Question

prove that the parallelogram circumscribing a circle is a rhombus​

Answer 1

Step-by-step explanation:

Expert Answer:

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.

Answer 2

Answer

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

To prove ABCD is a rhombus :-

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

Therefore :-

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

Adition 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 parellelogram so

AB = DC and AD = BC.

AB = BC

Therefore :-

• AB = BC = CD = AD

Hence ABCD is a rhombus ..

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