Show 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, \(2 AB = 2BC\)
\(AB = BC\)
If the of a parallelogram are equal, then it is a rhombus.
Hence, \(ABCD\) is a rhombus.