120
Prove that the parallelogram circumscribing a circle is a rhombus.
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Correct figure
$\therefore AP = AS$
$BP = BQ$
$CR = CQ$
$DR = DS$
Adding,
$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$
$\Rightarrow AB + CD = AD + BC$
Now $AB = CD$ and $AD = BC$
$\Rightarrow 2 AB = 2 BC$
$\Rightarrow AB = BC$
$\Rightarrow ABCD$ is a rhombus
$\therefore AP = AS$
$BP = BQ$
$CR = CQ$
$DR = DS$
Adding,
$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$
$\Rightarrow AB + CD = AD + BC$
Now $AB = CD$ and $AD = BC$
$\Rightarrow 2 AB = 2 BC$
$\Rightarrow AB = BC$
$\Rightarrow ABCD$ is a rhombus