Bulgaria

Denote by $k(I,r)$ the incircle of $△ABC$. It is clear that the image of $d$ is the interval $\Delta =(0,r]$.

Varying $D$ and $E$ on $AI$, it follows that $f$ is a mid-point concave function on $\Delta$, i.e. $f(2x)+f(2y)\le f(x+y)$.

Note that the points on given distance (from the boundary) form a triangle homothetic to $△ABC$ (with center of homothety $I$) and the mid-points of the segments with ends at these points run over the whole interior of this triangle. It follows that $f$ is an increasing function on $\Delta$.

Since $f$ is a mid-point concave and increasing function, then it is continuous. Indeed, if $f^{-}(x)$ and $f^{+}(x)$ are the left- and the right-hand limit of $f$ at $x$ ($f^{+}(r):=f(r)$), then $f^{-}(x)+f(x)\le 2f^{-}(x)$ and $f^{-}(x)+f^{+}(x)\le 2f(x)$ (why?) and hence $f^{-}(x)=f(x)=f^{+}(x)$. So $f$ is a convex and increasing function.

Then, since $d$ is a concave function, it is easy to see that $f\circ d$ is a concave function. This also follows from the fact that $f\circ d$ is a mid-point concave and continuous function (because of the continuity of $f$ and $d$).