Question

prove that the parallelogram circumscribing a circle is a rhombus

Answer 1

Hey mate..
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The figure of the given question is in the pic .

Restatement:--
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Let , ABCD be a parallelogram in which a circle with centre circumscribes. AB , BC , CD and AD touches the circle at the points E, F , G and H respectively.

Proof:---
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We know,

Lengths of the tangets drawn from an external point to a circle are equal.

So,

AE = AH..........(1)

BE= BF..............(2)

CG = CF.............(3)

DG = DH.............(4)

Now,

(1) + (2) + (3) + (4)

=> AE + BE + CG + DG = AH + BF + CF + DH

=> AB + CD = ( AH + DH ) + ( BF + CF )

=> AB + CD = AD + BC

We know,

Opposite sides of the parallelogram are equal.

Therefore ,

=> AB + AB = AD + AD

( Since, AB=CD and AD=BC)

=> 2 AB = 2 AD

=> AB = AD

But, They are adjacent sides of the parallelogram.

We also know ,

If the adjacent sides of the parallelogram are equal, then it is a rhombus.

Therefore,

The parallelogram circumscribing a circle is a rhombus.

( Hence , Proved )

#racks