It is easy to prove that △ AEF △ BDE △ CFD, the sum of the areas of these three triangles is equal to the difference of the areas of two regular triangles.

△ area of △ABC =√3a? /4, the area of △ def =√3b? /4,

So the area of △AEF is S=(√3a? /4 - √3b？ /4)/3=(√3/ 12)(a？ -B? )

△AEF, EF=b, AE+AF=a, and perimeter c = a+b.

Radius of inscribed circle r=2S/C=(√3/6)(a? -B? )/(a+b)=(√3/6)(a-b)