is a rectangle.
Given: ABCD is a parallelogram. ABC = 90^{0}
To Show: ABCD is a rectangle
Steps 
Statement

Reason 

1 
_{}ABC = 90^{0} 
Given 

2 
_{}ABC+_{}BCD = 180^{0} 
Sum of two consecutive angles is
180^{0} 

3 
_{}BCD = 90^{0} 
Substitution in step 2 for ABC 

4 
_{}CDA =_{}ABC 
Opposite angles of a
parallelogram 

5 
_{}_{}CDA = 90^{0} 


6 
_{}BAD= _{}BCD 
Opposite angles of a
parallelogram 

7 
_{}_{}BAD = 90^{0} 

Hence
ABCD is a rectangle
6.8.7 Corollary 2:
In a parallelogram, if all the sides are equal and all the angles are
equal, then it is a square.
Given: ABCD is a parallelogram and AB=BC=CD=DA.^{}
To Show: ABCD is a square
Steps 
Statement

Reason 

1 
_{}BAD+_{}ADC = 180^{0} 
Sum of two consecutive angles is
180^{0} 

2 
2_{}BAD = 180^{0} 
Given that all the angles are
equal 

3 
_{}_{}BAD=_{}ADC=90^{0} 


4 
_{}ABC+_{}BCD = 180^{0} 
Sum of two consecutive angles is
180^{0} 

5 
2_{}ABC = 180^{0} 
Given that all the angles are
equal 

6 
_{}_{}ABC=_{}BCD=90^{0} 


7 
_{}All are right angles 
Since it is also given that all
sides are equal, ABCD is a square.
6.8.7 Corollary 3:
The diagonals of a square are equal and bisect each other
perpendicularly.
Given: ABCD is a square hence AB=BC=CD=DA and _{}ABC =_{}BCD=_{}CDA=_{}DAC=90^{0}
To Show: AC=BD,AO=CO,BO=DO,_{}AOB=_{}BOC = 90^{}
Steps 
Statement 
Reason 


Consider _{}ABC and _{}BCD 

1 
AB=CD 
In a square, all sides are equal 

2 
BC is common 


3 
_{}ABC=_{}BCD = 90^{0} 
In a square all angles are right angles 

4 
_{}ABC _{} _{}BCD 
SAS Postulate 

5 
_{}AC=BD 
Corresponding sides are equal 


Consider _{}ABO and _{}OCD 

6 
AB=CD 
In a square, all sides are equal 

7 
_{}ABO =_{}ODC 
Alternate angles of ABCD, BD is transversal 

8 
_{}BAO=_{}OCD 
Alternate angles of ABCD, AC is transversal 

9 
_{}ABO _{} _{}OCD 
ASA Postulate 

10 
_{}AO=OC,BO=OD 
Corresponding sides are equal 


Consider _{}ABO and _{}OCB 

11 
AB=BC 
In a square, all sides are equal 

12 
AO=OC 
BO bisects AC (Step 9) 

13 
BO is
common 


14 
_{}ABO _{} _{}OCB 
SSS Postulate 

15 
_{}AOB=_{}BOC 
Corresponding angles are equal 

16 
_{}AOB+_{}BOC=180^{0} 
Sum of angles on a straight line is 180^{0} 

17 
2_{}AOB=180^{0} 


18 
_{}AOB=90^{0}=_{}BOC 
(Step 15) 
Hence the diagonals of a square
bisect each other at right angles.
6.8.7 Corollary 4:
The straight line segments joining the extremities of two equal and
parallel line segments on the same side are equal and parallel.
Given: AB=CD and ABDC and A is
joined with D and B is joined with C
To Show: AD=BC and ADBC
Construction: join AC.
Steps 
Statement 
Reason 

1 
AB=CD 
Given 

2 
_{}DAC=_{}ACB 
Alternate angles of ABDC, AC is transversal 

3 
AC is common 


4 
_{}ABC _{} _{}ACD 
SAS Postulate 

5 
_{}AD=BC 
Corresponding sides are equal 

6 
_{}DAC=_{}ACB 
(Step 2) 
Since DAC and ACB are alternate
angles with respect to the lines AD and BC with AC as a transversal, ADBC.
6.8.7 Problem 1: In the adjoining adjacent figure, ABCD is a
parallelogram. P is the mid point of BC. Prove that AB=BQ
Given: ABCD is a parallelogram
with P as mid point of BC
To Show: AB=BQ
Steps 
Statement

Reason 

1 
BP = PC 
Given that P is mid point of BC 

2 
_{}BPQ=_{}CPD 
Vertically opposite angles 

3 
_{}PBQ=_{}PCD 
Alternate angles as ABDC 

4 
_{}BPQ _{} _{}CDP 
ASA Postulate 

5 
BQ = DC 
Corresponding sides are equal 

6 
DC = AB 
Opposite sides of parallelogram
are equal 

7 
_{}BQ = AB 
From 5 and 6 
6.8.7 Problem 2: In the adjoining figure, the bisectors of angles of
a parallelogram ABCD enclose PQRS. Prove that PQRS is a rectangle.
Given: ABCD is a parallelogram.
AP, BP, CR and DR are the angular bisectors of angles A, B, C and D
respectively.
To Show: PQRS is a rectangle.
Steps 
Statement

Reason 

1 
_{}DAB+_{}ADC= 180^{0} 
Sum of consecutive angles of a parallelogram =180^{0} 

2 
_{}DAB=2_{}DAS 
AP is bisector of _{}DAB 

3 
_{}ADC=2_{}ADS 
DR is bisector of _{}ADC 

4 
_{} _{}DAS+_{}ADS = 90^{0} 
Substituting values of angles in
Step 1, from 2 and 3 

5 
_{}DAS+_{}ADS+ _{}ASD= 180^{0} 
Sum of all angles in a triangle (_{}ADS ) is 180^{0} 

6 
_{}_{}ASD= 180^{0 }(_{}DAS+_{}ADS) 
Transposition 

7 
= 180^{0}90^{0 }=
90^{0} 
Substitute value from step 4 

8 
_{}PSR =_{}ASD 
Vertically opposite angles are
equal 

9 
= 90^{0} 


10 
_{}DAB +_{}ABC =180^{0} 
Consecutive angles in a
parallelogram 

11 
1/2_{}DAB +1/2_{}ABC =90^{0} 


12 
_{}PAB +_{}ABP =90^{0} 
AP bisects A and BP bisects B 

13 
_{}_{}SPQ =90^{0
} 
Sum of all angles in _{}APB = 180^{0} 
Similarly we can show that _{}PQR=_{}QRS=90^{0}
Hence PQRS is a rectangle.
6.8.7 Problem 3: PQRS is a parallelogram. PS is extended (produced)
to M so that SM = SR and MR is extended to meet PQ extended at N.
Prove that QN=QR.
Given: PQRS is a parallelogram.
SM=SR
To Show: QN=QR
Steps 
Statement

Reason 

1 
_{}SRM =_{}SMR 
SM=SR and hence SRM is an
isosceles triangle 

2 
_{}MSR =_{}SPQ 
Corresponding angles(SRPQ) 

3 
_{}SPQ=_{}RQN 
Corresponding angles(PSQR) 

4 
_{}_{}MSR=_{}RQN 
Equating Step 

5 
_{} _{}SRM =_{}QNR 
Corresponding angles(SRPN) 

6 
_{}SMR=_{}QRN 
Since two angles in _{}MSR and _{}NQR are equal, third
angle also has to be equal (sum of 3 angles in a triangle = 180^{0}) 

7 
_{}QNR=_{}SRM=_{}SMR =_{}QRN 
From Step4, Step1, Step5 

8 
_{}QN=QR 
From Step 6, we conclude that _{}QNR is an isosceles triangle 
6.8.7 Problem 4: ABCD is a parallelogram. The bisectors of _{}A and _{}B meet BC and AD at X and Y respectively. Prove that XY=CD.
Given: ABCD is a parallelogram. AX
bisects_{}A and BY bisects_{}B. Let AX and BY meet at O
To Show: XY=CD
Consider _{}ABX and _{}AXY
Data : AD BC and hence AYBX,
AB=CD,
Steps 
Statement

Reason 

1 
_{}BAX = _{}AXY 
Alternate angles AYBX and AX
is transverse 

2 
_{}XAY = _{}AXB 
Alternate angles AYBX and AX
is transverse 


Consider _{}BAY and _{}XYB 

3 
_{}BAY = _{}YXB 
Sum of angles in step 1 and step
2 

4 
BY is common side 


5 
_{}AYB = _{}YBX 
Alternate angles AYBX and BY
is transverse 

6 
_{}BAY _{}_{}XYB 
ASA Postulate 

7 
AB =XY, AY=BX 
Corresponding sides, Step 6 

8 
XY =CD 
AB=CD(given), step 7 
6.8.7 Problem 5: ABCD is a parallelogram with ABCD. P is the mid
point of AB and CP bisects _{}BCD. Prove that _{}CPD =90^{0}
Hint: If we can prove that _{}PDC+ _{}PCD =90^{0} then it follows that _{}CPD =90^{0} ^{ } 1. Note that PBC is an isosceles triangle (_{}DCP = _{}CPB = _{}PCB as ABCD and CP is a transverse)^{} 2. Prove that ADP is an
isosceles triangle (AD=BC, AP=PB and PB=BC and hence AD=AP) 3. DP bisects _{}ADC (ABCD and DP is a transverse) 4. _{}ADC + _{}DCB = 180^{0} (Sum of 2 consecutive angles of a
parallelogram) 5. 2(_{}PDC+ _{}PCD) = 180^{0}(follows from step 4) 6. _{}CPD =90^{0 }(sum of all angles in a triangle = 180^{0}) 

6.8.7 Theorem 3: Parallelograms standing on the same
base and between same parallel lines have equal areas.
Data: ABCD and ABEF are two
parallelograms standing on the same base AB and between same parallel lines PQ
and RS
To prove: Area of ABCD = Area of
ABEF.
Proof:
Steps 
Statement 
Reason 


Consider _{}FAD and _{}EBC 

1 
AF = BE 
Opposite sides of ABEF. 

2 
_{}AFD = _{}BEC, _{}ADF = _{}BCE, 
Corresponding angles of
AFBE and ADBC, PQ is transversal 

3 
_{}FAD = _{}EBC 
When two angles of triangles are equal, third angle also has to be
equal 

4 
_{}FAD _{} _{}EBC 
ASA Postulate 

5 
Area of _{}FAD = Area of _{}EBC 
Congruent triangles have equal area. 

6 
_{}Area of _{}FAD+ Area of DEBA = Area of _{}EBC+ Area DEBA 
Adding area of quadrilateral DEBA to both sides. 

7 
i.e. Area of FEBA = Area of ABCD 
If equals are added to equals, then the resulting sums are also
equal. 
This proves the theorem.
Note: Earlier we have seen that the
area of parallelogram is product of base and its altitude (height). Since PQ
and RS are parallel lines, the altitudes
of FEBA and ABCD are equal. Since
the parallelograms, FEBA and ABCD have the same base AB, it follows that the
parallelograms have same area.
6.8.7 Corollary 1:
Parallelograms standing on equal base and between same parallel lines
have equal areas.
Note: We know that the area of parallelogram is product of base and its
altitude (height)(Refer 6.8.2) Since PQ and RS are parallel
lines, the altitudes of EFGH and ABCD
are equal. Since the parallelograms, EFGH and
ABCD have equal bases (AB=EF), it follows that the
parallelograms have same area. 

6.8.7 Corollary 2:
If a parallelogram and a triangle stand on the same base and between
same parallel lines, then the area of triangle is half the
area of the parallelogram.
Note:
We know that the area of parallelogram
(ABCD) is product of its base and its
altitude (height) (=AB*h).*(Refer 6.8.2) We have also proved that diagonal
divides a parallelogram in to two congruent triangles (refer 6.8.7 Theorem 2) _{}ADB _{} _{}CDB _{}Area of parallelogram ABCD = area of _{}ADB + area of _{}CDB = 2 *area of _{}ADB _{} Area of _{}ADB = half the area of parallelogram. =1/2( AB*h) = 1/2 base *height 

Note : So far we have been
using the formula for area of a triangle = 1/2 base *height, without proof.
In the process of
proving above corollary, we have also arrived at the formula for calculation
for the area of triangle.
6.8.7 Corollary 3:
Triangles standing on the same base and between same parallel lines are
equal in area.
We know that the area of a
triangle is half the product of its base and its altitude _{}Area of_{}ABC = (1/2)*base*height = (1/2)*AB*h In addition, we know that the
distance (h) between any two parallel lines is always fixed. Hence _{}ABD has the same height ‘h’ as _{}ABC. _{}Area of _{}ABD = 1/2(AB*h) _{}Area of _{}ABC= Area of _{}ABD 

Note: We can also prove the above by
constructing two parallelograms one  AC and another BD with AB as common
base and using theorem 6.8.7
Theorem 3 and 6.8.7 Corollary 2.
6.8.7 Corollary 4:
Triangles standing on equal base and between same parallel lines are
equal in area.
(By formula, area of triangle is
half of its base and height).
6.8.7 Problem 6: In the given figure DEBC, prove that the area of _{}BOD = Area of _{}COE
Given: DE BC
To Show: Area of _{}BOD= Area of _{}COE
Steps 
Statement

Reason 

1 
Area of _{}BCD= Area of _{}BCE 
BC is the common base for both _{}BCD and _{}BCE. Both triangles have same height as DEBC. 

2 
Area of_{}BCD Area of _{}BOC =Area of _{}BCE Area of _{}BOC 
The common area of triangle BOC
is subtracted from the areas of both
the triangles of Step 1 

3 
_{} Area of _{}BOD= Area of_{}COE 

6.8.7 Problem 7: In the given figure, D and E are the points on the
sides AB and AC respectively, such that the area of_{}BCE = Area of_{}BCD.
Show that DEBC
Given: Area of _{}BCE = Area of _{}BCD
To Show: DE  BC
Steps 
Statement

Reason 

1 
Area of_{}BCE = 1/2 BC*altitude of _{}BCE 
Formula for area of the triangle
with BC as base 

2 
Area of_{}BCD = 1/2 BC*altitude of_{}BCD 
Formula for area of the triangle
with BC as base (BC is common base) 

3 
_{}BC*altitude of_{}BCE= BC*altitude of_{}BCD 
It is given that areas of both
the triangles are same 

4 
Altitude of_{}BCE = Altitude of _{}BCD 
From step 3 

5 
DE  BC 
Distance between two lines are
same 
6.8.7
Data: In triangle ABC, D and E are
mid points of AB and AC respectively
To prove: DEBC and DE=1/2BC.
Construction: From C, draw a line parallel
to AB. Extend DE to meet this line at F
Proof :
Steps 
Statement

Reason 


Consider _{}ADE and _{}ECF 

1 
AE = CE 
E is midpoint of AC 

2 
_{}BAE = _{}ECF 
Alternate angles of ABCF 

3 
_{}ADE = _{}CEF 
Vertically opposite angles 

4 
_{}EDA _{} _{}EFC 
ASA Postulate 

5 
DE=EF,AD=CF 
Corresponding sides are equal 

6 
AD=DB 
Given that D is midpoint of BA 

7 
_{}DB=CF 
From step 5 and step 6 

8 
DBCF is a parallelogram 
Opposite sides CF and BD are
equal and parallel 

9 
DFBC,DF=BC 
From Step 8 

10 
DE=EF 
From Step 5 

11 
BC = 2DE 
From Step 9 and Step 10 
6.8.7 Converse of midpoint theorem: The straight line
drawn through the mid point on one side of a triangle and parallel to another
bisects the third side.
Data: In triangle ABC, D is mid
point of AB. DEBC
To prove: E is mid point of AC
Construction: From C draw a line parallel
to AB. Extend DE to meet this line at F
Proof:
Steps 
Statement 
Reason 

1 
BD = CF 
DBCF is parallelogram 

2 
AD = BD 
Given D is mid point of AB 


Consider _{}ADE and _{}ECF 

3 
AD=CF 


4 
_{}BAE = _{}ECF 
Alternate angles of ABCF 

5 
_{}AED = _{}CEF 
Vertically opposite angles 

6 
_{}ADE _{} _{}ECF 
ASA Postulate 

7 
AE=EC 
Step 6 
6.8.7 Problem 8: Prove that the figure
obtained by joining the mid points of adjacent sides of a quadrilateral is a
parallelogram.
Given: P, Q, R, S are the mid
points of AB, BC, CD, DA respectively, of quadrilateral ABCD.
To Show: PQRS is a parallelogram
Construction: Join BD.
Steps 
Statement

Reason 

1 
PSBD and 2PS=BD 
Midpoint theorem for_{}ABD with P and S as mid points of AB and AD respectively 

2 
BDRQ and 2RQ=BD 
Midpoint theorem for_{}CDB with Q and R as mid points of BC and CD 

3 
i.e. PSRQ
and 2PS=2RQ or PS=RQ 
From Step 2 

4 
PSRQ is a
parallelogram 
From Step 3 
6.8.7 Problem 9: Prove that the figure
obtained by joining mid points of the adjacent sides of a rhombus is a
rectangle
Given: P, Q, R, S are the mid
points of AB, BC, CD, DA respectively of quadrilateral ABCD
To Show: PQRS is a rectangle.
Construction: Join BD and AC
Proof:
Firstly, follow the steps
described above to show that PQRS is a parallelogram.
Steps 
Statement

Reason 

1 
AO=OC,OD=BO and AOD= 90^{0} 
In a rhombus diagonals bisect
each other perpendicularly 

2 
_{}DAO+_{} 
Sum of angles in a triangle 

3 
_{}DAO+_{} 
Step 1, Step 2 

4 
_{}DAO =_{}DSR 
SRAC and _{}DAO,_{}DSR are corresponding angles 

5 
PSBD 
(Mid point theorem) 

6 
_{} 
Corresponding angles(Step 5) 

7 
_{}DSR+_{}ASP =90^{0} 
Step3,Step 4 and Step 6 

8 
_{}PSR = 180^{0 }_{}DSR+_{}ASP = 180^{0 }90^{0 }=90^{0 } 
ASD is a straight line and step
7 

9 
_{}SRQ = _{}RQP =_{}QPS =90^{0} 
Similar steps 2 to Step 8 
6.8.7 Problem 10: In the following figure
with ABCD, prove that R is the mid point of BC and PR = 1/2(AB+DC)
Given: P and Q are midpoints of AD
and BD respectively, ABCD
To Show: R is mid point of BC, PR
= 1/2(AB+DC)
Steps 
Statement 
Reason 

1 
PQAB and PQ=1/2AB 
Midpoint
theorem for _{}ABD with P and Q as mid points of AD and BD 

2 
R is mid point of BC 
Converse
of midpoint theorem for _{}DBC with Q as mid point of DB and PRDC 

3 
QR=1/2CD 
Midpoint
theorem for _{}DBC with R and Q as mid points of BC and BD respectively 

4 
_{} PR
=PQ+QR = 1/2AB+1/2CD 
From Step 1 and Step 3 

5 
i.e. PR = 1/2(AB+DC) 

Note: From this we
conclude that the length of the line segment joining mid points of two non
parallel sides of a trapezium is half the sum
of lengths of the
parallel sides.
6.8.7 Problem 11: D, E and F are
midpoints of AB, AC and BC of an isosceles triangle ABC in which AB=BC. Prove
that _{}DEF is also isosceles.
Given: D, E and F are midpoints of
AB, AC and BC respectively, AB=BC
To Show: _{}DEF is an isosceles triangle (any two sides in _{}DEF are equal).
Solution:
Steps 
Statement

Reason 

1 
2DE = BC 
Midpoint theorem for_{}ABC with D and E as mid points of AB and AC 

2 
2FE = AB 
Midpoint theorem for_{}ABC with E and F as mid points of AC and BC 

3 
AB=BC 
Given 

4 
_{}2DE=2FE, DE=FE 
Substituting results of step 1 and step 2
in step 3 
6.8 Summary of learning
No 
Points
to remember 
1 
The diagonals of a parallelogram
bisect each other 
2 
Each diagonal divides the
parallelogram in to two congruent triangles 
3 
Parallelograms standing on the same
base and between the same parallel lines have equal areas 
4 
The line joining the mid points of
any two sides of a triangle is parallel to and half the third side 
Some useful formulae for calculation of areas:
Type 
Figure 
Area 
Triangle 

1/2 *ah 1/2* base*height OR _{} Where s = (1/2) (a+b+c) =1/2(sum of sides) 
Quadrilateral 

(1/2) *d(h1+h2) 1/2 * diagonal * sum of altitudes on the diagonal 
Parallelogram 

a*h (base*height) 
Trapezium 

(1/2) *h(a+b) (1/2) *product of height and sum of parallel sides 
Rectangle 

ab Product of sides 
Square 

a*a = a^{2} Square of sides 
Rhombus 

(1/2)*ab (1/2)* product of diagonals 
Additional Points:
6.8.7 Intercept Theorem: If three or more lines
make equal intercepts on one transversal, then they make equal intercepts on
any other transversal.
Given: Transversal p makes equal intercepts
(AB=BC) on three lines l, m and n.
(i.e. l  m  n) q is another
transversal which makes intercepts DE and EF.
To prove: DE=EF.
Construction:
Steps 
Statement 
Reason 


Consider _{}ABS and _{}BCT 

1 
_{}ABS = _{}BCT, _{}BAS = _{}CBT, 
Corresponding angles (It is given that line l  line m) 

2 
AB=BC 
Given 

3 
_{}ABS  _{}BCT 
ASA postulate on congruence. 

4 
AS=BT 
Corresponding sides are equal 



5 
ASED is a parallelogram 
ASDE(construction), ADBE(Given) 

6 
AS=DE 
Sides of a parallelogram 

7 
BTFE is a parallelogram 
BTEF(construction), BECF(Given) 

8 
BT=EF 
Sides of a parallelogram 

9 
DE=EF 
Steps 4,6 and 8 
6.8.7 Problem 12: D, E and F are the mid
points of the sides AB, BC and CA of the triangle ABC. AE meets DF at O. P and
Q are mid points of
Steps 
Statement

Reason 


Consider _{} ABC 

1 
DFBC and 2DF = BC 
Mid point theorem (D and F are
mid points) 


Consider _{} BOC 

2 
PQBC and 2PQ = BC 
Mid point theorem (P and Q are
mid points) 


Consider _{} ABO 

3 
DPAO and 2DP = AO 
Mid point theorem (D and P are
mid points) 


Consider _{} ACO 

4 
FQAO and 2FQ = AO 
Mid point theorem (F and Q are
mid points) 

5 
DF = PQ, DFPQ 
Steps 1,2 

6 
DP = FQ, DPFQ 
Steps 3,4 
6.8.7 Problem 13: In a trapezium ABCD,
ABDC, P and Q are mid points of AD and BC respectively. BP produced meets CD
produced at point E.
Prove that P bisects BE and PQAB.
Steps 
Statement

Reason 


Consider _{} BAP and _{}EDP 

1 
_{}EPD = _{}BPA 
Vertically opposite angles 

2 
PD = PA 
Given (P is mid point of AB) 

3 
_{}ABP = _{}PED 
Alternate angle ABDE 

4 
_{} BAP _{} _{} EDP 
ASA Postulate 

5 
PE=PB 
Step 4 


Consider _{} EBC 

6 
P and Q are mid points of sides
EB and BC respectively 
Step 5 and given 

7 
PQDC 
Mid point theorem 

8 
DCAB 
Given 

9 
PQAB 
Step 7 and 8 
6.8.7 Theorem 4:
In a parallelogram prove that opposite sides and opposite angles are
equal
Data: In ABCD ABCD and ADBC
To Prove: AD=BC, AB=CD and _{}ADC = _{}ABC and _{}DAB = _{}DCB
Construction: Join BD
Steps 
Statement 
Reason 


Consider _{} ADB and _{}BCD 

1 
_{}CDB = _{}DBA 
Alternate
angle ABCD 

2 
BD is common 
Construction 

3 
_{}ADB = _{}DBC 
Alternate angle ADBC 

4 
_{}ADB _{} _{}BCD 
ASA Postulate 

5 
AD=BC, AB=CD 
Step 4 

6 
_{}DAB = _{}BCD 
Step 4 

7 
_{}CDB + _{}ADB= _{}DBA+_{}DBC 
Addition
of Step 1 and 3 

8 
_{}ADC = _{}ABC 
Simplification
of step 7 
Exercise: Prove the following:
No 
Theorems: 
1 
A quadrilateral is a
parallelogram if their diagonals bisect each other 
2 
A quadrilateral is a
parallelogram if their opposite sides are equal 
3 
A quadrilateral is a
parallelogram if their opposite angles are equal 
4 
A quadrilateral is a
parallelogram if a pair of opposite sides is equal and parallel 
5 
A parallelogram is a rectangle
if their diagonals are equal 
6 
A parallelogram is a square if
their diagonals are equal and intersect at right angles 
7 
A diagonal of a square makes an
angle of 45^{o} with the sides of the square 
8 
A parallelogram is a rhombus if adjacent
sides are equal 
9 
A diagonal of a rhombus bisects
vertex angles 
10 
The diagonals of a rhombus
intersect each other at right angles 
11 
A parallelogram is a rhombus if
their diagonals intersect at right angles 
12 
A rhombus is a square if their
diagonals are equal 
Hint: Use the following properties/statements to prove all of
the above theorems after drawing diagonals if required.
1. SSS/ASA/SAS
postulates for congruency of triangles.
2. Alternate/Corresponding
angles are equal when lines are parallel.
3. In an isosceles
triangle angles opposite to equal sides are equal.
6.8.7 Problem 14: ABCD is a parallelogram.
A line through A cuts DC at point P and BC produced to Q. Prove that area of _{}BCP = area of _{}DPQ
Construction: Draw AE_{}BC and CF_{}AB
Steps 
Statement

Reason 

1 
Area of ABCD = AB*CF=AD*AE=BC*AE 
Formula for area and AD=BC 

2 
Area of _{}ADQ = (1/2)*AD*AE 
Formula for the area of _{} 

3 
Area of _{}ABP = (1/2)*AB*CF 
Formula for the area of _{} 

4 
Area of _{}ADQ = Area of _{}ABP 
AD*AE=AB*CF(step 1),Step2,3 

5 
Area of _{}ADQ = Area of _{}ADP + Area of _{}DPQ 
From figure 

6 
Area of _{}ADP + Area of _{}BCP= Area of ABCD  Area of _{}ABP= AB*CF – (1/2)*AB*CF = (1/2)AB*CF 
From figure Step 1,3 

7 
= Area
of _{}ABP 
Step 3 

8 
Area of _{}ADP + Area of _{}BCP = Area of _{}ADP + Area of _{}DPQ 
Step 4 and 5 

9 
Area of _{}BCP = Area of _{}DPQ 
Simplification of step 8 