If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
Answers
Answer:
Step-by-step explanation:
ANSWER
Given : Sides AB,BC,CD andDA of a parrallelogram ABCD touch a circle at P,Q,Rand S respectively.
To prove : Parallelogram ABCD is a rhombus.
Proof :
SinceTangents drawn from an external point to a circle are equal
AP=AS.....(1)(Tangent from point A )
BP=BQ....(2)(Tangent from point B )
CR=CQ....(3)(Tangent from point A )
DR=DS....(4)(Tangent from point A )
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==AD+AD [In a Parallelogram ABCD, opposite sides are equal]
2AB==2ADorAB==AD
But AB==CDandAD==BC [Opposite sides of a Parallelogram are equal]
AB==BC==CD==DA
Hence, Parallelogram ABCD is a rhombus.plz mark me.....
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Your answer
Since ABCD is a parallelogram,
AB = CD …(1)
BC = AD …(2)
It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC …(3)
Comparing equations (1), (2), and (3), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.