Question

prove that the parallalogram circumseribing a circle is a rhombus​

Answer 1

Answer:

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Step-by-step explanation:

Given: ABCD be a parallelogram circumscribing a circle with centre O.

To prove: ABCD is a rhombus.

We know that the tangents drawn to a circle from an exterior point are equal in length.

Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.

Adding the above equations,

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

2AB = 2BC

(Since, ABCD is a parallelogram so AB = DC and AD = BC)

AB = BC

Therefore, AB = BC = DC = AD.

Hence, ABCD is a rhombus.

Answer 2

Given :-

ABCD is a parallelogram

Therefore,

AB = CD [ Opposite sides of parallelogram ]

BC = AD [ Opposite sides of parallelogram ]

Solution :-

[ A rhombus is a type of parallelogram whose all sides are equal ]

It can be observed from the figure,

• DR = DS

[ Tangent of circle from point D] (1)

• CR = CQ

[ Tangent of circle from point C ] (2)

• BP = BQ

[ Tangent of a circle from point B ] (3)

• AP = AS

[ Tangent of a circle from point A ] (4)

Adding eq( 1 ) , ( 2 ) , ( 3 ) and ( 4 )

DR + CR + BP + AP = DS + CQ + BQ + AS

{ DR + CR } + { BP + AP} = { DS + AS} + { CQ + AQ}

CD + AB = AD + BC

[ ABCD is a parallelogram ]

2AB = 2BC [ AB = CD, BC = AD ]

AB = BC ( 5 )

CD = BC ( 6 )

CD = AD ( 7 )

AD = AB ( 8)

From ( 5) , ( 6) , ( 7 ) and ( 8 )

AB = BC = CD = AD

All sides of rhombus are equal .

Hence proved. $.$

[ Note :- Refer the attachment ]