Question
Prove that the parallelogram circumscribing a circle is a rhombus.
✓
Answer
Let $\ce{ABCD}$ is the parallelogram circumscribing a circle with centre $O$.
To Prove: $\ce{ABCD}$ is a rhombus.
In parallelogram $\ce{ABCD}$,
$A S=A P ($tangents drawn to a circle from an exterior point$)$
$B Q=B P ($tangents drawn to a circle from an exterior point$)$
$C Q=C R ($tangents drawn to a circle from an exterior point$)$
$D S=D R ($tangents drawn to a circle from an exterior point$)$
Thus,
$A S+B Q+C Q+D S=A P+B P+C R+D R$
$(A S+D S)+(B Q+C Q)=(A P+B P)+(C R+D R)$
$\therefore A D+B C=A B+C D$
$\because A D=B C $ and $ A B=C D ($Opposite sides of parallelogram$)$
$\therefore 2 B C=2 A B$
$\Rightarrow B C=A B$
$\therefore A B=B C=D C=A D$
$\therefore \ce{ABCD}$ is a rhombus.
Hence proved.
To Prove: $\ce{ABCD}$ is a rhombus.
In parallelogram $\ce{ABCD}$,
$A S=A P ($tangents drawn to a circle from an exterior point$)$
$B Q=B P ($tangents drawn to a circle from an exterior point$)$
$C Q=C R ($tangents drawn to a circle from an exterior point$)$
$D S=D R ($tangents drawn to a circle from an exterior point$)$
Thus,
$A S+B Q+C Q+D S=A P+B P+C R+D R$
$(A S+D S)+(B Q+C Q)=(A P+B P)+(C R+D R)$
$\therefore A D+B C=A B+C D$
$\because A D=B C $ and $ A B=C D ($Opposite sides of parallelogram$)$
$\therefore 2 B C=2 A B$
$\Rightarrow B C=A B$
$\therefore A B=B C=D C=A D$
$\therefore \ce{ABCD}$ is a rhombus.
Hence proved.
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