proove that a parallelogram circumscribing a circle is a rhombus with complete proof.
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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
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.
∴ 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.
jimmy14:
how can u say that ABCD is a rhombus but it can also be a square.
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.
thanks bro..
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