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Prove that the parallelogram circumscribing a circle is a rhombus.

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Please refer to video for the figure.
For parallelogram `ABCD`
`AB=CD and AD = BC`
We know, tangents drawn from a point to a circle are always equal.So,
`AE = AH,EB = BF,GC=CF,DG = HD`
`AE+EB+CG+DG = AH+HD+CF+BF`
`AB+CD = AD+BC`
`2AB = 2BC=>AB=BC`
Similarly, `AD=CD`
Thus, all sides are equal.So, ABCD is a rhombus.

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