prove that the parallelogram circumscribed a circle is a rhombus .
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Answer:
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.