Question

Prove that a parallelogram circumscribing a circle is a rhombus brainly

Answer 1

pls mark this as the brainliest answer..plss

Answer 2

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To Prove : A parallelogram circumscribing a circle is a rhombus.

PROOF :

Let, ABCD be a parallelogram.

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

AP  = AS      ------------- ( 1 )

BP = BQ      ------------- ( 2 )

CR = CQ      ------------- ( 3 )

DR = DS      -------------  ( 4 )

Now, we have to add ( 1 ) , ( 2 ) , ( 3 ) and ( 4 ) , then we get :

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

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

⇒ AB + CD = AD + BC

⇒ 2AB = 2BC    { ABCD is a parallelogram ∴ AB = CD and BC = AD }

⇒ AB = BC

Thus,

AB = BC = CD = AD

∴ ABCD is a rhombus

HENCE IT IS PROVED...!!!