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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 