Question

Prove logically the diagonals of a parallelogram bisect each other. Show conversely that a quadrilateral in which diagonals bisect each other is a parallelogram.

Answer 1

i ) The diagonals of a parallelogram

bisects each other .

Proof : Draw a parallelogram ABCD.

Draw both of it's diagonals AC and BD

to intersect at a point O .

In ∆OAB and ∆OCD

Mark the angles formed as <1,<2,<3,<4

<1 = <3

<2 = <4

[ AB//CD and AC transversal ,

Interior alternate angles ]

and

AB = CD [ property of parallelogram ]

∆OCD is congruent to ∆OAB

[ A.S.A congruence property ]

CO = OA ,

DO = OB

Or

Diagonals bisect each other
___________________________

Converse : If the diagonals of a

quadrilateral bisect each other then

it is a parallelogram.

Proof : ABCD is a quadrilateral

[ from figure ( 2 ) ]

AC and BD are the diagonals intersect

at O .

such that OA = OC

and OB = OD

In ∆AOB and ∆COD ,

OA = OC

<AOB = <COD

OB = OD

Therefore ,

∆AOB is congruent to ∆COD

AB = CD [ C.P.C.T ]

<OAB = <OCA [ Alternate angles ]

AB//CD

Similarly ,

AD = BC , AD // BC

Therefore ,

ABCD is a parallelogram.

•••••

Answer 2

Let ABCD be a rectangle.

Then AC and BD are the diagonals.

To Prove: AC=BD

Proof : In △ABC and △ABD

∠ABC = ∠BAD [90⁰]

BC=AD [Opp. Sides]

AB=AB [Common side]

△ABC ≅ △ABD[SAS]

AC=BD

Hence proved✅.