If all sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.

Given: A parallelogram whose all sides touching a circle.
To do: To prove that the parallelogram is a rhombus.
Here we draw a parallelogram PQRS circumscribing a circle and all its sides touches the circle at points A, B, C and D. 

As it is a parallelogram 
$PS=QR$ and $PQ\ =RS$
As we know tangents to a circle from the same external point are always equal in length.
By adding above four equations 
$( QD+RD) +( PB+SC) =( PA+QA) +( RC+SB)$
$QR+( PB+SB) =PQ+( RC+SC)$                                      $( as\  known\ SB=SC\ )$
$QR+PS=PQ+RS$                                              $( PB+SB=PS\ and\ RC+SC=RS)$
$QR+QR=PQ+PQ$                                       $( \because \ PS=QR\ and PQ=RS)$
$\Rightarrow\ PQ=QR=RS=SP$
$\therefore$ parallelogram $PQRS$ is a rhombus.


Simply Easy Learning

Updated on: 10-Oct-2022


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