Using similitudes to prove Pythagoras' Theorem Right triangle collage
(See above link for the image I refer to in this problem.)
I'm teaching myself about similitudes, and it's been bothering me that I've been struggling to come up with the similitude that represents the map from the bigger right triangle to the smaller blue right triangle. The original problem asked to use similitudes to prove Pythagoras' Theorem, but I want to make sure I first understand similitudes.
I'm trying this out with a specific example, by letting the big triangle be a 3-4-5 right triangle $(b=3, a=4, c=5)$ where I set the right angle of the triangle at the origin where the $x$- and $y$-axis intersect. I worked with the general similitude form:
$
   s\left( {\begin{array}{c}
   x \\
   y \\
  \end{array} } \right)=
  \left( {\begin{array}{cc}
   r\cos(\theta) & -r\sin(\theta) \\
   r\sin(\theta) & r\cos(\theta) \\
  \end{array} } \right)\left( {\begin{array}{c}
   x \\
   y \\
  \end{array} } \right) + \left( {\begin{array}{c}
   e \\
   f \\
  \end{array} } \right)
$
Since I want to map to the small blue triangle, I let $r = 3/5$. So, mapping from the big triangle to the small triangle, we go from the point $(0,0)$ to $(\frac{12}{5}\cos(53.13^\circ),\frac{12}{5}\sin(53.13^\circ))$ - I did this through some triangle similarity & right-triangle trig. The point $(0,3)$ stays at $(0,3)$, and $(4,0)$ goes to $(0,0)$.
I said $e=\frac{12}{5}\cos(53.13^\circ)$ and $f=\frac{12}{5}\sin(53.13^\circ)$. Intuitively I thought we would need to rotate the larger triangle $223.18^\circ$, but solving for $\theta$ leaves me with $53.18^\circ$. Either way, I cannot get the transformation from $(4,0)$ to $(0,0)$ to work out with my value of $\theta, e,$ and $f$.
Is there something glaringly wrong about how I'm thinking about similitudes? (Teaching myself)

It seems like the approach you're taking is by composing basic affine transformations with each other.
$
   s\left( {\begin{array}{c}
   x \\
   y \\
  \end{array} } \right)=
  \color{red}{\left( {\begin{array}{cc}
   r\cos \theta & -r\sin\theta \\
   r\sin\theta & r\cos\theta \\
  \end{array} } \right)}\left( {\begin{array}{c}
   x \\
   y \\
  \end{array} } \right) + \color{green}{\left( {\begin{array}{c}
   e \\
   f \\
  \end{array} } \right)}
$
Multiplying by the red is a scaling and a rotation, and adding the green is a translation.
You overlooked one detail: You cannot transform the big right triangle to either of the smaller blue and red triangles with only scaling, rotation and translation. You need a reflection as well.
Since any reflection works, let's use a basic one where we flip over the $y$-axis first.
$
   s\left( {\begin{array}{c}
   x \\
   y \\
  \end{array} } \right)= \color{blue}{\left( {\begin{array}{cc}
   -1 & 0 \\
   0 & 1 \\
  \end{array} } \right)}
  \color{red}{\left( {\begin{array}{cc}
   r\cos \theta & -r\sin\theta \\
   r\sin\theta & r\cos\theta \\
  \end{array} } \right)}\left( {\begin{array}{c}
   x \\
   y \\
  \end{array} } \right) + \color{green}{\left( {\begin{array}{c}
   e \\
   f \\
  \end{array} } \right)}
$
$
   s\left( {\begin{array}{c}
   x \\
   y \\
  \end{array} } \right)= 
  \color{purple}{\left( {\begin{array}{cc}
   -r\cos \theta & r\sin\theta \\
   r\sin\theta & r\cos\theta \\
  \end{array} } \right)}\left( {\begin{array}{c}
   x \\
   y \\
  \end{array} } \right) + \color{green}{\left( {\begin{array}{c}
   e \\
   f \\
  \end{array} } \right)}
$
This may also be why your intuition of what $\theta$ should be doesn't match the value you solved for.
