A circle of radius $r$ is inscribed in a right isosceles triangle, and a circle of radius $R$ is circumscribed about the triangle. Then $R/r$ equals
$\mathrm{(A)\ } 1+\sqrt{2} \qquad \mathrm{(B) \ }\frac{2+\sqrt{2}}{2} \qquad \mathrm{(C) \  } \frac{\sqrt{2}-1}{2} \qquad \mathrm{(D) \  } \frac{1+\sqrt{2}}{2} \qquad \mathrm{(E) \  }2(2-\sqrt{2})$
Label the points as in the figure above. Let the side length $AB=AC=s$. Therefore, $BC=s\sqrt{2}$. Since the circumradius of a right triangle is equal to half of the length of the hypotenuse, we have $R=\frac{s\sqrt{2}}{2}$.
Now to find the inradius. Notice that $IFAE$ is a square with side length $r$, and also $AD=R$. Therefore, $s=AD=AI+ID=r\sqrt{2}+r$, and so $r=\frac{R}{\sqrt{2}+1}$.
Finally, $\frac{R}{r}=1+\sqrt{2}, \boxed{\text{A}}$.