Let the radius of circle O be a, and the extension line of BO intersects with AC at point E, so it is easy to know that it is ⊥ AC.

Let AC = 2y (0 < Y ≤ A), OE=x, AB=l, then a2=x2+y2, S=y(a+x),

L2 = y2+(a+x)2 = y2+a2+2ax+x2 = 2 a2+2ax = 2a(a+x)= 2aS/y

Because ∠ABC=2∠OBA=2∠OAB=∠BDO, AB=BC, DB=DO.

So △BDO∽△ABC, so BD/AB =BO /AC, which means r/l =a/2y, so R = Al/2Y. .

So R2 = a2l2/4y2 = a2/4y2-2as/y = s/2-(a/y) 3 ≥ s/2.

That is, r≥ 2S /2, where the equal sign holds when a=y and AC is the diameter of the circle O. Therefore, the minimum value of the radius r of the circle D is (root sign 2S)/2.