Prove that the least perimeter of an isosceles triangle in which a circle of radious `r` can be inscribed is `6sqrt(3)r`

Explanation

Let `ABC` be isosceles triangle with `AB=AC` and `r` is radius of circle.

`AF^2+BF^2 =AB^2`

`implies(3r)^2+y^2=x^2ldots`(1)

From `triangle AOD,`

`(2r)^2=r^2+(AD)^2`

`implies 3r^2=AD^2`

`implies AD=sqrt3r`

Now, `BD=BF` & `EC=FC`(as tangents are equal from an external point)

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