Question

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer 1

STEP 1 : Assumption

Consider a parallelogram $ABCD$ which is circumscribing a circle with a centre $O$.

Since $ABCD$ is a parallelogram, $AB=CD$ and $BC=AD$.

STEP 2 : Proving that $ABCD$ is a rhombus

We know that the tangents drawn to a circle from an exterior point are equal is length.

$\Rightarrow AP=AS$
$\Rightarrow BP=BQ$,
$\Rightarrow CR=CQ$ and
$\Rightarrow DR=DS$

Adding all the above equations, we get

$AP+BP+CR+DR=AS+BQ+CQ+DS$

$(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)$

$\Rightarrow AB+CD=AD+BC$

$\Rightarrow 2AB=2AD$ (Since $ABCD$ is a parallelogram, $AB=CD$ and $BC=AD$)

$\Rightarrow AB=AD$

Since, $AB=AD$, $AB=CD$ and $AD=BC$

$\Rightarrow AB=BC=CD=AD$

Since all the sides of a parallelogram are equal

Hence, $ABCD$ is a rhombus.