Question

13. Prove that the parallelogram circumscribing a circle is a rhombus.​

Answer 1

$\huge\bold\red{solution}$

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