How can I transform these triangles to $30-60-90$ triangles? I'm trying to transform the triangles below subject to a few constraints, and cannot figure out how (algebraically) to do so. For ease of notation, I'll refer to points like $E(C)$ as $C$. 
I want $CEF$ and $BEF$ to be $30-60-90$. I want to leave $E$ and $C$ fixed. Right now $ECF < 30$ and $EBF > 30$. 
How can I continuously transform the points of these triangles so the end results is to slide $F$ down (so that its $y$-coordinate is the same as $C$'s) and slide $B$ down and to the right so that $EBF = 30$, while preserving line segments inside each triangle? Bonus points if you can preserve line segments that cross between the two triangles (I think it's impossible). 
Ideally the solution would look something line $x' = f(x,y)$, $y' = g(x,y)$. I rotated and translated to get to the orientation below, because I suspected polar coordinates would help, but if it's easier to use some other starting point, that's fine.
Thanks!

 A: In general you can map $\triangle EBC$ to $\triangle EBC^\prime$ with a linear transformation $f(\mathbb{R}^2)\mapsto\mathbb{R}^2$ which leaves every point of line $EB$ fixed and maps $B$ to $C^\prime$. However, it will not map $F$ to $F^\prime$ unless $F$ is the midpoint of segment $CB$.
For each point $X$ in $\triangle CEB$ the ray $BX$ intersects side $EC$ at some unique point $P$ and there is a unique number $s\in[0,1]$ such that
\begin{equation}
X=(1-s)B+sP \tag{1}
\end{equation}
For each $P$ of segment $EC$ there is a unique number $t\in[0,1]$ such that 
\begin{equation}
P=(1-t)E+tC \tag{2}
\end{equation}
Thus for each point $X\in\triangle CEB$ there are unique $\triangle CEB$ coordinates $(s,t)$ for $X$ satisfying
\begin{equation}
X=(1-s)B+s(1-t)E+stC \tag{3}
\end{equation}
In  $\triangle CEC^\prime$ the diagram below, segment $CF^\prime$ is congruent to segment $F^\prime C^\prime$.
The transformation
\begin{equation}
f(X(s,t))=(1-s)C^\prime+s(1-t)E+stC \tag{4}
\end{equation}
maps $\triangle CEB$ to $\triangle CEC^\prime$ and maps each line segment in $\triangle CEB$ to a line segment in $\triangle CEC^\prime$. However, it does not map $F$ to $F^\prime$ (unless $F$ is the midpoint of $CB$). Instead it maps $F^\prime$ to some point $F^{\prime\prime}$ on segment $CC^\prime$.
There is no linear transformation $g\,:\triangle CEC^\prime\to\triangle CEC^\prime$ leaving the vertices of $\triangle CEC^\prime$ fixed while mapping $F^{\prime\prime}$ to $F^\prime$.

