60
If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
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As lengths of tangents drawn from an external point to a circle are equal
$AP = AS \dots (i)$
$BP = BQ \dots (ii)$
$DR = DS \dots (iii)$
$CR = CQ \dots (iv)$
Adding the above equations we get, $AB + CD = AD + BC$
$\implies 2AB = 2 AD$ [$\because$ Opp. Sides of a Parallelogram are equal]
$\implies AB = AD \implies \parallel gm ABCD$ is a rhombus.
$AP = AS \dots (i)$
$BP = BQ \dots (ii)$
$DR = DS \dots (iii)$
$CR = CQ \dots (iv)$
Adding the above equations we get, $AB + CD = AD + BC$
$\implies 2AB = 2 AD$ [$\because$ Opp. Sides of a Parallelogram are equal]
$\implies AB = AD \implies \parallel gm ABCD$ is a rhombus.