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Hint: Use the property that tangents from an external point to a circle are equal in length. Hence show that all the sides of the parallelogram are equal.A quadrilateral (parallelogram) in which all four sides have equal in length then it is said to be rhombus.Finally show that the parallelogram is a rhombus.

Complete step-by-step answer:

Given: A parallelogram ABCD circumscribing a circle.

To prove: ABCD is a rhombus

Proof:

We know that tangents drawn from an external point to a circle are equal in length.

We have B is an external point, and BP and BQ are tangents to the circle. So, we have

BP = BQ.

Similarly, C is an external point, and CQ and CR are external tangents.

Hence CQ = CR.

Similarly DR = DS and AS = AP

We have

AP = AS

Adding BP on both sides, we get

AP+BP = AS+BP

But BP = BQ and AP+BP = AB, so we have

AB = AS+ BQ

Adding CQ on both sides, we get

AB + CQ = AS+ BQ +CQ

But CQ = CR and BQ +CQ = BC, so we have

AB + CR = AS + BC

Adding DR on both sides, we get

AB + CR + DR = AS + BC + DR.

But DR = DS and CR + CR = CD, so we have

AB + CD = AS+DS + BC

But, AS + DS = AD, so we have

AB + CD = AD + BC.

Since ABCD is a parallelogram, we have AB = CD and AD = BC (Because opposite sides of a parallelogram are equal).

Hence, we have AB + AB = AD +AD

i.e 2AB = 2AD

Hence AB = AD.

But AB = CD and AD = BC

Hence AB = BC = CD = DA

Hence ABCD is a rhombus.

Note: [1] A quadrilateral in which all four sides have equal in length then it is said to be rhombus.

[2] Lengths of tangents from an external point are equal. This can be shown by joining the point to the centre of the circle and connecting points of contact to the centre. Hence show that the two triangles so formed are congruent.

[3] The circle which is circumscribed by the rhombus is called the incircle of the rhombus.

Complete step-by-step answer:

Given: A parallelogram ABCD circumscribing a circle.

To prove: ABCD is a rhombus

Proof:

We know that tangents drawn from an external point to a circle are equal in length.

We have B is an external point, and BP and BQ are tangents to the circle. So, we have

BP = BQ.

Similarly, C is an external point, and CQ and CR are external tangents.

Hence CQ = CR.

Similarly DR = DS and AS = AP

We have

AP = AS

Adding BP on both sides, we get

AP+BP = AS+BP

But BP = BQ and AP+BP = AB, so we have

AB = AS+ BQ

Adding CQ on both sides, we get

AB + CQ = AS+ BQ +CQ

But CQ = CR and BQ +CQ = BC, so we have

AB + CR = AS + BC

Adding DR on both sides, we get

AB + CR + DR = AS + BC + DR.

But DR = DS and CR + CR = CD, so we have

AB + CD = AS+DS + BC

But, AS + DS = AD, so we have

AB + CD = AD + BC.

Since ABCD is a parallelogram, we have AB = CD and AD = BC (Because opposite sides of a parallelogram are equal).

Hence, we have AB + AB = AD +AD

i.e 2AB = 2AD

Hence AB = AD.

But AB = CD and AD = BC

Hence AB = BC = CD = DA

Hence ABCD is a rhombus.

Note: [1] A quadrilateral in which all four sides have equal in length then it is said to be rhombus.

[2] Lengths of tangents from an external point are equal. This can be shown by joining the point to the centre of the circle and connecting points of contact to the centre. Hence show that the two triangles so formed are congruent.

[3] The circle which is circumscribed by the rhombus is called the incircle of the rhombus.