Question

prove that the parallelogram circumscribe a circle is a rhombus​

Answer 1

Given:

ABCD is a parallelogram.

To prove:

ABCD is a rhombus.

Proof:

We know that,

tangents drawn from an external point to a circle are equal in length.

AP=AS. ---(i)

BP=BQ. ---- (ii)

CR=CQ. ---- (iii)

DR=DS. -----(iv)

Adding (I) (II) (III) and (IV)

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

AB+CD=AD+BC

Since,ABCD is a parallelogram

Ab=CD and BC=AD ---- (v)

So

AB+AB=AD+AD

2AB=2AD

AB=AD. -----(vi)

From (v) and (vi) we get

AB=CD=BC=AD

So, ABCD is a rhombus

Answer 2

Answer:

A circle with centre O

A parallelogram ABCD is touching the circle at points P,Q,R and S

To prove: ABCD is a rhombus

Proof:

A rhombus is a parallelogram with all sides equal,So,we have to prove all sides equal

In parallelogram ABCD,

AB=CD and AD = BC

Hence,

AP = AS

BP = BQ

CR = CQ

DR = DS

Addying 2 + 3 + 4 + 5

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

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

AB + CD = AD + BC

AB + AB = AD + AD

2AB = 2AD

AB = AD

So AB = AD

and AB = CD and AD = BC

So, ABCD is a parallelogram with all sides equal

ABCD is a Rhombus

Hence Proved

Hope it helps u

Good afternoon