Math, asked by gamingdevil861, 5 months ago

PROVE THAT THE PARALLELOGRAM CIRCUMSCRIBING A CIRCLE IS A RHOMBUS​

Answers

Answered by Anonymous
0

Answer:

iven: 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.

Answered by swatiyadav10
1

Answer:

Since ABCD is a parallelogram circumscribed in a circle

AB=CD........(1)

BC=AD........(2)

DR=DS (Tangents on the circle from same point D)

CR=CQ(Tangent on the circle from same point C)

BP=BQ (Tangent on the circle from same point B )

AP=AS (Tangents on the circle from same point A)

Adding all these equations we get

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

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

CD+AB=AD+BC

Putting the value of equation 1 and 2 in the above equation we get

2AB=2BC

AB=BC...........(3)

From equation (1), (2) and (3) we get

AB=BC=CD=DA

∴ABCD is a Rhombus

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