Two right-angle marks coded to be congruent but look different

I have two right-angle marks drawn in a right triangle - one for the right angle in the given triangle and one for the altitude.
They are coded to have an edge length of $(1/\sqrt{2})*3mm$. They do not look to be congruent, though. Is this an optical illusion?

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\usepackage{mathtools}

\begin{document}

\begin{tikzpicture}

\path (0,0) coordinate (A) ({(1/5)*16},0) coordinate (B) ({(1/5)*12},{(1/5)*(4*sqrt(3))}) coordinate (C);
\draw (A) -- (B) -- (C) -- cycle;

%The vertices are typeset.
\node[anchor=north, inner sep=0, font=\footnotesize] at (0,-0.15){$A$};
\node[anchor=north, inner sep=0, font=\footnotesize] at ($(B) +(0,-0.15)$){$B$};
%
%A pin is drawn to C.
\draw let \p1=($(A)-(C)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in ($(C) +({0.5*(\n1+(\n2+180))}:0.05)$) -- ($(C) +({0.5*(\n1+(\n2+180))}:0.45)$);
\draw let \p1=($(A)-(C)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1+(\n2+180))+180}, inner sep=0, font=\footnotesize] at ($(C) +({0.5*(\n1+(\n2+180))}:0.5)$){$C$};

%A right-angle mark is drawn at C. (Since the "calc" package was not drawing the right-angle mark
%correctly, the code for drawing the right-angle mark at C avoids the "calc" package.)
\draw ($(C)!{(1/sqrt(2))*3mm}!(A)$) -- ($($(C)!{(1/sqrt(2))*3mm}!(A)$)!{(1/sqrt(2))*3mm}!-90:(C)$) coordinate (T) -- ($(T)!{(1/sqrt(2))*3mm}!-90:($(C)!{(1/sqrt(2))*3mm}!(A)$)$);

%The foot of the altitude is labeled F.
\coordinate (F) at ({(1/5)*(12)},0);
\node[anchor=north, inner sep=0, font=\footnotesize] at ($(F) +(0,-0.15)$){$F$};
\draw[dashed] (F) -- (C);

%A right-angle mark is drawn at F.
\coordinate (V) at ($(F)!3mm!-45:(A)$);
\draw ($(A)!(V)!(B)$) -- (V) -- ($(C)!(V)!(F)$);

%The lengths of the two parts of the hypotenuse that are partitioned by F are typeset.
\node[anchor=south, inner sep=0, font=\scriptsize] at ($($(A)!0.5!(F)$) +(0,0.1)$){$12$};
\node[anchor=south, inner sep=0, font=\scriptsize] at ($($(B)!0.5!(F)$) +(0,0.1)$){$4$};

\path[fill=gray!50] (A) -- (C) -- ($(C)!{(1/5)*8*sqrt(3)*1cm}!-90:(A)$) coordinate (P) -- ($(P)!{(1/5)*8*sqrt(3)*1cm}!-90:(C)$) -- (A) -- cycle;
\draw (A) -- (C) -- ($(C)!{(1/5)*8*sqrt(3)*1cm}!-90:(A)$) -- ($(P)!{(1/5)*8*sqrt(3)*1cm}!-90:(C)$) -- cycle;

\draw (B) -- (C) -- ($(C)!{(1/5)*8*1cm}!90:(B)$) coordinate (Q) -- ($(Q)!{(1/5)*8*1cm}!90:(C)$) -- (B) -- cycle;

\end{tikzpicture}

\end{document}


• @gernot I originally scaled the diagram by a factor of 1/4. Before posting the code, I decided that a factor of 1/5 would be more appropriate for my file. I must have been in the process of changing all these factors when I posted the code. My apologies. cfr edited the code for me. (He changed all the factors to 1/5.) – A gal named Desire Oct 15 '16 at 21:55
• @AgalnamedDesire Note that tikz has all kinds of scaling options. If you start with \begin{tikzpicture}{scale=0.5}, the diagram will be scaled down to half its size. If you use the label option instead of placing labels as separate nodes, then this will do exactly what you try to achieve with your factors. – gernot Oct 15 '16 at 22:08
• @gernot If I were to use scale=0.5, does that decrease the font of the labels for the vertices and edge lengths? – A gal named Desire Oct 15 '16 at 22:25
• @gernot Don't you mean that to be an optional argument? @A gal named Desire assumptions are dangerous online .,... – cfr Oct 15 '16 at 23:15
• @AgalnamedDesire As @cfr points out it should read \begin{tikzpicture}[scale=0.5] (square brackets instead of braces). This will not scale done labels, only distances will shrink. It's meant to scale drawings but not written stuff. If you want to scale everything, including fonts, you have to use canvas scaling, another option. Its effect is what you achieve in pdf or html viewers when magnifying or shrinking a picture. – gernot Oct 16 '16 at 7:02

\draw [red] (C) circle ({3mm/(sqrt(2))}) (F) circle ({3mm/(sqrt(2))});