Question

If all the sides of a parallelogram touches a circle, then the parallelogram is necessarily a

• A. rhombus.
• B. rectangle
• C. square
• D. None of the above.

Answer 1

The correct option is A rhombus.

Sides $AB,BC,CD$ and $DA$ of a $|{|}^{gm}ABCD$ touches a circle at $P,Q,R$ and $S$ respectively.
To prove: $|{|}^{gm}ABCD$ is a rhombus.
$AP=AS$ ....$\left(i\right)$
$BP=BQ$ ....$\left(ii\right)$
$CR=CQ$ ....$\left(iii\right)$
$DR=DS$ ....$\left(iv\right)$
[Tangents drawn from an external point to a circle are equal ]
Adding $\left(i\right),\left(ii\right),\left(iii\right)$ and $\left(iv\right)$, we get
$\Rightarrow AP+BP+CR+DR=AS+BQ+CQ+DS$
$\Rightarrow (AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)$
$\Rightarrow AB+CD=AD+BC$
$\Rightarrow AB+AB=AD+AD$
[In a $|{|}^{gm}$, opposite sides are equal]
$\Rightarrow 2AB=2AD$ or $AB=AD$
But $AB=CD$ and $AD=BC$ [Opposite side of a $|{|}^{gm}$]
$\therefore AB=BC=CD=DA$
Hence, $|{|}^{gm}ABCD$ is a rhombus.