Question

Prove that a parallelogram circumscribing a circle is a rhombus​

Answer 1

$\huge{\pink {\sf\underline{Step\: by\: step\: explanation}}}$

Given ABCD is a ||gm such that its sides touch a circle with centre O.

∴ AB = CD and AB || CD,

AD = BC and AD || BC

Now, P, Q, R and S are the touching point of both the circle and the ||gm

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

∴ AP = AS [Tangents from point A] ... (1)

BP = BQ [Tangents from point B] ... (2)

CR = CQ [Tangents from point C] ... (3)

DR = DS [Tangents from point D] ... (4)

On adding (1), (2), (3) and (4), we get

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

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

⇒ AB + CD = AD + BC

⇒ AB + AB = BC + BC [∵ ABCD is a ||gm . ∴ AB = CD and AD = BC]

⇒ 2AB = 2BC

⇒ AB = BC

Therefore, AB = BC implies

AB = BC = CD = AD

Hence, ABCD is a rhombus.

In rhombus, it is not necessary that diagonals are equal. If they are equal, then rhombus is considered as a square whose diagonals are always equal. So, there isn't any use of proving that the diagonals of a rhombus are equal.

Answer 2

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Given ABCD is a ||gm such that its sides touch a circle with centre O.

∴ AB = CD and AB || CD,

AD = BC and AD || BC

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

∴ AP = AS [Tangents from point A] ... (1)

BP = BQ [Tangents from point B] ... (2)

CR = CQ [Tangents from point C] ... (3)

DR = DS [Tangents from point D] ... (4)

On adding (1), (2), (3) and (4), we get

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

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

⇒ AB + CD = AD + BC

⇒ AB + AB = BC + BC [∵ ABCD is a ||gm . ∴ AB = CD and AD = BC]

⇒ 2AB = 2BC

⇒ AB = BC

Therefore, AB = BC implies

AB = BC = CD = AD.

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