Question

prove that the parallelogram circumscribing a circle is a rhombus..​

Answer 1

Answer:

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Step-by-step explanation:

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