Question

If all the sides of a parallelogram touches a circle, show that the parallelogram is a rhombus.

Answer 1

Given : Sides AB,BC,CD andDA of a parrallelogram ABCD touch a circle at P,Q,Rand S respectively.
To prove : Parallelogram ABCD is a rhombus.
Proof :
SinceTangents drawn from an external point to a circle are equal
AP=AS.....(1)(Tangent from point A )
BP=BQ....(2)(Tangent from point B )
CR=CQ....(3)(Tangent from point A )
DR=DS....(4)(Tangent from point A )
Adding (1), (2), (3) and (4), we get
AP+BP+CR+DR$=$AS+BQ+CQ+DS
(AP+BP)+(CR+DR)$=$(AS+DS)+(BQ+CQ)
AB+CD$=$AD+BC
AB+AB$=$AD+AD [In a Parallelogram ABCD, opposite sides are equal]
2AB$=$2ADorAB$=$AD
But AB$=$CDandAD$=$BC [Opposite sides of a Parallelogram are equal]
AB$=$BC$=$CD$=$DA
Hence, Parallelogram ABCD is a rhombus.