Quadrilateral - Practically Study Material (2024)

9.1. INTRODUCTION
If there are four points in a plane in such a way that no three of them are collinear, then the figure obtained by joining four points, in order is called a quadrilateral. We can say that a quadrilateral is a closed figure with four sides :

e.g. ABCD is a quadrilateral which has four sides AB, BC, CD and DA, four angles $\angle \mathrm{A},\angle \mathrm{B},\angle \mathrm{C}\text{and}\angle \mathrm{D}$ and four vertices A, B, C and D and also has two diagonals AC and BD. i.e. A quadrilateral has four sides, four angles, four vertices and two diagonals.
9.2. ANGLES SUM PROPERTY OF A QUADRILATERAL
Theorem : The sum of the angles of a quadrilateral is 360^o
Given : Let ABCD be a quadrilateral and AC be its one diagonal

To prove : $\angle$A + $\angle$B + $\angle$C + $\angle$D = 360^o
Solution : In $△$ACD, we have
$\angle$DAC + $\angle$ACD + $\angle$D = 180^o … (i)
[Angles sum property]
In $△$ABC, we have
$\angle$BAC + $\angle$ACB + $\angle$B = 180^o … (ii)
[Angles sum property]
Adding (i) and (ii), we get
$\angle$DAC + $\angle$ACD + $\angle$D + $\angle$BAC + $\angle$ACB + $\angle$B
= 180^o + 180^o
$\Rightarrow$ ($\angle$DAC + $\angle$BAC) + ($\angle$ACD + $\angle$ACB) + $\angle$D + $\angle$B = 360^o
$\therefore$ A + $\angle$C + $\angle$D + $\angle$B = 360^o
Hence, $\angle$A + $\angle$B + $\angle$C + $\angle$D = 360^o Proved.
9.3. TYPES OF QUADRILATERALS
I. A Trapezium : In a quadrilateral if one pair of opposite sides is parallel, then it is called a trapezium i.e. If AB || CD then quadrilateral ABCD is a trapezium.

II. A parallelogram : In a quadrilateral, if both pairs of opposite sides are parallel and equal, then it is called a parallelogram. i.e., AB || CD and AB = CD; AD || BC and AD = BC, then ABCD is a parallelogram.

III. A Rectangle : In a quadrilaterals (parallelogram) if all angles are right angles, then it is called a rectangle. i.e. AB || CD, AB = CD, AD || BC; AD = BC and $\angle$A = $\angle$B = $\angle$C = $\angle$D = 90^o, then ABCD is a rectangle.

IV A Rhombus : In a quadrilaterals (parallelogram) if all sides are equal, then it is called a rhombus , i.e., AB || CD, AD || BC and AB = BC = CD = DA, then ABCD is a rhombus.

V A Square : In a quadrilateral (parallelogram) if all sides are equal and all angles are 90^o, then it is called a square. i.e. AB || CD, AD || BC, AB = BC = CD = DA and $\angle$A = $\angle$B = $\angle$C = $\angle$D = 90^o

VI A Kite : In a quadrilateral ABCD, if AD = CD and AB = CB, then it is called a kite; i.e., two pairs of adjacent sides are equal but it is not a parallelogram.

LEARN IT
(i) A square is a rectangle and also a rhombus
(ii) A parallelogram is a trapezium
(iii) A trapezium is not a parallelogram
(iv) A trapezium is not a parallelogram
(v) A rectangle or a rhombus is not a square.
9.4. PROPERTIES OF A PARALLELOGRAM
Theorem : A diagonal of a parallelogram divides it into two congruent triangles.
Given : ABCD is a parallelogram and AC be diagonal.

To prove : $△$ABC$\cong$ $△$CDA,
Solution : In $△$ABC and $△$CDA, we have
BC || AD and AC is a transversal
$\therefore$ $\angle$BCA = $\angle$DAC [Pair of alternate angles]
Similarly, AB || DC and AC is a transversal
$\therefore$ $\angle$BAC = $\angle$DCA [Pair of alternate angles]
and AC = AC [Common]
Hence, $△$ABC$\cong$ $△$CDA [By ASA congruence rule]
i.e. Diagonal AC divides parallelogram ABCD into two congruent triangles ABC and CDA.
Theorem : If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Given : Let ABCD be a quadrilateral in which AB = CD and BC = AD .

To prove : ABCD is a parallelogram.
Construction : Joined AC
Solution: In ${△}^s$ ABC and CDA, we have
AB = CD [Given]
BC = AD [Given]
and AC = AC [Common]
$\therefore △\mathrm{ABC}\cong △\mathrm{CDA}$ [By SSS congruence rule]
$\therefore$ $\angle$CAB = $\angle$ACD … (i)
and $\angle$ACB = $\angle$CAD … (ii) [By CPCT]
Now, line AC intersects AB and DC at A and C, such that
$\angle$CAB = $\angle$ACD [From (i)]
But these are alternate interior angles
$\therefore$ AB || DC … (iii)
Similarly, line AC intersects BC and AD at C and A such that
$\angle$ACB = $\angle$CAD [from (ii)]
But these are alternate interior angles
$\therefore$ BC || AD … (iv)
From (iii) and (iv), we have
AB || DC and BC || AD
Hence, ABCD is a parallelogram. Proved.
Theorem : In a parallelogram, opposite angles are equal.
If ABCD is a parallelogram,

Then, $\angle$A = $\angle$C and $\angle$B = $\angle$D
Theorem : If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Given : ABCD is a quadrilateral in which $\angle$A = $\angle$C and $\angle$B = $\angle$D .

To prove : ABCD is a parallelogram
Solution : In quadrilateral ABCD, we have
$\angle$A = $\angle$C … (i) [Given]
$\angle$B = $\angle$D …(ii) [Given]
Now, adding (i) and (ii) we get
$\angle$A + $\angle$B = $\angle$C + $\angle$D … (iii)
We know that the sum of the angles of a quadrilateral is 360^o.
$\therefore$$\angle$A + $\angle$B + $\angle$C + $\angle$D = 360^o
$\Rightarrow$$\angle$A + $\angle$B) + ($\angle$A + $\angle$B) = 360^o [From (iii)]
$\Rightarrow 2(\angle \mathrm{A}+\angle \mathrm{B})=360°$
$\therefore$$\angle$A + $\angle$B = 180^o … (iv)
i.e. $\angle$A + $\angle$B = $\angle$C + $\angle$D = 180^o … (v)
[From (iii) and (iv)]
line AB intersects AD and BC at A and B respectively, such that
$\angle$A + $\angle$B = 180^o
[The sum of consecutive interior angles is 180^o]
$\therefore$ AD || BC … (vi)
Again, $\angle$A + $\angle$B = 180^o
$\Rightarrow$$\angle$C + $\angle$B = 180^o [ $\because$ $\angle$A = $\angle$C given]
Line BC intersects AB and DC at A and C respectively, such that
$\angle$B + $\angle$C = 180^o
$\therefore$ AB || DC [The sum of consecutive interior angles is 180^o] … (vii)
From (vi) and (vii), we get
AD || BC and AB || DC
Hence, ABCD is a parallelogram.
Theorem : The diagonals of a parallelogram bisect each other.
Given : ABCD is a parallelogram in which AB = CD, AB || CD and BC = AD; BC || AD diagonals AC and BD intersect each other at O.

To prove :
OA = OC and OB = OD
Solution :
In $△$OAB and $△$OCD
AB || CD and BD is a transversal [Given]
$\therefore$ $\angle$ABD = $\angle$CDB [Pair of alternate angles]
$\Rightarrow$$\angle$BAO = $\angle$CDO
Again, $\because$ BC || AD and AC is a transversal [Given]
$\therefore$$\angle$BAC = $\angle$DCA [Pair of alternate angles]
$\Rightarrow$ $\angle$BAO = $\angle$DCO
and AB = CD [Given]
$\therefore △\mathrm{OAB}\cong △\mathrm{OCD}$ [By ASA congruence rule]
Hence, OA = OC [By CPCT] and OB = OD Proved.
Converse of the above Theorem
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Given :
Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at O;
i.e. OA = OC and OB = OD.

To prove :
ABCD quadrilateral is a parallelogram
Solution:
In$△S$ AOD and COB, we have
AO = CO [Given]
OD = OB [Given]
and $\angle$AOD = $\angle$BOC [Vertically opposite angles]
$\therefore △\mathrm{AOD}\cong △\mathrm{COB}$ [By SAS congruence rule]
$\therefore$ $\angle$OAD = $\angle$OCB [By CPCT] … (i)
But these are alternate interior angles
$\therefore$ AD || BC. Similarly, AB || CD
Hence, ABCD is a parallelogram. Proved.
Theorem :
A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel
Given :
ABCD is a quadrilateral in which AB = CD and AB || CD .

To prove :
ABCD is a parallelogram.
Construction :
Joined AC.
Solution:
In ${△}^s$ ABC and CDA, we have
AB = CD [Given]
AC = AC [Common]
and $\angle$BAC = $\angle$DCA
[$\because$ AB||CD and AC is a transversal. $\therefore$ Alternate interior angles are equal]
$\therefore △\mathrm{ABC}\cong △\mathrm{CDA}$ [SAS congruence rule]
$\therefore$ $\angle$BCA = $\angle$DAC [By CPCT]
But these are alternate angles.
$\therefore$ AD || BC
Now, AB || CD and AD || BC. Hence, quadrilateral ABCD is a parallelogram. Proved.
9.5. MIDPOINT THEOREM
The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Given :
ABC is a triangle in which D and E are the mid-points of sides AB and AC respectively DE is joined.

To prove :
DE || BC and DE = 1/2 BC
Construction : Line segment DE, is produced to F, such that DE = EF joined FC.
Proof : In ${△}^s$ AED and CEF, we have
AE = CE [ $\because$ E is the midpoint of AC Given]
$\angle$AED = $\angle$CEF [vertically opposite angles]
and DE = EF [By construction]
$\therefore △\mathrm{AED}\cong △\mathrm{CEF}$ [By SAS congruence rule]
$\therefore$ AD = CF … (i) [By CPCT]
and $\angle$ADE = $\angle$CFE … (ii) [By CPCT]
Now, D is the mid-point of AB
$\therefore$ AD = BD
$\Rightarrow$ DB = CF … (iii) [From (i) AD = CF]
DF intersects AD and FC at D and F respectively such that
$\angle$ADE = $\angle$CFE … [From (ii)]
i.e. alternate interior angles are equal
$\therefore$ AD || FC
$\Rightarrow$ DB || CF … (iv)
From (iii) and (iv), we get that DBCF is a quadrilateral such that one pair side is equal and parallel
$\therefore$ DBCF is a parallelogram
$\therefore$ DF || BC and DF = BC [ $\because$ opposite sides of a || gm are equal are parallel]
But, D, E, F are collinear and DE = EF
Hence, DE || BC and DE = ½ BC. Proved.
Converse of above Theorem : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
Given : Let ABC be a triangle in which E is the mid-point of AB, line l is passing through E and is parallel to BC, and CM || BA intersects AC at point F.

To prove : AF = CF
Construction : CM || BE and intersecting line l at D is drawn
Proof : $\because$CM || BE and l || BC
$\therefore$ BCDE is a parallelogram.
$\therefore$ BE = CD [opposite sides of a ||gm]
But BE = AE [Given]
$\therefore$ AE = CD … (i)
Now, in $△$AEF and $△$CDF, we have
$\angle$AFE = $\angle$CFD [vertically opposite angles]
$\angle$A = $\angle$FCD [Alternate angles]
AE = CD … [from (i)]
$\therefore △\mathrm{AEF}\cong △\mathrm{CDF}$ [AAS congruence rule]
Hence, AF = CF [CPCT].