Question

Prove that parallelogram of a circumcircle of a circle is a rhombus.

Answer 1

Since ABCD is a parallelogram,

AB = CD …(1)

BC= AD ....2

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

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

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

Adding all these equations, we obtain

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

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

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain

2AB = 2BC

AB = BC …(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.

Answer 2

Ilet ABCD be a parallelogram circumscribing a circle and P, Q, R and S be points touching the sides AB, BC, CD, and AD respectively.
here
PB is equal to BQ - - - - - i
AP is equal to AS - - - - - ii
CR is equal to CQ - - - - -iii
DRis equal to DS - - - - - iv
Tangents from external point
Now, i + ii + iii + iv
=> BP + AP + CR + DR= BQ + CQ + AS + DS
=> AB + CD = AD + BC
=> 2AB = 2 BC. ( Opposite sides of parallelogram )
=> AB= BC
Thus, CD=AB=BC=AD
AB=BC=CD=AD
Therefore ABCD is a rhombus