Question

prove that the parallelogram circumscribing a circle is a rhombus​

Answer 1

Step-by-step explanation:

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

(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.

Answer 2

Prove that the parallelogram circumscribing a circle is a rhombus.

Solution :-

ABCD is a parallelogram

AB = CD ________eq (1)

BC = AD ________eq (2)

(Opposite sides are equal)

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 get

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

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

CD + AB = AD + BC

Putting These values in eqn (1) & (2)

We Obtain

2AB = 2BC

AB = BC _______(3)

Comparing eqn (1) ,(2) & (3) we obtain

AB = BC = CD = DA

So, ABCD is rhombus

Hence Proved

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