Question

prove that the parallelogram circumscribing a circle is a rhombus​

Answer 1

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

Answer 2

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