Question

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer 1

Given, a parallelogram ABCD circumscribes a circle with centre O.
$AB=BC=CD=AD$
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
$$\begin{gathered} \therefore AP=AS\dots \dots .\left(i\right)\left[\mathrm{tangents}\mathrm{}\mathrm{from}A\right] \\ BP=BQ\dots \dots \dots .\left(ii\right)\left[\mathrm{tangents}\mathrm{}\mathrm{from}B\right] \\ CR=CQ\dots \dots \dots .\left(iii\right)\left[\mathrm{tangents}\mathrm{}\mathrm{from}C\right] \\ DR=DS\dots \dots \dots ..\left(iv\right)\left[\mathrm{tangents}\mathrm{}\mathrm{from}D\right] \\ \therefore AB+CD=AP+BP+CR+DR \\ =AS+BQ+CQ+DS[\mathrm{from}\left(i\right),\left(ii\right),\left(iii\right)\mathrm{and}\left(iv\right)] \\ =\left(AS+DS\right)+(BQ+CQ) \\ =AD+BC \\ \mathrm{Thus},\left(AB+CD\right)=\left(AD+BC\right) \\ \Rightarrow 2AB=2AD\left[\because \mathrm{opposite}\mathrm{}\mathrm{sides}\mathrm{}\mathrm{of}\mathrm{}\mathrm{a}\mathrm{}\mathrm{parallelogram}\mathrm{}\mathrm{are}\mathrm{}\mathrm{equal}\right] \\ \Rightarrow AB=AD \\ \therefore CD=AB=AD=BC \end{gathered}$$
Hence, ABCD is a rhombus.