Is the limit of this infinite step construction an equilateral triangle?
Think about what happens to the maximum difference between angles over time.
For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $\vert y-x\vert$. Then when we move one of the $y$-angled points, our new triangle will have angles
$$y, {x+y\over 2}, {x+y\over 2}$$
since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is
$$\left\vert {y\over 2}-{x\over 2}\right\vert={1\over 2}\vert y-x\vert.$$
So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $\vert y-x\vert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.
$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, ${1\over 2}$): if $r\in(-1,1)$ then for any $a$ we have
$$\lim_{n\rightarrow\infty}ar^n=0.$$
Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!
By the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $\theta_i$, so that the base angles are $\frac12(\pi - \theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $\theta_{i} = \frac12(\pi-\theta_{i-1})$. Thus, $$\begin{align}\theta_n &= -\frac12\theta_{n-1} + \frac12\pi \\[6pt] &=\frac12\left(-\frac12(\pi-\theta_{n-2})+\pi\right) = \frac14\theta_{n-2}+\frac12\pi-\frac14\pi \\[6pt] &= \cdots \\[6pt] &= \left(-\frac12\right)^{n}\theta_0 \;-\; \sum_{i=1}^n\left(-\frac12\right)^{n}\pi \\[6pt] \lim_{n\to\infty}\theta_n &= 0\cdot\theta_0 \;-\; \frac{(-1/2)}{1-(-1/2)}\pi \\ &=\frac{\pi}{3} \end{align}$$
Thus, in the limit, the triangle becomes equilateral. $\square$
Assume WLOG that the initial triangle is isoceles. Let $\alpha$ be the apical angle, and let $\beta$ be a remaining angle. Then the transformation in question sends
$$\begin{bmatrix}\alpha \\ \beta \end{bmatrix}\mapsto \begin{bmatrix}0 & 1 \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} \begin{bmatrix}\alpha \\ \beta \end{bmatrix}\text{.}$$ Let $\mathsf{X}$ be the $2\times 2$ transformation matrix on the rhs. $\mathsf{X}$ has characteristic polynomial $x^2-\tfrac{1}{2}x-\tfrac{1}{2}=0.$ By the Cayley–Hamilton theorem, $$\mathsf{X}^2=\tfrac{1}{2}\mathsf{X}+\tfrac{1}{2}\text{.}$$ Therefore we have a Sylvester formula $$f(\mathsf{X})=f(1)\left(\frac{1+2\mathsf{X}}{3}\right)+f(-\tfrac{1}{2})\left(\frac{2-2\mathsf{X}}{3}\right)$$ for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus, $$\mathsf{X}^n=\frac{1+2\mathsf{X}}{3}+(-\tfrac{1}{2})^n\left(\frac{2-2\mathsf{X}}{3}\right)\text{.}$$ The second term converges to zero, so $$\begin{split} \lim_{n\to\infty}\mathsf{X}^n&=\frac{1+2\mathsf{X}}{3}\\ &=\frac{1}{3}\begin{bmatrix}1 & 2 \\ 1 & 2\end{bmatrix}\\ &=\frac{1}{3}\begin{bmatrix} 1\\ 1\end{bmatrix}\begin{bmatrix}1&2\end{bmatrix}\text{,} \end{split}$$ $$\lim_{n\to\infty} \begin{bmatrix}0 & 1 \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}^n \begin{bmatrix}\alpha \\ \beta \end{bmatrix}=\begin{bmatrix}\tfrac{\alpha+2\beta}{3}\\ \tfrac{\alpha+2\beta}{3}\end{bmatrix}\text{.}$$ i.e., the apical and side angles approach equality as the operation is repeated.