Question

If all the sides of a parallelogram touch a circle, show that tha parallelogram is a rhombus.
Prove that a parallelogram circumscribing a circle is a rhombus.

Answer 1

Let $ABCD$ be a parallelogram such that its sides touch a circle with centre $O$.

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

$\therefore AP=AS\left[FromA\right].\left(i\right)$

$BP=BQ\left[FromB\right]\left(ii\right)$

$CR=CQ\left[FromC\right]...\left(iii\right)$

and, $DR=DS\left[FromD\right]...\left(iv\right)$

Adding $\left(i\right),\left(ii\right),\left(iii\right)$ and $\left(iv\right),$ we get

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

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

$\Rightarrow AB+CD=AD+BC$

$\Rightarrow 2AB=2BC$ $[$ $\because ABCD$ is a parallelogram $\Rightarrow AB=CD$ and $BC=AD$ $]$

$\Rightarrow AB=BC$

Thus, $AB=BC=CD=AD$

Hence, $ABCD$ is a rhombus.