5.Prove that the Ilgm circumscribing a circle is a rhombus.
Answers
Step-by-step explanation:
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
solution
Given: 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.
