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Prove that the parallelogram circumscribing a circle is a rhom

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Prove that the parallelogram circumscribing a circle is a rhombus.

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Text solutionVerified

Since ABCD is a parallelogram circumscribed in a circle

A B = C D ........ (1)

BC = A D ........ (2)

D R = D S

(Tangents on the circle from same point

D

)

CR = CQ

(Tangent on the circle from same point

C

)

BP = BQ

(Tangent on the circle from same point

B

)

A P = A S

(Tangents on the circle from same point

A

)

Adding all these equations we get

D R + CR + BP + A P = D S + CQ + BQ + A S

(D R + CR) + (BP + A P) = (CQ + BQ) + (D S + A S)

C D + A B = A D + BC

Putting the value of equation

1

and

2

in the above equation we get

2 A B = 2 BC

A B = BC ........... (3)

From equation

(1)

(2)

and

(3)

we get

A B = BC = C D = D A

∴ A BC D

is a Rhombus

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Question Text	Prove that the parallelogram circumscribing a circle is a rhombus.
Answer Type	Text solution:1
Upvotes	150

Prove that the parallelogram circumscribing a circle is a rhombus.

Text solutionVerified

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