Prove that the parallelogram circumscribing a circle is a rhombus.

Let ABCD be a parallelogram such that its sides touch a circle with centre O.



We know that the tangents to a circle from an exterior point are equal in length.

\(AP \ = \ AS \) , \(BP \ = \ BQ \) , \(DR \ = \ DS \) , and \(CR \ = \ CQ \)

Adding these all, we get

\( (AP \ + \ BP) \ + \ (CR \ + \ RD) \ = \ (BQ \ + \ QC) \ + \ (DS \ + \ SA) \)

\( \Rightarrow \ AB \ + \ CD \ = \ BC \ + \ DA \)

\( \Rightarrow \ 2AB \ = \ 2BC \) [\(\because \) ABCD is a parallelogram ]
\( \Rightarrow AB \ = \ CD \) and \( BC \ = \ AD \)]

\(\Rightarrow AB \ = \ BC \)

\(\therefore \) \( AB \ = \ BC \ = \ CD \ = \ AD \)

\(\therefore \) ABCD is a rhombus.