Question

Prove that the parallelogram circumscribing a circle is a rhombus. ​

Answer 1

Answer:

Given: ABCD be a parallelogram circumscribing a circle with centre O.

To prove: ABCD is a rhombus.

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

Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.

Adding the above equations,

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

Answer 2

Step-by-step explanation:

⭐Question:-

Prove that the parallelogram circumscribing a circle is a rhombus.

⭐Answer:-

Solution:

We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle) with centre D.

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

∴ AP = AS

BP = BQ

CR = CQ

DR = DS

Adding, we get,

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

⇒ AB + CD = AD + BC

But AB = CD [opposite sides of ABCD]

and BC = AD

∴ AB + CD = AD + BC ⇒ 2 AB = 2 BC

⇒ AB = BC

Similarly AB = DA and DA = CD

Thus, AB = BC = CD = AD

Hense ABCD is a rhombus.