Math, asked by Rayman123, 1 year ago

Prove that a parallelogram circumscribing a circle is a rhombus.

Answers

Answered by millybhatia

Here, is ur ans.... Hope it helps

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Rayman123: Thanx❤

millybhatia: Wlcm

millybhatia: If u find any problem plz let me know

Rayman123: No problem

Answered by snehchudasma

Given:- Quadrilateral ABCD is a parallelogram

T.P.:- quadrilateral ABCD is rhombus

Proof:-

According to theorem,

AB + CD = BC + AD

so, 2AB = 2BC (because ABCD is II^gm)

so, AB = BC = CD= AD

SO, quadrilateral ABCD is Rhombus

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millybhatia: U have written this according to which theorem????

snehchudasma: according to the theorem that the two tangents drawn from the external point are equal.

snehchudasma: so AP = AS, PB = BQ, cr = cq, dr = ds. so adding them we will get AB + CD = AD + BC

millybhatia: but u should give that in explaination

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