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Q prove that parallelogram circumscribing a …

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Q prove that parallelogram circumscribing a circle is a rhombus??
  • 3 answers

Gaurav Seth 6 years, 11 months ago

Since ABCD is a parallelogram,

AB = CD ---- i)

BC = AD ---- ii) 

It can be observed that

DR = DS (Tangents on the circle from point D)

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

AP = AS (Tangents on the circle from point A)

Adding all these equations, we obtain

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

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

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain

2AB = 2BC

AB = BC …(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.

Baap Of All. 6 years, 11 months ago

Kaku bhai i am not satisfied on your answer. Plz solve it in detail with proper explaination.

Yaar 6 years, 11 months ago

ab + cd = bc + ad == ab + ab = bc + bc.....(ab = cd, bc = ad) == 2ab = 2bc == ab = bc
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