Question

If the sides of a parallelogram touch a circle. Prove that the parallelogram is a rhombus.

Answer 1

In the above drawn figure, we observe the following:

$PQ+SR=PA+AQ+SC+CR$

But $PA=PD,$ $AQ=BQ,$ $SC=SD$ and $CR=RB$

On rearranging we get,

$PQ+SR=PD+SD+BQ+RB$

But $PD+SD=PS$ and $BQ+RB=QR$

Therefore, $PQ+SR=PS+QR$

But $PQ=SR$ and $PS=QR$ (Opposite sides of parallelogram)

Thus,

$PQ+PQ=QR+QR\Rightarrow 2PQ=2QR\Rightarrow PQ=QR\Rightarrow PQ=QR=RS=SP$

Hence, $PQRS$ is a rhombus.