Class 8: Quadrilaterals – Lecture Notes

$\displaystyle \text{QUADRILATERAL}$

$\displaystyle \text{Any closed plane figure bounded by four line segments is called}$
$\displaystyle \text{a quadrilateral.}$

$\displaystyle \text{In the given quadrilateral }ABCD\text{:}$

$\displaystyle \text{(i) The four vertices are }A,\ B,\ C\text{ and }D.$

$\displaystyle \text{(ii) The four sides are }AB,\ BC,\ CD\text{ and }DA.$

$\displaystyle \text{(iii) The four angles are }\angle A,\ \angle B,\ \angle C \text{ and }\angle D.$

$\displaystyle \text{(iv) The two diagonals are }AC\text{ and }BD.$

$\displaystyle \text{ADJACENT SIDES}$

$\displaystyle \text{Two sides of a quadrilateral having a common vertex are called adjacent sides.}$
$\displaystyle \text{For example, }AD\text{ and }DC\text{ are adjacent sides, with }D \text{ as their common vertex.}$

$\displaystyle \text{Similarly, }AB\text{ and }BC\text{ are adjacent sides, with }B \text{ as their common vertex.}$

$\displaystyle \text{OPPOSITE SIDES}$

$\displaystyle \text{Two sides of a quadrilateral that do not have a common vertex are called opposite sides.}$
$\displaystyle \text{For example, }AB\text{ and }DC\text{ are opposite sides.}$

$\displaystyle \text{Similarly, }AD\text{ and }BC\text{ are opposite sides.}$

$\displaystyle \text{ADJACENT ANGLES}$

$\displaystyle \text{Two angles of a quadrilateral having a common side are called adjacent angles.}$
$\displaystyle \text{For example, }\angle A\text{ and }\angle B\text{ are adjacent angles}$
$\displaystyle \text{and have }AB\text{ as their common side.}$

$\displaystyle \text{Similarly, }\angle B\text{ and }\angle C\text{ are adjacent angles}$
$\displaystyle \text{and have }BC\text{ as their common side.}$

$\displaystyle \text{OPPOSITE ANGLES}$

$\displaystyle \text{Two angles of a quadrilateral that do not have a common side are called opposite angles.}$
$\displaystyle \text{For example, }\angle A\text{ and }\angle C\text{ are opposite angles.}$

$\displaystyle \text{Similarly, }\angle B\text{ and }\angle D\text{ are opposite angles.}$

$\displaystyle \text{IMPORTANT THEOREM}$

$\displaystyle \text{The sum of the interior angles of a polygon with }n \text{ sides is}$
$\displaystyle (n-2)\times180^{\circ}.$

$\displaystyle \text{For a quadrilateral, }n=4.$
$\displaystyle \therefore (4-2)\times180^{\circ}=360^{\circ}.$
$\displaystyle \text{Hence, the sum of the interior angles of a quadrilateral is } 360^{\circ}.$

$\displaystyle \text{PROOF}$

$\displaystyle \text{Draw the diagonal }AC\text{ of quadrilateral }ABCD.$

$\displaystyle \text{The diagonal }AC\text{ divides the quadrilateral into two} \text{ triangles, }\Delta ABC\text{ and }\Delta ADC.$

$\displaystyle \text{In }\Delta ABC,$
$\displaystyle \angle 2+\angle B+\angle 4=180^{\circ}. \qquad\text{...(i)}$

$\displaystyle \text{In }\Delta ADC,$
$\displaystyle \angle 1+\angle D+\angle 3=180^{\circ}. \qquad\text{...(ii)}$

$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle (\angle1+\angle2)+\angle B+(\angle3+\angle4) +\angle D=360^{\circ}.$

$\displaystyle \text{But }\angle1+\angle2=\angle A \text{ and }\angle3+\angle4=\angle C.$

$\displaystyle \therefore \angle A+\angle B+\angle C+\angle D =360^{\circ}.$

$\displaystyle \text{Hence, the sum of the interior angles of a quadrilateral is } 360^{\circ}.$

$\displaystyle \text{TYPES OF QUADRILATERALS}$

$\displaystyle \text{PARALLELOGRAM}$

$\displaystyle \text{Definition: A parallelogram is a quadrilateral in which both pairs of opposite sides}$
$\displaystyle \text{are parallel to each other.}$

$\displaystyle \text{In the given parallelogram }ABCD,$
$\displaystyle AB\parallel DC\quad\text{and}\quad AD\parallel BC.$

$\displaystyle \text{The diagonal }BD\text{ is a transversal to }AB\text{ and }DC.$
$\displaystyle \text{It is also a transversal to }AD\text{ and }BC.$

$\displaystyle \text{PROPERTIES OF A PARALLELOGRAM}$

$\displaystyle \text{1. Opposite sides are equal and parallel to each other.}$

$\displaystyle AB\parallel DC\quad\text{and}\quad AB=DC.$
$\displaystyle AD\parallel BC\quad\text{and}\quad AD=BC.$

$\displaystyle \text{2. Opposite angles are equal.}$

$\displaystyle \angle A=\angle C\quad\text{and}\quad\angle B=\angle D.$

$\displaystyle \text{3. Adjacent angles are supplementary.}$

$\displaystyle \angle A+\angle B=180^{\circ}.$
$\displaystyle \angle B+\angle C=180^{\circ}.$
$\displaystyle \angle C+\angle D=180^{\circ}.$
$\displaystyle \angle D+\angle A=180^{\circ}.$

$\displaystyle \text{4. The diagonals bisect each other.}$

$\displaystyle AO=OC\quad\text{and}\quad BO=OD.$

$\displaystyle \text{5. Each diagonal divides the parallelogram into two congruent triangles.}$
$\displaystyle \text{Diagonal }AC:\quad \Delta ABC\cong\Delta CDA.$

$\displaystyle \text{Diagonal }BD:\quad \Delta ABD\cong\Delta CDB.$

$\displaystyle \text{THEOREM}$

$\displaystyle \text{In a parallelogram:}$
$\displaystyle \text{(i) Opposite sides are equal.}$
$\displaystyle \text{(ii) Opposite angles are equal.}$
$\displaystyle \text{(iii) Each diagonal divides the parallelogram into two congruent triangles.}$
$\displaystyle \text{Given: Parallelogram }ABCD\text{ such that}$
$\displaystyle AB\parallel DC\quad\text{and}\quad AD\parallel BC.$

$\displaystyle \text{To prove:}$

$\displaystyle \text{(i) }AB=DC\quad\text{and}\quad AD=BC.$
$\displaystyle \text{(ii) }\angle A=\angle C\quad\text{and}\quad \angle B=\angle D.$
$\displaystyle \text{(iii) }\Delta ABD\cong\Delta CDB.$

$\displaystyle \text{Proof:}$

$\displaystyle \text{Consider }\Delta ABD\text{ and }\Delta CDB.$

$\displaystyle \angle ABD=\angle CDB$
$\displaystyle \text{(alternate interior angles, since }AB\parallel DC).$

$\displaystyle \angle ADB=\angle CBD$
$\displaystyle \text{(alternate interior angles, since }AD\parallel BC).$

$\displaystyle BD=DB\qquad\text{(common side)}.$

$\displaystyle \therefore \Delta ABD\cong\Delta CDB \qquad\text{(ASA congruence criterion)}.$

$\displaystyle \text{Hence, by corresponding parts of congruent triangles,}$

$\displaystyle AB=DC\quad\text{and}\quad AD=BC.$

$\displaystyle \text{Also, }\angle BAD=\angle DCB.$

$\displaystyle \therefore \angle A=\angle C.$

$\displaystyle \text{Further,}$
$\displaystyle \angle B=\angle ABD+\angle DBC$
$\displaystyle \text{and}\quad \angle D=\angle CDB+\angle BDA.$

$\displaystyle \text{Since }\angle ABD=\angle CDB \text{ and }\angle DBC=\angle BDA,$

$\displaystyle \therefore \angle B=\angle D.$

$\displaystyle \text{Hence, the opposite angles of a parallelogram are equal.}$

$\displaystyle \text{Also, since }\Delta ABD\cong\Delta CDB,$
$\displaystyle \text{diagonal }BD\text{ divides the parallelogram into two congruent triangles.}$
$\displaystyle \text{Similarly, by drawing diagonal }AC,$
$\displaystyle \Delta ABC\cong\Delta CDA.$

$\displaystyle \text{Thus, each diagonal divides a parallelogram into two congruent triangles.}$
$\displaystyle \text{RHOMBUS}$

$\displaystyle \text{Definition: A parallelogram in which all four sides are equal is called a rhombus.}$
$\displaystyle \text{In rhombus }ABCD,$
$\displaystyle AB\parallel DC\quad\text{and}\quad AD\parallel BC.$
$\displaystyle AB=BC=CD=DA.$

$\displaystyle \text{PROPERTIES OF A RHOMBUS}$

$\displaystyle \text{1. Opposite sides are parallel to each other.}$

$\displaystyle AB\parallel DC\quad\text{and}\quad AD\parallel BC.$

$\displaystyle \text{2. All four sides are equal.}$

$\displaystyle AB=BC=CD=DA.$

$\displaystyle \text{3. Opposite angles are equal.}$

$\displaystyle \angle A=\angle C\quad\text{and}\quad\angle B=\angle D.$

$\displaystyle \text{4. Adjacent angles are supplementary.}$

$\displaystyle \angle A+\angle B=180^{\circ},\quad \angle B+\angle C=180^{\circ}.$

$\displaystyle \text{5. The diagonals bisect each other at right angles.}$

$\displaystyle AO=OC\quad\text{and}\quad BO=OD.$

$\displaystyle \angle AOB=\angle BOC=\angle COD=\angle DOA=90^{\circ}.$

$\displaystyle \text{6. Each diagonal bisects a pair of opposite angles.}$

$\displaystyle \text{Diagonal }BD\text{ bisects }\angle B\text{ and }\angle D.$

$\displaystyle \angle ABD=\angle DBC\quad\text{and}\quad \angle ADB=\angle BDC.$

$\displaystyle \text{Diagonal }AC\text{ bisects }\angle A\text{ and }\angle C.$

$\displaystyle \angle BAC=\angle CAD\quad\text{and}\quad \angle BCA=\angle ACD.$

$\displaystyle \text{THEOREM}$

$\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}$

$\displaystyle \text{Given: }ABCD\text{ is a rhombus. The diagonals }AC\text{ and }BD$
$\displaystyle \text{intersect at }O.$

$\displaystyle \text{To prove:}$

$\displaystyle AO=OC,\quad BO=OD$
$\displaystyle \text{and}\quad AC\perp BD.$

$\displaystyle \text{Proof:}$

$\displaystyle \text{Since a rhombus is a parallelogram, its diagonals bisect each other.}$
$\displaystyle \therefore AO=OC\quad\text{and}\quad BO=OD.$

$\displaystyle \text{Now consider }\Delta AOB\text{ and }\Delta COB.$

$\displaystyle AB=BC\qquad\text{(all sides of a rhombus are equal)}.$

$\displaystyle AO=OC\qquad\text{(diagonals bisect each other)}.$

$\displaystyle OB=OB\qquad\text{(common side)}.$

$\displaystyle \therefore \Delta AOB\cong\Delta COB \qquad\text{(SSS congruence criterion)}.$

$\displaystyle \therefore \angle AOB=\angle COB \qquad\text{(corresponding parts of congruent triangles)}.$

$\displaystyle \text{But }A,\ O,\ C\text{ are collinear.}$

$\displaystyle \therefore \angle AOB+\angle COB=180^{\circ}.$

$\displaystyle \therefore \angle AOB=\angle COB=90^{\circ}.$

$\displaystyle \therefore AC\perp BD.$

$\displaystyle \text{Hence, the diagonals of a rhombus bisect each other at right angles.}$
$\displaystyle \text{RECTANGLE}$

$\displaystyle \text{Definition: A parallelogram in which all four angles are right}$
$\displaystyle \text{angles is called a rectangle.}$

$\displaystyle \text{PROPERTIES OF A RECTANGLE}$

$\displaystyle \text{1. Opposite sides are equal and parallel to each other.}$

$\displaystyle AB\parallel DC\quad\text{and}\quad AB=DC.$

$\displaystyle AD\parallel BC\quad\text{and}\quad AD=BC.$

$\displaystyle \text{2. Each angle measures }90^{\circ}.$

$\displaystyle \angle A=\angle B=\angle C=\angle D=90^{\circ}.$

$\displaystyle \text{3. The diagonals are equal.}$

$\displaystyle AC=BD.$

$\displaystyle \text{4. The diagonals }AC\text{ and }BD\text{ bisect each other.}$

$\displaystyle AO=OC\quad\text{and}\quad BO=OD.$

$\displaystyle \text{THEOREM}$

$\displaystyle \text{The diagonals of a rectangle are equal.}$

$\displaystyle \text{Given: }ABCD\text{ is a rectangle and }AC\text{ and }BD \ \text{are its diagonals.}$
$\displaystyle \text{To prove: }AC=BD.$

$\displaystyle \text{Proof:}$

$\displaystyle \text{Consider }\Delta ADC\text{ and }\Delta BCD.$

$\displaystyle AD=BC\qquad\text{(opposite sides of a rectangle)}.$

$\displaystyle DC=DC\qquad\text{(common side)}.$

$\displaystyle \angle ADC=\angle BCD=90^{\circ}.$

$\displaystyle \therefore\Delta ADC\cong\Delta BCD \qquad\text{(SAS congruence criterion)}.$

$\displaystyle \therefore AC=BD \qquad\text{(corresponding parts of congruent triangles)}.$

$\displaystyle \text{Hence proved.}$

$\displaystyle \text{SQUARE}$

$\displaystyle \text{Definition: A parallelogram in which all four sides are equal and}$
$\displaystyle \text{all four angles are right angles is called a square.}$

$\displaystyle \text{A square can also be defined as a rectangle in which all four sides are equal.}$
$\displaystyle \text{PROPERTIES OF A SQUARE}$

$\displaystyle \text{1. All four sides are equal.}$

$\displaystyle AB=BC=CD=DA.$

$\displaystyle \text{2. Opposite sides are parallel.}$

$\displaystyle AB\parallel DC\quad\text{and}\quad AD\parallel BC.$

$\displaystyle \text{3. Each angle measures }90^{\circ}.$

$\displaystyle \angle A=\angle B=\angle C=\angle D=90^{\circ}.$

$\displaystyle \text{4. The diagonals are equal.}$

$\displaystyle AC=BD.$

$\displaystyle \text{5. The diagonals bisect each other.}$

$\displaystyle AO=OC\quad\text{and}\quad BO=OD.$

$\displaystyle \text{6. The diagonals intersect at right angles.}$

$\displaystyle AC\perp BD.$

$\displaystyle \text{7. Each diagonal bisects a pair of opposite angles.}$

$\displaystyle \angle BAC=\angle CAD\quad\text{and}\quad \angle BCA=\angle ACD.$

$\displaystyle \angle ABD=\angle DBC\quad\text{and}\quad \angle ADB=\angle BDC.$

$\displaystyle \text{THEOREM}$

$\displaystyle \text{The diagonals of a square are equal and perpendicular to each other.}$

$\displaystyle \text{Given: }ABCD\text{ is a square. Its diagonals }AC\text{ and }BD$
$\displaystyle \text{intersect at }O.$

$\displaystyle \text{To prove: }AC=BD\quad\text{and}\quad AC\perp BD.$

$\displaystyle \text{Proof:}$

$\displaystyle \text{First, consider }\Delta ADC\text{ and }\Delta BCD.$

$\displaystyle AD=BC\qquad\text{(all sides of a square are equal)}.$

$\displaystyle DC=DC\qquad\text{(common side)}.$

$\displaystyle \angle ADC=\angle BCD=90^{\circ}.$

$\displaystyle \therefore\Delta ADC\cong\Delta BCD \qquad\text{(SAS congruence criterion)}.$

$\displaystyle \therefore AC=BD \qquad\text{(corresponding parts of congruent triangles)}.$

$\displaystyle \text{Now, since a square is a parallelogram, its diagonals bisect each other.}$
$\displaystyle \therefore OA=OC\quad\text{and}\quad OB=OD.$