Consider the following construction of a triangle center: (The method could also be easily generalized to any shape with finite perimeter)
For each point $X$ on the triangle, find point $X'$ such that $X$ and $X'$ split the triangle into two sections of equal length (each of length equal to the triangle semiperimeter). Then, let $Y$ be the midpoint of $X$ and $X'$. The locus of all such $Y$ is also triangular, so we can repeat the process on the triangle created by the locus of all such $Y$, ad infinitum until the triangles converge to a point.
Through some computer simulation, I was able to determine that this convergence point isn't one of the common triangle centers. Searching through the Encyclopedia of Triangle Centers, I also wasn't able to find this triangle center listed. Does this center have an established name, or is there literature available on the topic?
Not an answer, but information that may help lead to one.
With the help of Mathematica to push some symbols, I've determined that the coordinates of the "Draksis Triangle" $\triangle A^\prime B^\prime C^\prime$ is obtained from those of $\triangle ABC$ thusly:
$$\begin{align} A^\prime &= \frac{1}{2}\;A + \frac{\cos(\beta/2) \sin(\gamma/2)}{2\cos(\alpha/2)}\;B + \frac{\sin(\beta/2)\cos(\gamma/2)}{2\cos(\alpha/2)}\;C \\[4pt] B^\prime &= \frac{\sin(\gamma/2)\cos(\alpha/2)}{2\cos(\beta/2)}\;A + \frac{1}{2}\;B + \frac{\cos(\gamma/2) \sin(\alpha/2)}{2\cos(\beta/2)}\;C \\[4pt] C^\prime &= \frac{\cos(\alpha/2) \sin(\beta/2)}{2\cos(\gamma/2)}\;A + \frac{\sin(\alpha/2)\cos(\beta/2)}{2\cos(\gamma/2)}\;B + \frac{1}{2}\;C \end{align}$$ where $\alpha := \angle BAC$, $\beta := \angle CBA$, $\gamma := \angle ACB$.
Here are some metrics of the Draksis Triangle: $$a^\prime = a\sin(\beta/2) \sin(\gamma/2) \qquad b^\prime = b\sin(\gamma/2) \sin(\alpha/2) \qquad c^\prime = c\sin(\alpha/2) \sin(\beta/2)$$
$$\alpha^\prime = \frac{\pi - \alpha}{2} \qquad \beta^\prime = \frac{\pi - \beta}{2} \qquad \gamma^\prime = \frac{\pi - \gamma}{2}$$
where $a := |\overline{BC}|$ and $a^\prime := |\overline{B^\prime C^\prime}|$, etc. Note that the angle formula confirms @MvG's assertion that iterated Draksis triangles converge on an equilateral: If $\alpha = \frac13\pi + \theta$, then $\alpha^\prime = \frac13\pi - \frac12\theta$.
Iterating the construction to get at the "Draksis Point" seems daunting, but given the relations above, there may yet be hope.