QUADRILATERLS

Properties of Parallelogram

Opposite sides of a parallelogram are equal

in $\Delta ABCand\Delta CDA$
$AC=AC$ [Common / transversal]
$\angle BCA=\angle DAC$ [alternate angles] 
$\angle BAC=\angle DCA$ [alternate angles] 
$\Delta ABC\cong \Delta CDA$  [ASA rule]
Hence,
$AB=DC$ and $AD=BC$ [ C.P.C.T.C]

Opposite angles in a parallelogram are equal

In parallelogram $ABCD$
$AB\Vert CD$; and $AC$ is the transversal
Hence, $\angle 1=\angle 3$....(1) (alternate interior angles)

$BC\Vert DA$; and $AC$ is the transversal
Hence, $\angle 2=\angle 4$....(2) (alternate interior angles)

Adding (1) and (2)
$\angle 1+\angle 2=\angle 3+\angle 4$
$\angle BAD=\angle BCD$
Similarly,
$\angle ADC=\angle ABC$

Properties of diagonal of a parallelogram

- Diagonals of a parallelogram bisect each other.

In  $\Delta AOBand\Delta COD,$
$\angle 3=\angle 5$ [alternate interior angles]
$\angle 1=\angle 2$  [vertically opposite angles]
$AB=CD$ [opp. Sides of parallelogram]
$\Delta AOB\cong \Delta COD$ [AAS rule]
$OB=OD$ and $OA=OC$ [C.P.C.T]
Hence, proved

Conversly,
- If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

- Diagonal of a parallelogram divides it into two congruent triangles.

In $\Delta ABC$ and $\Delta CDA,$
$AB=CD$ [Opposite sides of parallelogram]
$BC=AD$ [Opposite sides of parallelogram]
$AC=AC$ [Common side]
$\Delta ABC\cong \Delta CDA$ [by SSS rule]
Hence, proved
 

Diagonals of a rhombus bisect each-other at right angles

Diagonals of a rhombus bisect each - other at right angles

In $\Delta AODand\Delta COD,$
$OA=OC$ [Diagonals of parallelogram bisect each other]
$OD=OD$ [Common side]
$AD=CD$ [Adjacent sides of a rhombus]
$\Delta AOD\cong \Delta COD$ [SSS rule]
$\angle AOD=\angle DOC$ [C.P.C.T]
$\angle AOD+\angle DOC=180$ [$\because$ AOC is a straight line]
Hence, $\angle AOD=\angle DOC=90$
Hence proved

Diagonals of a rectangle bisect each-other and are equal

Rectangle ABCD

In $\Delta ABCand\Delta BAD,$
$AB=BA$ [Common side]
$BC=AD$ [Opposite sides of a rectangle]
$\angle ABC=\angle BAD$ [Each = ${90}^0\because$ ABCD is a Rectangle]
$\Delta ABC\cong \Delta BAD$ [SAS rule]
$\therefore AC=BD$  [C.P.C.T]

Consider $\Delta OADand\Delta OCB,$
$AD=CB$  [Opposite sides of a rectangle]
$\angle OAD=\angle OCB$  [$\because$ AD||BC and transversal AC intersects them]
$\angle ODA=\angle OBC$  [$\because$ AD||BC and transversal BD intersects them]
$\Delta OAD\cong \Delta OCB$ [ASA rule]
$\therefore OA=OC$  [C.P.C.T]
Similarly we can prove $OB=OD$

Diagonals of a square bisect each-other at right angles and are equal

Square ABCD

In $\Delta ABCand\Delta BAD,$
$AB=BA$ [Common side]
$BC=AD$ [Opposite sides of a Square]
$\angle ABC=\angle BAD$ [Each = ${90}^0\because$ ABCD is a Square]
$\Delta ABC\cong \Delta BAD$ [SAS rule]
$\therefore AC=BD$  [C.P.C.T]

Consider $\Delta OADand\Delta OCB,$
$AD=CB$  [Opposite sides of a Square]
$\angle OAD=\angle OCB$  [$\because$ AD||BC and transversal AC intersects them]
$\angle ODA=\angle OBC$  [$\because$ AD||BC and transversal BD intersects them]
$\Delta OAD\cong \Delta OCB$ [ASA rule]
$\therefore OA=OC$  [C.P.C.T]
Similarly we can prove $OB=OD$

In $\Delta OBAand\Delta ODA,$
$OB=OD$   [ proved above]
$BA=DA$  [Sides of a Square]
$OA=OA$  [ Common side]
$\Delta OBA\cong \Delta ODA,$  [ SSS rule]
$\therefore \angle AOB=\angle AOD$  [ C.P.C.T]
But, $\angle AOB+\angle AOD={180}^0$  [ Linear pair]
$\therefore \angle AOB=\angle AOD={90}^0$ 
 

Important results related to parallelograms

Parallelogram ABCD

Opposite sides of a parallelogram are parallel and equal.
$AB||CD,AD||BC,AB=CD,AD=BC$

Opposite angles of a parallelogram are equal adjacent angels are supplementary.
$\angle A=\angle C,\angle B=\angle D,$
$\angle A+\angle B={180}^0,\angle B+\angle C={180}^0,\angle C+\angle D={180}^0,\angle D+\angle A={180}^0$

A diagonal of parallelogram divides it into two congruent triangles.
$\Delta ABC\cong \Delta CDA$ [With respect to AC as diagonal]
$\Delta ADB\cong \Delta CBD$   [With respect to BD as diagonal]

The diagonals of a parallelogram bisect each other.
$AE=CE,BE=DE$

$\angle 1=\angle 5$ (alternate interior angles)
$\angle 2=\angle 6$ (alternate interior angles)
$\angle 3=\angle 7$ (alternate interior angles)
$\angle 4=\angle 8$ (alternate interior angles)
$\angle 9=\angle 11$ (vertically opp. angles)
$\angle 10=\angle 12$ (vertically opp. angles)

The Mid-Point Theorem

The mid-point theorem

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of the third side


In $\Delta ABC,E$ - midpoint of $AB;F$ - midpoint of $AC$
Construction: Produce $EF$ to $D$ such that $EF=DF.$ 
In $\Delta AEF$ and $\Delta CDF$,
$AF=CF$  [ F is midpoint of AC]
$\angle AFE=\angle CFD$  [ V.O.A]
$EF=DF$ [ Construction]
$\therefore \Delta AEF\cong \Delta CDF$ [SAS rule]
Hence,
$\angle EAF=\angle DCF$....(1) 
$DC=EA=EB$  [ E is the midpoint of AB]
$DC\Vert EA\Vert AB$ [Since, (1), alternate interior angles]
$DC\Vert EB$
So $EBCD$ is a parallelogram 
Therefore, $BC=ED$ and $BC\Vert ED$
Since, $ED=EF+FD=2EF=BC$  [ $\because$ EF=FD]
We have,$EF=\frac{1}{2}BC$ and $EF||BC$
Hence proved

Introduction to Quadrilaterals

Quadrilaterals

Any four points in a plane, of which three are non collinear are joined in order results in to a four sided closed figure called 'quadrilateral'
 

Quadrilateral

 

Angle sum property of a quadrilateral

Angle sum property - Sum of angles in a quadrilateral is 360

In $△ADC,$
$\angle 1+\angle 2+\angle 4=180$ (Angle sum property of triangle)................(1)

In $△ABC,$
$\angle 3+\angle 5+\angle 6=180$ (Angle sum property of triangle)..................(2)

(1) + (2):
$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6=360$
I.e, $\angle A+\angle B+\angle C+\angle D=360$
Hence proved

Types of Quadrilaterals

Trapezium

A trapezium is a quadrilateral with any one pair of opposite sides parallel.
 

Trapezium

$PQRS$ is trapezium in which $PQ||RS$

Parallelogram

A parallelogram is a quadrilateral, with both pair of opposite sides parallel and equal. In parallelogram, diagonals bisect each other.
 

Parallelogram ABCD

Parallelogarm ABCD in which $AB||CD,BC||AD$  and $AB=CD,BC=AD$

Rhombus

A rhombus is a parallelogram with all sides equal. In rhombus, diagonals bisect each other perpendicularly
 

Rhombus ABCD

A rhombus $ABCD$ in which $AB=BC=CD=AD$ and  $AC\perp BD$

Rectangle

A rectangle is a parallelogram with all angles as right angles.
 

Rectangle ABCD

A rectangle $ABCD$ in which, $\angle A=\angle B=\angle C=\angle D={90}^0$
 

Square

A square is a special case of parallelogram with all angles as right angles and all sides equal.
 

Square ABCD

A square $ABCD$ in which $\angle A=\angle B=\angle C=\angle D={90}^0$ and $AB=BC=CD=AD$

 

Kite

A kite is a quadrilateral with adjacent sides equal.
 

Kite ABCD

A kite $ABCD$ in which $AB=BC$ and $AD=CD$

Venn diagram for different types of quadrilaterals