Prove that the parallelogram circumscribing a circle is rhombus.

Ans   Given : ABCD is a parallelogram circumscribing a circle.

To prove : - ABCD is a rhombus or

AB=BC=CD=DA

Proof: Since the length of tangents from external are equal in length

∴AS = AR       .....(1)

BQ = BR      .....(2)

QC = PC      .....(3) SD = DP           .....(4)

Adding (1), (2), (3) & (4).

AS + SD + BQ + QC = AR + BR + PC + DP AD + BC = AB + DC

AD + AD = AB + AB

Since BC = AD & DC = AB (opposite sides of a parallelogram are equal)

2AD = 2AB

∴AD = AB      .....(5)

BC = AD (opposite sides of a parallelogram)

DC = AB        .....(6) From (5) and (6)

AB = BC = CD = DA Hence proved