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Parallelogram: Opposite sides of a parallelogram are equal

In ΔABC and ΔCDA

AC=AC [Common / transversal]

∠BCA=∠DAC [alternate angles]

∠BAC=∠DCA [alternate angles]

ΔABC≅ΔCDA  [ASA rule]

Hence,

AB=DC and AD=BC [ C.P.C.T.C]Opposite angles in a parallelogram are equal

In parallelogram ABCD

AB‖CD; and AC is the transversal

Hence, ∠1=∠3….(1) (alternate interior angles)

BC‖DA; and AC is the transversal

Hence, ∠2=∠4….(2) (alternate interior angles)

∠1+∠2=∠3+∠4

Similarly,

Properties of diagonal of a parallelogram

– Diagonals of a parallelogram bisect each other.

In  ΔAOB and ΔCOD,

∠3=∠5 [alternate interior angles]

∠1=∠2 AB=CD [opp. Sides of parallelogram]

ΔAOB≅ΔCOD [AAS rule]

OB=OD and OA=OC [C.P.C.T]

Hence, proved

Conversely,
– 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 ΔABC and ΔCDA,

AB=CD [Opposite sides of parallelogram]

AC=AC [Common side]

ΔABC≅ΔCDA [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 ΔAOD and ΔCOD,

OA=OC [Diagonals of parallelogram bisect each other]

OD=OD [Common side]

ΔAOD≅ΔCOD [SSS rule]

∠AOD=∠DOC [C.P.C.T]

∠AOD+∠DOC=180 [∵ AOC is a straight line]

Hence, ∠AOD=∠DOC=90

Hence proved

Diagonals of a rectangle bisect each other and are equal

Rectangle ABCD

AB=BA [Common side]

BC=AD [Opposite sides of a rectangle]

∠ABC=∠BAD [Each = 900 ∵ ABCD is a Rectangle]

∴AC=BD  [C.P.C.T]

AD=CB  [Opposite sides of a rectangle]

∠ODA=∠OBC  [∵ AD||BC and transversal BD intersects them]

∴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

AB=BA [Common side]

BC=AD [Opposite sides of a Square]

∠ABC=∠BAD [Each = 900 ∵ ABCD is a Square]

∴AC=BD  [C.P.C.T]

AD=CB  [Opposite sides of a Square]

∠ODA=∠OBC  [∵ AD||BC and transversal BD intersects them]

∴OA=OC  [C.P.C.T]

Similarly we can prove OB=OD

In ΔOBA and ΔODA,

OB=OD   [ proved above]

BA=DA  [Sides of a Square]

OA=OA  [ Common side]

ΔOBA≅ΔODA,  [ SSS rule]

∴∠AOB=∠AOD  [ C.P.C.T]

But, ∠AOB+∠AOD=1800  [ Linear pair]

∴∠AOB=∠AOD=900

Important results related to parallelograms

Parallelogram ABCD

Opposite sides of a parallelogram are parallel and equal.

Opposite angles of a parallelogram are equal adjacent angels are supplementary.

∠A=∠C,∠B=∠D,

∠A+∠B=1800,∠B+∠C=1800,∠C+∠D=1800,∠D+∠A=1800

diagonal of parallelogram divides it into two congruent triangles.

ΔABC≅ΔCDA [With respect to AC as diagonal]

ΔADB≅ΔCBD   [With respect to BD as diagonal]

The diagonals of a parallelogram bisect each other.

AE=CE,BE=DE

∠1=∠5 (alternate interior angles)

∠2=∠6 (alternate interior angles)

∠3=∠7 (alternate interior angles)

∠4=∠8 (alternate interior angles)

∠9=∠11 (vertically opp. angles)

∠10=∠12 (vertically opp. angles)

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 ΔABC, E – the midpoint of AB; F – the midpoint of AC

Construction: Produce EF to D such that EF=DF.

In ΔAEF and ΔCDF,

AF=CF  [ F is the midpoint of AC]

∠AFE=∠CFD  [ V.O.A]

EF=DF [ Construction]

∴ΔAEF≅ΔCDF [SAS rule]

Hence,

∠EAF=∠DCF….(1)

DC=EA=EB  [ E is the midpoint of AB]

DC‖EA‖AB [Since, (1), alternate interior angles]

DC‖EB

So EBCD is a parallelogram

Therefore, BC=ED and BC‖ED

Since, ED=EF+FD=2EF=BC  [ ∵ EF=FD]

We have,EF=12BC and EF||BC

Hence proved

Any four points in a plane, of which three are non-collinear are joined in order results into a four-sided closed figure called ‘quadrilateral’

Angle sum property of a quadrilateral

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

∠1+∠2+∠4=180 (Angle sum property of triangle)…………….(1)

In △ABC,

∠3+∠5+∠6=180 (Angle sum property of triangle)………………(2)

(1) + (2):

∠1+∠2+∠3+∠4+∠5+∠6=360

I.e, ∠A+∠B+∠C+∠D=360

Hence proved

Trapezium

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

Trapezium

PQRS is a trapezium in which PQ||RS

Parallelogram

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

Parallelogram ABCD

Rhombus

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

Rhombus ABCD

A rhombus ABCD in which AB=BC=CD=AD and  AC⊥BD

Rectangle

rectangle is a parallelogram with all angles as right angles.

Rectangle ABCD

A rectangle ABCD in which, ∠A=∠B=∠C=∠D=900

Square

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

Square ABCD

Asquare ABCD in which ∠A=∠B=∠C=∠D=900 and AB=BC=CD=AD

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