Prove that the parallelogram circumscribing a circle is a rhombus.

Correct figure
$$\begin{aligned}& AP = AS -----(i) \\ & BP = BQ -----(ii) \\ & CR = CQ -----(iii) \\ & DR = DS -----(iv) \\ & \text{Adding (i), (ii), (iii) \& (iv)} \\ & AP + BP + CR + DR = AS + BQ + CQ + DS \\ & \Rightarrow AB + CD = AD + BC \\ & \text{But ABCD is a parallelogram } \Rightarrow AB = CD \text{ and } AD = BC \\ & \therefore 2AB = 2AD \text{ or } AB = AD \\ & \text{Hence, ABCD is a rhombus.}\end{aligned}$$