Prove please soon rhombus is a parallelogram circumscribing circle
Answers
Given:
A parallelogram ABCD circumscribing a circle
To Prove :
It's a rhombus.
Proof:
ABCD is a ||gm such that its sides touch a circle with centre O.
Therefore,
AB = CD and AB || CD,
also,
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.
Therefore,
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)
Now,
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
Thus,
we get,
AB = BC = CD = AD
Hence,
ABCD is a rhombus.
Answer:
Step-by-step explanation: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
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
2AB = 2BC
(Since, ABCD is a parallelogram so AB = DC and AD = BC)
AB = BC
Therefore, AB = BC = DC = AD.
Hence, ABCD is a rhombus.
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