Question
Prove that the parallelogram circumscribing a circle is a rhombus.
✓
Answer
$
\left.\begin{array}{rl}
\therefore AP & =AS \\
BP & =BQ \\
CR & =CQ \\
DR & =DS
\end{array}\right]
$
Adding,
$\ce{(AP + BP) + (CR + DR)=(AS + DS) + (BQ + CQ)}$
$ \Rightarrow \ce{AB + CD=AD + BC}$
Now $\ce{AB=CD}$ and $\ce{AD=BC}$
$\Rightarrow \ce{2 AB=2 BC}$
$\Rightarrow \ce{AB=BC}$
$\Rightarrow \ce{ABCD} \text { is a rhombus }$
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