prove that a prallelogram comprising a circle is a rhombus
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Answer: answer in the picture
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Step-by-step explanation:
Let ABCD be a parallelogram which circumscribes the circle.
AP = AS [Since tangents drawn from an external point to a circle are equal in length]
BP = BQ [Since tangents drawn from an external point to a circle are equal in length]
CR = CQ [Since tangents drawn from an external point to a circle are equal in length]
DR = DS [Since tangents drawn from an external point to a circle are equal in length]
Consider, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
But AB = CD and BC = AD [Since opposite sides of parallelogram ABCD]
AB + CD = AD + BC
Hence 2AB = 2BC
Therefore, AB = BC
Similarly, we get AB = DA and DA = CD
Thus ABCD is a rhombus.
Diagram Draw A CIRCLE enclosed by a parallelogram by touching the four sides of a parallelogram. And name the vertices and the joints where the circle touches the parallelogram, also name the centre of the circle as O.
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