Chapter 8
Chapter 8
Chapter 8
Chapter 8
Quadrilaterals
Introduction:
● We're all familiar with planar figures with sides defined by straight line
segments, which are known as Polygons.
● The word polygon comes from the Greek language.
● It refers to a figure with a lot of angles, meaning a lot of sides.
● Quadrilaterals are squares, rectangles, and other four-sided geometric shapes
produced by the union of four line segments.
● A quadrilateral is a polygon with four sides.
Examples of Quadrilaterals:
Parallelograms:
● A parallelogram is a quadrilateral with parallel and equal opposite sides.
● Parallelograms include a rectangle, a rhombus, and a square.
● A trapezium is a quadrilateral with one pair of opposite sides that are parallel
to one other. As a result, it isn't a parallelogram.
● The opposite sides of each pair are equal and parallel.
In the diagram,
OD OB and OA OC
● Each diagonal divides the parallelogram into two congruent triangles.
In the diagram,
ABC CDA
ABD CDB
Theorem 1 :
Statement:
o The diagonals of a parallelogram bisect each other.
o If two sides of a triangle are unequal, the longer side has the greater angle
To prove:
The diagonals AC and BD bisect each other that is,
AO OC and
BO DO .
Proof:
AB CD (By definition of parallelogram)
AC is a transversal.
OAB OCB ..... i (Alternate angles are equal in a parallelogram)
Also,
AB DC (Opposite sides are equal in a parallelogram)
Now in AOB and COD ,
AB DC (Opposite sides of parallelogram are equal)
OAB OCD (Proved by i )
AOB COD (Vertically opposite angles are equal)
Therefore,
AOB COD ( AAS Congruency condition)
Therefore,
AO OC and BO OD (corresponding parts of congruent triangles are congruent)
that is the diagonals of a parallelogram bisect each other.
Converse:
"If both pairs of opposite sides of a quadrilateral are equal, then the
Theorem 2 :
Statement:
If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a
parallelogram.
Given:
ABCD is a quadrilateral in which diagonals AC and BD intersect at O such that
AO OC and BO OD .
To prove:
ABCD is a parallelogram.
Proof:
In triangles AOB and COD ,
AO CO (Given)
BO OD (Given)
AOB COD (Vertically opposite angles are equal)
Therefore,
AOB COD ( SAS Congruency condition)
Therefore,
OAB OCD ( cpct )
Since these are alternate angles made by the transversal AC intersecting AB and
CD
Class IX Maths www.vedantu.com 5
Therefore,
AB CD
Similarly,
AD BC
Hence, ABCD is a parallelogram.
Theorem 3 :
Statement:
A quadrilateral is a parallelogram if one pair of opposite sides are equal and
parallel.
Given:
ABCD is a quadrilateral in which AB CD and AB CD .
To prove:
ABCD is a parallelogram.
Construction:
Join AC .
Proof:
In triangles ABC and ADC ,
AB CD (Given)
BAC ACD (Alternate angles are equal)
AC AC (Common side)
Therefore,
ABC CDA ( SAS Congruency condition)
BCA DAC (Corresponding parts of corresponding triangles)
Since these are alternate angles,
AB CD
Thus, in the quadrilateral ABCD , AB CD and AD BC
Therefore, ABCD is a parallelogram.
Special Parallelograms:
Rectangle:
A rectangle is a parallelogram with one of its angles as a right angle.
Rhombus:
A rhombus is a parallelogram with a pair of its consecutive sides equal.
Square:
A square is a rectangle with a pair of its consecutive sides equal.
Theorem 4 :
Statement:
The diagonals of a rectangle are equal in length.
Given:
ABCD is a rectangle.
AC and BD are diagonals.
To prove:
AC BD
Proof:
Let, A 90 (By definition of rectangle)
A B 180 (Consecutive interior angle)
A B 90
Now in triangles, ABD and ABC ,
AB AB (Common side)
A B 90 (Each angle is a right angle)
AD BC (Opposite sides of parallelogram)
Therefore,
ABD BAC
Therefore,
BD AC (Corresponding parts of corresponding triangles)
Hence the theorem is proved.
Converse of Theorem 4 :
Statement:
If two diagonals of a parallelogram are equal, it is a rectangle.
Theorem 5 :
Statement:
The diagonals of a rhombus are perpendicular to each other.
Converse of Theorem 5 :
Statement:
If the diagonals of a parallelogram are perpendicular then it is a rhombus.
Given:
ABCD is a parallelogram in which AC and BD are perpendicular to each other.
To prove:
ABCD is a rhombus.
Proof:
Let AC and BD intersect at right angles at O .
AOB 90
In triangles AOD and COD ,
AO OC (Diagonals bisect each other)
OD OD (Common side)
AOD COD 90 (Given)
Therefore,
AOD COD ( SAS congruency condition)
AD DC
That is, the adjacent sides are equal.
Therefore, by definition, ABCD is a rhombus.
Hence the theorem is proved.
Theorem 6 :
Given:
ABCD is a square.
AC and BD are diagonals intersecting at O .
To prove:
AC BD and AC BD
Proof:
AB AD (Sides of a square are equal)
AB DC (Opposite sides of a square are parallel)
Therefore,
ABCD is parallelogram with consecutive sides equal.
Hence, ABCD is a rhombus. (By definition)
Since, the diagonals of a rhombus are perpendicular to each other,
AC BD
Therefore,
ABCD is a parallelogram.
AB AD and A 90
Therefore, ABCD is a rectangle with a pair of its consecutive sides equal.
Since the diagonals of a rectangle are equal, AC BD .
Therefore,
Diagonal AC = Diagonal BD and AC BD
Hence, the theorem is proved.
Converse of Theorem 6 :
Statement:
If in a parallelogram, the diagonals are equal and perpendicular, then it is a square.
Given:
ABC in which D is the mid-point of AB and DE BC .
To prove:
E is the mid-point of AC . That is, to prove AE=EC .
Construction:
Since DE BC , we can complete a parallelogram with DB and BC as
consecutive sides.
Hence draw EF BD to meet DE produced at F .
Proof: