How much do we have to do this action to make this triangle equilateral? 
How much do we have to do this action to make this triangle equilateral:

$P(A): $ Move $A$ to the meet of the perpendicular bisector of $\overline{BC}$ and parallel line of $\overline{BC}$ which includes $A$.

So, I tried some:
Original:

$P(A)$:

$P(B)$:

$P(C)$:

And so on...
Can we make this triangle equilateral with these actions? If we can, how much do I have to do these actions($P(A), P(B), P(C), P(A_1), P(B_1), P(C_1), \cdots$) to make $\triangle A_{\square}B_{\square}C_{\square}$ equilateral?
 A: Notice that after the first step, you always have an isosceles triangle. So we want to get from an arbitrary isosceles triangle to the equilateral triangle with the same area (as area is preserved by that action). Or, ignoring area and using similarity rather than congruence, we want to get from an arbitrary isosceles triangle to the equilateral triangle.
Working backwards, we want to get from the equilateral triangle to all isosceles triangles, by applying the inverse action: Given an isosceles triangle, take the vertex $A$ on the line of symmetry, and move it along the line $l$ (containing $A$) parallel to the opposite edge $BC$, until you have another isosceles triangle. There are (up to congruence) at most two ways to do this, since the circle of radius $|BC|$ centred at $C$ intersects the line $l$ in at most two points.
So we have a binary tree, where each node is an isosceles (possibly equilateral) triangle. The tree may be finite or countably infinite. But the set of all isosceles triangles is uncountably infinite, as an angle may vary in a continuum.
Hence, for almost all isosceles triangles, the equilateral triangle cannot be reached in a finite number of steps. It might be reached as some kind of limit, though.

Let $\theta_n$ be one angle in the $n$th isosceles triangle, with the other two angles being $(\pi-\theta_n)/2$. The initial value $\theta_1$ can be anything between $0$ and $\pi$. Your action gives the next angle
$$\cos\theta_{n+1}=\frac{3-4\cos^2\theta_n}{5-4\cos^2\theta_n}.$$
(I won't derive this formula here, unless someone asks, maybe.)
We want to know whether $\lim_{n\to\infty}\theta_n=\pi/3$, or equivalently whether $\lim_{n\to\infty}\cos\theta_n=1/2$, given $-1<\cos\theta_1<1$.
The fixed points of the function $x\mapsto(3-4x^2)/(5-4x^2)$ are $-1$, $1/2$, and $3/2$. The third is irrelevant here. The first is repulsive; letting $\cos\theta_n=-1+\varepsilon_n$, we have
$$\varepsilon_{n+1}=\frac{8\,\varepsilon_n(2-\varepsilon_n)}{1+8\,\varepsilon_n-4\,\varepsilon_n^2}\approx16\,\varepsilon_n$$
when $\varepsilon_n$ is small. (In other words, applying your action to a nearly flat isosceles triangle gives a less flat triangle.) The second is attractive; letting $\cos\theta_n=1/2+\varepsilon_n$, we have
$$\varepsilon_{n+1}=\frac{-\varepsilon_n(1+\varepsilon_n)}{2(1-\varepsilon_n-\varepsilon_n^2)}\approx-\frac12\,\varepsilon_n$$
when $\varepsilon_n$ is small. (In other words, applying your action to a nearly equilateral triangle gives a more nearly equilateral triangle.) There is still some work to be done, to show that $\cos\theta_n$ eventually gets close enough to $1/2$ that this approximation is valid, even when $\theta_1$ is far from $1/2$....
A: Inspired by mr_e_man, I work as follows
Using the fact that the triangles are area preserving and isosceles for $n \ge 1$, I arrive at a different formula, namely
$$t_{n+1}= \frac {1+{t_n}^2}{4t_n}$$
where $$t_n=\tan \frac {\theta_n}{2}$$
and $\theta_n$ is the vertical angle of the $n$th isosceles triangle.
From the formula
$$t_{n+1}= \frac {1+{t_n}^2}{4t_n}$$
and considering the fixed point $t=\frac{1}{\sqrt 3}$.
We can deduce that $$ \left \vert \frac {t_{n+1}-\frac {1}{\sqrt 3}}{t_{n}-\frac {1}{\sqrt 3}} \right \vert = \frac {1}{4}\left \vert \frac {t_n- \sqrt 3}{t_n} \right \vert$$
Our job is done if we can prove that $\frac {1}{4}\left \vert \frac {t_n- \sqrt 3}{t_n}  \right \vert \le c$ for some $0 \le c \lt 1$.
Put $c=\frac {3}{4}$
$$\frac {1}{4}\left \vert \frac {t_n- \sqrt 3}{t_n} \right \vert \le\frac {3}{4} $$
$$ \iff -3 \le\frac {t_n- \sqrt 3}{t_n}  \le 3$$
$$ \iff \frac {\sqrt 3}{4} \le t_n $$
But this is always true for $n \ge 2$ because $$t_{n+1}= \frac {1+{t_n}^2}{4t_n} \ge \frac {2t_n}{4t_n}=\frac {1}{2} \ge \frac {\sqrt 3}{4} $$
for $n \ge 1$.
