Question

If all the sides of a parallelogram touches a circle, show that the parallelogram is a rhombus.
                                

Answer 1

Hint : This question is based on the concept of geometry of circles. And its type is based on the result that the tangents drawn from an exterior point to a circle are equal in length. By using some properties of tangent to a circle we can prove this statement.

Complete step by step answer:
Given that ,$ABCD$ is a parallelogram with sides $AB,BC,CD,AD$ touching a circle with centre$O$ at $P,Q,R,S$ respectively.
We have to prove that $ABCD$ is a rhombus .
Proof:
We know that the tangents to a circle from an exterior point are always equal in length.
$\therefore$$AP=AS$ [Tangent From A] …..(i)
$BP=BQ$ [Tangent From B] …….(ii)
$CR=CQ$ [Tangent From C] …….(iii)
And $DR=DS$ [Tangent From D] ……(iv)
Now Adding equation (i),(ii),(iii) and (iv), we get
$AP+BP+CR+DR=AS+BQ+CQ+DS$
Now we separate those sides together which are making a complete side ,
$\Rightarrow$$\left(AP+BP\right)+\left(CR+DR\right)=\left(AS+DS\right)+\left(BQ+CQ\right)$
As given in figure, $\left(AP+BP\right)=AB$ and $\left(CR+DR\right)=CD$ and $\left(AS+DS\right)=AD$ and $\left(BQ+CQ\right)=BC$,we get
$\Rightarrow$$AB+CD+AD+BC$
We know that in a parallelogram, opposite sides are equal so,
$\Rightarrow$$2AB=2BC$ [$\therefore$$ABCD$ is a parallelogram $\therefore$$AB=CD$ and $BC=AD$ ]
$\Rightarrow$$AB=BC$
Thus ,$AB=BC=CD=AD$
Hence ,all sides of $ABCD$ are equal so we can say that $ABCD$ is a rhombus.

Note : In this question ,first we have to identify what is given and what we have to prove, then by using some properties of tangent to a circle we will make some equations based on the sides of the parallelogram then by adding those equations and again using some properties we made a statement which is a property of a rhombus , hence we proved the given statement.