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