Many incircles

Since $D$ is the point of tangency of the incircle of $ABC$ with $BC$ , we have $BD = s-b$ and $CD = s-c$ where $s = \tfrac12(a+b+c)$ is the semiperimeter of $ABC$ . Similarly, if $E_1,E_2$ are the points of tangency of the incircles of $ABD$ and $ACD$ with $AD$ , then $DE_1 = s_1 - c$ and $DE_2 = s_2-b$ , where $s_1,s_2$ are the semiperimeters of $ABD$ and $ACD$ . But $s_1 = \tfrac12(c + AD + s-b) \hspace{2cm} s_2 = \tfrac12(b + AD + s - c)$ and hence $s_1 - c = s_2 - b$ , and hence $E_1 = E_2$ .

The line perpendicular to $AD$ passing through $E_1=E_2$ must be a radius to both incircles, and hence must pass through both $I_1$ and $I_2$ . Thus it is clear that $I_1I_2 = r_1+r_2$ , making the answer $\boxed{0}$ .