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Prove that the parallelogram circumscribing a circle is a rhombus.
Proof:
We know that, the lengths of tangents drawn from an external point to a circle are equal.
\(AP = \) ----(1)
\(BP = \) ----(2)
\(CR = \) -----(3)
\(DR = \) -----(4)
By, adding (1), (2), (3) and (4) RHS = LHS, we get
\(AP + BP + CR + DR = AS + BQ + CQ + DS (AP + BP) + (CR + DR)\)
\(= (AS + DS) + (BQ + CQ)\)
\( + CD = AD +\)
As \(ABCD\) is a, \(AB = CD\) and \(AD = BC\)
Hence, \(2 AB = 2 BC\)
\(AB = BC\)
If the of a parallelogram are equal, then it is a rhombus.
Hence, \(ABCD\) is a rhombus.
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