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# 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.

**Solution:**

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.

$PB=PA$

$QD=QA$

$RD=RC$

$SC=SB$

By adding above four equations

$PB+QD+RD+SC=PA+QA+RC+SB$

$( QD+RD) +( PB+SC) =( PA+QA) +( RC+SB)$

$QR+( PB+SB) =PQ+( RC+SC)$ $( as\ known\ SB=SC\ )$

$QR+PS=PQ+RS$ $( PB+SB=PS\ and\ RC+SC=RS)$

$QR+QR=PQ+PQ$ $( \because \ PS=QR\ and PQ=RS)$

$2QR=2PQ$

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

$\therefore$ parallelogram $PQRS$ is a rhombus.

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