question_answer

The parallelogram circumscribing a circle is a

A)  Square                        

B)  Rectangle

C)  Rhombus     

D)  Cyclic quadrilateral

Correct Answer: C

Solution :

(c): Let ABCD be a parallelogram circumscribing a circle. Using the property that the tangents to a circle from an exterior point are equal in length. \[AM=AP\] and \[BM=BN\] \[\therefore \]\[\text{ }CO=CN\]and \[DO=DP\] On adding all equations, we get \[\left( AM+BM \right)+\left( CO+DO \right)\]\[=AP+BN+CN+DP\left\{ \begin{align}   & \because \,AB=AM+MB, \\  & \,\,\,\,BC=BN+NC, \\  & \,\,\,\,CD=CO+OD, \\  & \,\,\,\,AD=AP+PD \\ \end{align} \right\}\] \[\Rightarrow \]\[AB+CD=\left( AP+PD \right)+\left( BN+NC \right)\]\[=AD+BC\] \[\Rightarrow \]\[2AB=2BC\] (\[\because \] ABCD is a  parallelogram \[\Rightarrow \] \[AB=CD\] and \[BC=AD\]) \[\Rightarrow \]\[AB=BC\]\[\therefore \] \[AB=BC=CD=DA\] Hence, ABCD is a rhombus.