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# Prove that a parallelogram circumscribing a circle is a rhombus.

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.

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).