Let $D_{1}^{\prime}$ and $D_{2}^{\prime}$ be the points diametrically opposite $D_{1}$ and $D_{2}$ on the incircle and $A$-excircle, respectively. As $I_{x}$ is the midpoint of $D_{x}$ and $D_{x}^{\prime}$, we have $$\frac{\left[A I_{1} D_{1}\right]}{\left[A I_{2} D_{2}\right]}=\frac{\left[A D_{1} D_{1}^{\prime}\right]}{\left[A D_{2} D_{2}^{\prime}\right]}$$ Now, $\triangle A D_{1} D_{1}^{\prime}$ and $\triangle A D_{2} D_{2}^{\prime}$ are homothetic with ratio $\frac{r}{r_{A}}=\frac{s-a}{s}$, where $r$ is the inradius, $r_{A}$ is the $A$-exradius, and $s$ is the semiperimeter. Our answer is thus $$\left(\frac{s-a}{s}\right)^{2}=\left(\frac{4}{12}\right)=\frac{1}{9}$$