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Mathematics in Daily Life
9th Grade
Theorems on Parallelograms
Objective
After learning this chapter, you should be able to
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•
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Prove the properties of parallelograms logically.
Explain the meaning of corollary.
State the corollaries of the theorems.
Solve problems and riders based on the theorem.
2
Flowchart on Procedure to Prove a Theorem
Let us recall the procedure of proving a theorem logically.
Observe the following flow chart.
Consider/take a statement or the
Enunciation of the theorem
For example, in any triangle the sum of
three angles is 180˚
A
Draw the appropriate figure and name it.
B
C
Write the data using symbols.
ABC is a triangle
1
2
3
Flowchart on Procedure to Prove a Theorem
1
2
Write what is to be proved using
symbols
BAC  ABC  BCA  180
Analyze the statement of the theorem
and write the hypothetical construction
if needed and write it symbolically
Through the Vertex A draw
EF || BC
E
A
F
Write the
reason for
construction
Draw the appropriate figure and name it.
Use postulates, definitions and
previously proved theorems along with
what is given and construction
4
Theorems on Parallelograms
Theorem 1:
The diagonals of a parallelogram bisect each other.
Theorem 2:
Each diagonal divides the parallelogram into two congruent
triangles.
5
Theorem 1 Proof
Theorem:
The diagonals of a parallelogram bisect each other.
D
C
O
A
B
Given: ABCD is a parallelogram.
AC and BD are the diagonals intersecting at O.
To Prove: AO = OC
BO = OD
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Theorem 1 Proof Contd..
Proof:
Statement
Reasons
Process of Analysis
1) In ∆AOB and ∆COD,
AB = DC
2) AOB  COD
Opposite sides of
the parallelogram
Vertically opposite
angles
Recognise the ∆s
which contain the
sides AO, BO, CO,
DO.
3) OAB  OCD
Alternate angles
AB || DC and BD is a
transversal.
ASA Postulate
Use the data to
prove the
congruency of
these two ∆s
ABO  ODC
ΔAOB  ΔCOD
 AO  OC and BO  OD Corresponding sides
of congruent ∆s
i.e., The diagonals of parallelogram bisect each other.
7
Theorem 2 Proof
Theorem:
Each diagonal divides the parallelogram into two congruent
triangles. D
C
O
A
B
Given: ABCD is a parallelogram in which AC is a diagonal.
AC = DC, AD = BC
To Prove:
ΔABC  ΔADC
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Theorem 2 Proof Contd..
Proof:
Statement
1) AB = DC
Reasons
Opposite sides of the parallelogram
2) BC = AD
Opposite sides of the parallelogram
3) AC is common
ΔABC  ΔADC
S.S.S. postulate
 Diagonal AC divides the parallelogram ABCD into two
congruent triangles.
Similarly, we can prove that ΔABD  ΔBDC.
 Each diagonal divides the parallelogram into two congruent
triangles.
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Corollary
A corollary is a proposition that follows directly from a
theorem or from accepted statements such as definitions.
Corollaries of the Theorems
There are four corollaries for the theorems explained in the
previous slides. They are,
Corollary-1:
In a parallelogram, if one angle is a right angle, then it is a
rectangle.
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Corollaries of the Theorems Contd…
Corollary-2:
In a parallelogram, if all the sides are equal and all the angles
are equal, then it is a square.
Corollary-3:
The diagonals of a square are equal and bisect each other
perpendicularly.
Corollary-4:
The straight line segments joining the extremities of two
equal and parallel line segments on the same side are equal
and parallel.
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Corollary 1 Proof
Corollary:
In a parallelogram, if one angle is a right angle, then it is a
rectangle.
R
S
90˚
P
Q
o

Q

90
.
Given: PQRS is a parallelogram. Let
To Prove: PQRS is a rectangle.
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Corollary 1 Proof Contd..
Proof:
Statement
Q  90o
Given
---- (1) Opposite angles of parallelogram PQRS
Q  S  90o
Q  R  180
o
90o  R  180o
 R  180o  900
 R  90 o
 R  P  90o
Reasons
Sum of the consecutive angles of a
parallelogram PQRS is equal to 180˚
By substitution
By transposing
----(2)
----(3) Opposite angles of parallelogram PQRS
 P  Q  R  S  90o From the equations (1), (2) and (3)
Hence, PQRS is a rectangle.
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Corollary 2 Proof
Corollary:
In a parallelogram, if all the sides are equal and all the angles
are equal, then it is a square.
D
A
C
90˚
90˚
90˚
90˚
B
Activity!!!
Prove this corollary logically
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Corollary 3 Proof
Corollary:
The diagonals of a square are equal and bisect each other
perpendicularly. D
C
90˚
90˚
90˚
90˚
A
B
Given: ABCD is a square. AB = BC = CD = DA.
A  B  C  D  90o
To Prove: 1) AC = BD
2) AO = CO, BO = DO.
3) AOB  DOC  90o
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Proof:
Corollary 3 Proof Contd..
Statement
Reasons
1) Consider the ∆ABD and ∆ABC
AB = BC
Sides of the square are equal
Angles of the square are equal
BAD  ABC  90o
AB is common.
S.A.S postulate
ΔABD  ΔABC
Congruent parts of congruent ∆s
AC  BD
2) Consider the ∆AOB and ∆DOC
AB = DC
Opposite sides of the square
OAB  OCD
Alternate angles AB || DC
ABO  CDO
Alternate angles AB || DC
A.S.A postulate
ΔAOB  ΔDOC
Congruent parts of congruent ∆s
 AO  CO, BO  DO
Hence, diagonals of a square bisect each other
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Corollary 3 Proof Contd..
Statement
3) Consider the ∆AOD and ∆COD
AD = CD
AO = CO
DO is common
ΔAOD  ΔCOD
AOD  COD
AOD  COD  180o
Reasons
Sides of the square are equal
The diagonals bisect each other
S.S.S postulate
Congruent parts of congruent ∆s
Linear pair
 AOD  COD  90o
Hence, the digonals bisect each other at right angles.
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Corollary 4 Proof
Corollary:
The straight line segments joining the extremities of two
equal and parallel line segments on the same side are equal
and parallel.
Activity!!!
Prove this corollary logically
Hint :- S.A.S. Postulate of congruency triangles
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Examples
Example-1: In the given figure, ABCD is a parallelogram in which
DAB  70o and DBC  80o . Calculate the angles CDB and ADB .
D
Given: ABCD is a parallelogram
AB = DC, AD = BC
AB || DC, AD || BC
DAB  70
o
C
80˚
and DBC  80 .
o
70˚
A
B
To Find: CDB and ADB
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Examples Contd...
Solution:
Statement
DAB  70 o  BCD
Reasons
Opposite angles of the
parallelogram ABCD.
In ∆BDC,
DBC  BCD  CDB  180o
80o  70o  CDB  180o
 CDB  180o  150o
 CDB  30 o
DBC  ADB
Sum of three angles of a triangle
By substitution
By transposing
Alternate angles, AD || BC.
 ADB  800
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Examples Contd…
Example-2: In the figure, ABCD is a parallelogram. P is the mid
point of BC. Prove that AB = BQ.
D
Given: ABCD is a parallelogram
‘P’ is the mid point of BC.
BP = PC
A
C
P
Q
B
To Prove: AB = BQ
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Examples Contd...
Solution:
Statement
Consider the ∆BPQ and ∆CPD,
BP = PC
BPQ  CPD
PBQ  PCD
ΔBPQ  ΔCPD
 BQ  DC
------(1)
But DC = AB
 BQ  AB
------(2)
Reasons
Given
Vertically opposite angles
Alternate angles, AB || DC
A.S.A. postulate
C.P.C.T
Opposite sides of the parallelogram
From (1) and (2)
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Exercises
1. In a parallelogram ABCD,A =60⁰. If the bisectors of A and B
meet at P on DC. Prove that APB  90o.
2. In a parallelogram ABCD, X is the mid-point of AB and Y is the
mid-point of DC. Prove that BYDX is a parallelogram.
3. If the diagonals PR and QS of a parallelogram PQRS are equal,
prove that PQRS is a rectangle.
4. PQRS is a parallelogram. PS is produced to M so that SM = SR
and MR is produced to meet PQ produced at N. prove that
QN = QR.
5. ABCD is a parallelogram. If AB = 2 x AD and P is the mid-point
of AB, prove that CPD  90 o.
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