Question

Prove that the parallelogram circumscribing a circle is a square

Answer 1

the question is wrong. it is to actually prove it is a rhombus

Answer 2

$\bf\huge\underline{Question}$

Prove that the parallelogram circumscribing a circle is a square.

$\bf\huge\underline{Answer}$

We have ABCD, a parallelogram which circumscribes a circle (i.e., it's sides touch the circle) with centre O.

Since, tangents to a circle form an external point are equal in length

Therefore, ⠀AP = AS

⠀⠀⠀⠀⠀⠀⠀⠀BP = BQ

⠀⠀⠀⠀⠀⠀⠀⠀CR = CQ

⠀⠀⠀⠀⠀⠀⠀⠀DR = DS

On adding, we get

(AP + BP) + (CR + DR)

(AS + DS) + (BQ + CQ)

=> AB + CD = AD + BC

But AB = CD

⠀⠀⠀⠀⠀[Opposite sides of parallelogram]

and BC = AD

Therefore, AB + CD = AD + BC

=> 2AB = 2BC

=> AB = BC

Similarly, AB = DA and DA = CD

Thus, AB = BC = CD = DA

Hence, ABCD is a rhombus.