Question

prove that a parallelogram circumscribing a circle is rhombus

Answer 1

ABCD is a parallelogram which touches a circle at P, Q, R, S. But AP= AS.... i, &
PB=PQ..... ii( tangent from an external point are always equal)
So from.i and ii
We get
AB=CD=AD=BC.

Hope it help you

Answer 2

Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.

To Proof : ABCD is a rhombus.

As ABCD is a parallelogram

AB = CD and BC = AD …[1]

[opposite sides of a parallelogram are equal]

Now, As tangents drawn from an external point are equal.

We have

AP = AS

[tangents from point A]

BP = BQ

[tangents from point B]

CR = CQ

[tangents from point C]

DR = DS

[tangents from point D]

Add the above equations

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

AB + CD = AS + DS + BQ + CQ

AB + CD = AD + BC

AB + AB = BC + BC [From 1]

AB = BC …[2]

From [1] and [2]

AB = BC = CD = AD

And we know,

A parallelogram with all sides equal is a rhombus

So, ABCD is a rhombus.

Hence Proved.