Question

If all the sides of a parallelogram touch a circle, then the parallelogram is?

Answer 1

Find the possible quadrilateral.

Given: All the sides of a parallelogram touch a circle.

Let's draw a figure,

Sides $AB,BC,CD,DA$ of the parallelogram $ABCD$ touch the circle at $P,Q,R,S$.

Since, tangents drawn from an external point to a circle are equal.

Thus,

$AP=AS$ …………$\left(1\right)$ ( Tangent from point $A$ )

$BP=BQ$ …………$\left(2\right)$ ( Tangent from point $B$ )

$CR=CQ$…………$\left(3\right)$ ( Tangent from point $C$ )

$DR=DS$…………$\left(4\right)$ ( Tangent from point $D$ )

Add equation $\left(1\right)$,$\left(2\right)$,$\left(3\right)$,$\left(4\right)$

$$\begin{gathered} AP+BP+CR+DR=AS+BQ+CQ+DS \\ \Rightarrow \left(AP+BP\right)+\left(CR+DR\right)=\left(AS+DS\right)+\left(BQ+CQ\right) \\ \Rightarrow AB+CD=AD+BC \end{gathered}$$

$\Rightarrow AB+AB=AD+AD$ [Opposite side of the parallelogram]

$\Rightarrow 2AB=2AD$

$\Rightarrow AB=AD$

Since, $AB=CD$ and $AD=BC$ [Opposite side of the parallelogram]

Thus, $AB=BC=CD=DA$

We know that in a rhombus opposite side are parallel and all sides are equal.

Therefore the given parallelogram is a rhombus.

Hence, if all the sides of a parallelogram touch a circle, then the parallelogram is rhombus.