% Copyright 2019 by Till Tantau
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\section{Tutorial: Euclid's Amber Version of the \emph{Elements}}

In this third tutorial we have a look at how \tikzname\ can be used to draw
geometric constructions.

Euclid is currently quite busy writing his new book series, whose working title
is ``Elements'' (Euclid is not quite sure whether this title will convey the
message of the series to future generations correctly, but he intends to change
the title before it goes to the publisher). Up to now, he wrote down his text
and graphics on papyrus, but his publisher suddenly insists that he must submit
in electronic form. Euclid tries to argue with the publisher that electronics
will only be discovered thousands of years later, but the publisher informs him
that the use of papyrus is no longer cutting edge technology and Euclid will
just have to keep up with modern tools.

Slightly disgruntled, Euclid starts converting his papyrus entitled ``Book I,
Proposition I'' to an amber version.


\subsection{Book I, Proposition I}

The drawing on his papyrus looks like this:\footnote{The text is taken from the
wonderful interactive version of Euclid's Elements by David E. Joyce, to be
found on his website at Clark University.}

\bigskip
\noindent
\begin{tikzpicture}[thick,help lines/.style={thin,draw=black!50}]
  \pgfmathsetseed{1}
  \def\A{\textcolor{input}{$A$}}
  \def\B{\textcolor{input}{$B$}}
  \def\C{\textcolor{output}{$C$}}
  \def\D{$D$}
  \def\E{$E$}

  \colorlet{input}{blue!80!black}
  \colorlet{output}{red!70!black}
  \colorlet{triangle}{orange}

  \coordinate [label=left:\A]
    (A) at ($ (0,0) + .1*(rand,rand) $);
  \coordinate [label=right:\B]
    (B) at ($ (1.25,0.25) + .1*(rand,rand) $);

  \draw [input] (A) -- (B);

  \node [name path=D,help lines,draw,label=left:\D] (D) at (A) [circle through=(B)] {};
  \node [name path=E,help lines,draw,label=right:\E] (E) at (B) [circle through=(A)] {};

  \path [name intersections={of=D and E,by={[label=above:\C]C}}];

  \draw [output] (A) -- (C);
  \draw [output] (B) -- (C);

  \foreach \point in {A,B,C}
    \fill [black,opacity=.5] (\point) circle [radius=2pt];

  \begin{pgfonlayer}{background}
    \fill[triangle!80] (A) -- (C) -- (B) -- cycle;
  \end{pgfonlayer}

  \node [below right,text width=10cm,align=justify] at (4,3)
  {
    \small
    \textbf{Proposition I}\par
    \emph{To construct an \textcolor{triangle}{equilateral triangle}
      on a given \textcolor{input}{finite straight line}.}
    \par
    \vskip1em
    Let \A\B\ be the given \textcolor{input}{finite straight line}. It
    is required to construct an \textcolor{triangle}{equilateral
      triangle} on the \textcolor{input}{straight line}~\A\B.

    Describe the circle \B\C\D\ with center~\A\ and radius \A\B. Again
    describe the circle \A\C\E\ with center~\B\ and radius \B\A. Join the
    \textcolor{output}{straight lines} \C\A\ and \C\B\ from the
    point~\C\ at which the circles cut one another to the points~\A\ and~\B.

    Now, since the point~\A\ is the center of the circle \C\D\B,
    therefore \A\C\ equals \A\B. Again, since the point \B\ is the
    center of the circle \C\A\E, therefore \B\C\ equals \B\A. But
    \A\C\ was proved equal to \A\B, therefore each of the straight
    lines \A\C\ and \B\C\ equals \A\B. And
    things which equal the same thing also equal one another,
    therefore \A\C\ also equals \B\C. Therefore the three straight
    lines \A\C, \A\B, and \B\C\ equal one another.
    Therefore the \textcolor{triangle}{triangle} \A\B\C\ is
    equilateral, and it has been  constructed on the given finite
    \textcolor{input}{straight line}~\A\B.
  };
\end{tikzpicture}
\bigskip

Let us have a look at how Euclid can turn this into \tikzname\ code.


\subsubsection{Setting up the Environment}

As in the previous tutorials, Euclid needs to load \tikzname, together with
some libraries. These libraries are |calc|, |intersections|, |through|, and
|backgrounds|. Depending on which format he uses, Euclid would use one of the
following in the preamble:
%
\begin{codeexample}[code only]
% For LaTeX:
\usepackage{tikz}
\usetikzlibrary{calc,intersections,through,backgrounds}
\end{codeexample}

\begin{codeexample}[code only]
% For plain TeX:
\input{tikz.tex}
\usetikzlibrary{calc,intersections,through,backgrounds}
\end{codeexample}

\begin{codeexample}[code only]
% For ConTeXt:
\usemodule[tikz]
\usetikzlibrary[calc,intersections,through,backgrounds]
\end{codeexample}


\subsubsection{The Line \emph{AB}}

The first part of the picture that Euclid wishes to draw is the line $AB$. That
is easy enough, something like |\draw (0,0) -- (2,1);| might do. However,
Euclid does not wish to reference the two points $A$ and $B$ as $(0,0)$ and
$(2,1)$ subsequently. Rather, he wishes to just write |A| and |B|. Indeed, the
whole point of his book is that the points $A$ and $B$ can be arbitrary and all
other points (like $C$) are constructed in terms of their positions. It would
not do if Euclid were to write down the coordinates of $C$ explicitly.

So, Euclid starts with defining two coordinates using the |\coordinate|
command:
%
\begin{codeexample}[]
\begin{tikzpicture}
  \coordinate (A) at (0,0);
  \coordinate (B) at (1.25,0.25);

  \draw[blue] (A) -- (B);
\end{tikzpicture}
\end{codeexample}

That was easy enough. What is missing at this point are the labels for the
coordinates. Euclid does not want them \emph{on} the points, but next to them.
He decides to use the |label| option:
%
\begin{codeexample}[]
\begin{tikzpicture}
  \coordinate [label=left:\textcolor{blue}{$A$}]  (A) at (0,0);
  \coordinate [label=right:\textcolor{blue}{$B$}] (B) at (1.25,0.25);

  \draw[blue] (A) -- (B);
\end{tikzpicture}
\end{codeexample}

At this point, Euclid decides that it would be even nicer if the points $A$ and
$B$ were in some sense ``random''. Then, neither Euclid nor the reader can make
the mistake of taking ``anything for granted'' concerning these position of
these points. Euclid is pleased to learn that there is a |rand| function in
\tikzname\ that does exactly what he needs: It produces a number between $-1$
and $1$. Since \tikzname\ can do a bit of math, Euclid can change the
coordinates of the points as follows:
%
\begin{codeexample}[code only]
\coordinate [...] (A) at (0+0.1*rand,0+0.1*rand);
\coordinate [...] (B) at (1.25+0.1*rand,0.25+0.1*rand);
\end{codeexample}

This works fine. However, Euclid is not quite satisfied since he would prefer
that the ``main coordinates'' $(0,0)$ and $(1.25,0.25)$ are ``kept separate''
from the perturbation $0.1(\mathit{rand},\mathit{rand})$. This means, he would
like to specify that coordinate $A$ as ``the point that is at $(0,0)$ plus one
tenth of the vector  $(\mathit{rand},\mathit{rand})$''.

It turns out that the |calc| library allows him to do exactly this kind of
computation. When this library is loaded, you can use special coordinates that
start with |($| and end with |$)| rather than just |(| and~|)|. Inside these
special coordinates you can give a linear combination of coordinates. (Note
that the dollar signs are only intended to signal that a ``computation'' is
going on; no mathematical typesetting is done.)

The new code for the coordinates is the following:
%
\begin{codeexample}[code only]
\coordinate [...] (A) at ($ (0,0) + .1*(rand,rand) $);
\coordinate [...] (B) at ($ (1.25,0.25) + .1*(rand,rand) $);
\end{codeexample}

Note that if a coordinate in such a computation has a factor (like |.1|), you
must place a |*| directly before the opening parenthesis of the coordinate. You
can nest such computations.


\subsubsection{The Circle Around \emph{A}}

The first tricky construction is the circle around~$A$. We will see later how
to do this in a very simple manner, but first let us do it the ``hard'' way.

The idea is the following: We draw a circle around the point $A$ whose radius
is given by the length of the line $AB$. The difficulty lies in computing the
length of this line.

Two ideas ``nearly'' solve this problem: First, we can write |($ (A) - (B) $)|
for the vector that is the difference between $A$ and~$B$. All we need is the
length of this vector. Second, given two numbers $x$ and $y$, one can write
|veclen(|$x$|,|$y$|)| inside a mathematical expression. This gives the value
$\sqrt{x^2+y^2}$, which is exactly the desired length.

The only remaining problem is to access the $x$- and $y$-coordinate of the
vector~$AB$. For this, we need a new concept: the \emph{let operation}. A let
operation can be given anywhere on a path where a normal path operation like a
line-to or a move-to is expected. The effect of a let operation is to evaluate
some coordinates and to assign the results to special macros. These macros make
it easy to access the $x$- and $y$-coordinates of the coordinates.

Euclid would write the following:
%
\begin{codeexample}[preamble={\usetikzlibrary{calc}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (1.25,0.25);
  \draw (A) -- (B);

  \draw (A) let
              \p1 = ($ (B) - (A) $)
            in
              circle [radius={veclen(\x1,\y1)}];
\end{tikzpicture}
\end{codeexample}

Each assignment in a let operation starts with |\p|, usually followed by a
\meta{digit}. Then comes an equal sign and a coordinate. The coordinate is
evaluated and the result is stored internally. From then on you can use the
following expressions:
%
\begin{enumerate}
    \item |\x|\meta{digit} yields the $x$-coordinate of the resulting point.
    \item |\y|\meta{digit} yields the $y$-coordinate of the resulting point.
    \item |\p|\meta{digit} yields the same as
        |\x|\meta{digit}|,\y|\meta{digit}.
\end{enumerate}
%
You can have multiple assignments in a let operation, just separate them with
commas. In later assignments you can already use the results of earlier
assignments.

Note that |\p1| is not a coordinate in the usual sense. Rather, it just expands
to a string like |10pt,20pt|. So, you cannot write, for instance,
|(\p1.center)| since this would just expand to |(10pt,20pt.center)|, which
makes no sense.

Next, we want to draw both circles at the same time. Each time the radius is
|veclen(\x1,\y1)|. It seems natural to compute this radius only once. For this,
we can also use a let operation: Instead of writing |\p1 = ...|, we write
|\n2 = ...|. Here, ``n'' stands for ``number'' (while ``p'' stands for
``point''). The assignment of a number should be followed by a number in curly
braces.
%
\begin{codeexample}[preamble={\usetikzlibrary{calc}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (1.25,0.25);
  \draw (A) -- (B);

  \draw let \p1 = ($ (B) - (A) $),
            \n2 = {veclen(\x1,\y1)}
        in
          (A) circle [radius=\n2]
          (B) circle [radius=\n2];
\end{tikzpicture}
\end{codeexample}
%
In the above example, you may wonder, what |\n1| would yield? The answer is
that it would be undefined -- the |\p|, |\x|, and |\y| macros refer to the same
logical point, while the |\n| macro has ``its own namespace''. We could even
have replaced |\n2| in the example by |\n1| and it would still work. Indeed,
the digits following these macros are just normal \TeX\ parameters. We could
also use a longer name, but then we have to use curly braces:
%
\begin{codeexample}[preamble={\usetikzlibrary{calc}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (1.25,0.25);
  \draw (A) -- (B);

  \draw let \p1        = ($ (B) - (A) $),
            \n{radius} = {veclen(\x1,\y1)}
        in
          (A) circle [radius=\n{radius}]
          (B) circle [radius=\n{radius}];
\end{tikzpicture}
\end{codeexample}

At the beginning of this section it was promised that there is an easier way to
create the desired circle. The trick is to use the |through| library. As the
name suggests, it contains code for creating shapes that go through a given
point.

The option that we are looking for is |circle through|. This option is given to
a \emph{node} and has the following effects: First, it causes the node's inner
and outer separations to be set to zero. Then it sets the shape of the node to
|circle|. Finally, it sets the radius of the node such that it goes through the
parameter given to |circle through|. This radius is computed in essentially the
same way as above.
%
\begin{codeexample}[preamble={\usetikzlibrary{through}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (1.25,0.25);
  \draw (A) -- (B);

  \node [draw,circle through=(B),label=left:$D$] at (A) {};
\end{tikzpicture}
\end{codeexample}


\subsubsection{The Intersection of the Circles}

Euclid can now draw the line and the circles. The final problem is to compute
the intersection of the two circles. This computation is a bit involved if you
want to do it ``by hand''. Fortunately, the |intersections| library allows us
to compute the intersection of arbitrary paths.

The idea is simple: First, you ``name'' two paths using the |name path| option.
Then, at some later point, you can use the option |name intersections|, which
creates coordinates called |intersection-1|, |intersection-2|, and so on at all
intersections of the paths. Euclid assigns the names |D| and |E| to the paths
of the two circles (which happen to be the same names as the nodes themselves,
but nodes and their paths live in different ``namespaces'').
%
\begin{codeexample}[preamble={\usetikzlibrary{intersections,through}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (1.25,0.25);
  \draw (A) -- (B);

  \node (D) [name path=D,draw,circle through=(B),label=left:$D$]  at (A) {};
  \node (E) [name path=E,draw,circle through=(A),label=right:$E$] at (B) {};

  % Name the coordinates, but do not draw anything:
  \path [name intersections={of=D and E}];

  \coordinate [label=above:$C$] (C) at (intersection-1);

  \draw [red] (A) -- (C);
  \draw [red] (B) -- (C);
\end{tikzpicture}
\end{codeexample}

It turns out that this can be further shortened: The |name intersections| takes
an optional argument |by|, which lets you specify names for the coordinates and
options for them. This creates more compact code. Although Euclid does not need
it for the current picture, it is just a small step to computing the bisection
of the line $AB$:
%
\begin{codeexample}[preamble={\usetikzlibrary{intersections,through}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (1.25,0.25);
  \draw [name path=A--B] (A) -- (B);

  \node (D) [name path=D,draw,circle through=(B),label=left:$D$]  at (A) {};
  \node (E) [name path=E,draw,circle through=(A),label=right:$E$] at (B) {};

  \path [name intersections={of=D and E, by={[label=above:$C$]C, [label=below:$C'$]C'}}];

  \draw [name path=C--C',red] (C) -- (C');

  \path [name intersections={of=A--B and C--C',by=F}];
  \node [fill=red,inner sep=1pt,label=-45:$F$] at (F) {};
\end{tikzpicture}
\end{codeexample}


\subsubsection{The Complete Code}

Back to Euclid's code. He introduces a few macros to make life simpler, like a
|\A| macro for typesetting a blue $A$. He also uses the |background| layer for
drawing the triangle behind everything at the end.
%
\begin{codeexample}[pre={\pgfmathsetseed{1}},preamble={\usetikzlibrary{backgrounds,calc,intersections,through}}]
\begin{tikzpicture}[thick,help lines/.style={thin,draw=black!50}]
  \def\A{\textcolor{input}{$A$}}     \def\B{\textcolor{input}{$B$}}
  \def\C{\textcolor{output}{$C$}}    \def\D{$D$}
  \def\E{$E$}

  \colorlet{input}{blue!80!black}    \colorlet{output}{red!70!black}
  \colorlet{triangle}{orange}

  \coordinate [label=left:\A]  (A) at ($ (0,0) + .1*(rand,rand) $);
  \coordinate [label=right:\B] (B) at ($ (1.25,0.25) + .1*(rand,rand) $);

  \draw [input] (A) -- (B);

  \node [name path=D,help lines,draw,label=left:\D]   (D) at (A) [circle through=(B)] {};
  \node [name path=E,help lines,draw,label=right:\E]  (E) at (B) [circle through=(A)] {};

  \path [name intersections={of=D and E,by={[label=above:\C]C}}];

  \draw [output] (A) -- (C) -- (B);

  \foreach \point in {A,B,C}
    \fill [black,opacity=.5] (\point) circle [radius=2pt];

  \begin{pgfonlayer}{background}
    \fill[triangle!80] (A) -- (C) -- (B) -- cycle;
  \end{pgfonlayer}

  \node [below right, text width=10cm,align=justify] at (4,3) {
    \small\textbf{Proposition I}\par
    \emph{To construct an \textcolor{triangle}{equilateral triangle}
      on a given \textcolor{input}{finite straight line}.}
    \par\vskip1em
    Let \A\B\ be the given \textcolor{input}{finite straight line}.  \dots
  };
\end{tikzpicture}
\end{codeexample}


\subsection{Book I, Proposition II}

The second proposition in the Elements is the following:

\bigskip\noindent
\begin{tikzpicture}[thick,help lines/.style={thin,draw=black!50}]
  \pgfmathsetseed{1}
  \def\A{\textcolor{orange}{$A$}}   \def\B{\textcolor{input}{$B$}}
  \def\C{\textcolor{input}{$C$}}    \def\D{$D$}
  \def\E{$E$}                       \def\F{$F$}
  \def\G{$G$}                       \def\H{$H$}
  \def\K{$K$}                       \def\L{\textcolor{output}{$L$}}

  \colorlet{input}{blue!80!black}    \colorlet{output}{red!70!black}

  \coordinate [label=left:\A]  (A) at ($ (0,0) + .1*(rand,rand) $);
  \coordinate [label=right:\B] (B) at ($ (1,0.2) + .1*(rand,rand) $);
  \coordinate [label=above:\C] (C) at ($ (1,2) + .1*(rand,rand) $);

  \draw [input] (B) -- (C);
  \draw [help lines] (A) -- (B);

  \coordinate [label=above:\D] (D) at ($ (A)!.5!(B) ! {sin(60)*2} ! 90:(B) $);

  \draw [help lines] (D) -- ($ (D)!3.75!(A) $) coordinate [label=-135:\E] (E);
  \draw [help lines] (D) -- ($ (D)!3.75!(B) $) coordinate [label=-45:\F] (F);

  \node (H) at (B) [name path=H,help lines,circle through=(C),draw,label=135:\H] {};
  \path [name path=B--F] (B) -- (F);
  \path [name intersections={of=H and B--F}]
    coordinate [label=right:\G] (G) at (intersection-1);

  \node (K) at (D) [name path=K,help lines,circle through=(G),draw,label=135:\K] {};

  \path [name path=A to E line] (A) -- (E);
  \path [name intersections={of=K and A to E line}]
    coordinate [label=below:\L] (L) at (intersection-1);

  \draw [output] (A) -- (L);

  \foreach \point in {A,B,C,D,G,L}
    \fill [black,opacity=.5] (\point) circle [radius=2pt];

  \node [below right, text width=9cm,align=justify] at (4,4) {
    \small\textbf{Proposition II}\par
    \emph{To place a \textcolor{output}{straight line} equal to a
      given \textcolor{input}{straight line} with
      one end at a \textcolor{orange}{given point}.}
    \par\vskip1em
    Let \A\ be the given point, and \B\C\ the given
    \textcolor{input}{straight line}.
    It is required to place a \textcolor{output}{straight line} equal
    to the given \textcolor{input}{straight line} \B\C\ with one end
    at the point~\A.

    Join the straight line \A\B\ from the point \A\ to the point \B, and
    construct the equilateral triangle \D\A\B\ on it.

    Produce the straight lines \A\E\ and \B\F\ in a straight line with
    \D\A\ and \D\B. Describe the circle \C\G\H\ with center \B\ and
    radius \B\C, and  again, describe the circle \G\K\L\ with center
    \D\ and radius \D\G.

    Since the point \B\ is the center of the circle \C\G\H, therefore
    \B\C\ equals \B\G. Again, since the point \D\ is the center of the
    circle \G\K\L, therefore \D\L\ equals \D\G. And in these \D\A\
    equals \D\B, therefore the remainder \A\L\ equals the remainder
    \B\G. But \B\C\ was also proved  equal to \B\G, therefore each of
    the straight lines \A\L\ and \B\C\ equals \B\G. And things which
    equal the same thing also equal one another, therefore \A\L\ also
    equals \B\C.

    Therefore the \textcolor{output}{straight line} \A\L\ equal to the
    given \textcolor{input}{straight line} \B\C\  has been placed with
    one end at the \textcolor{orange}{given point}~\A.
  };
\end{tikzpicture}


\subsubsection{Using Partway Calculations for the Construction of \emph{D}}

Euclid's construction starts with ``referencing'' Proposition~I for the
construction of the point~$D$. Now, while we could simply repeat the
construction, it seems a bit bothersome that one has to draw all these circles
and do all these complicated constructions.

For this reason, \tikzname\ supports some simplifications. First, there is a
simple syntax for computing a point that is ``partway'' on a line from $p$
to~$q$: You place these two points in a coordinate calculation -- remember,
they start with |($| and end with |$)| -- and then combine them using
|!|\meta{part}|!|. A \meta{part} of |0| refers to the \emph{first} coordinate,
a \meta{part} of |1| refers to the second coordinate, and a value in between
refers to a point on the line from $p$ to~$q$. Thus, the syntax is similar to
the |xcolor| syntax for mixing colors.

Here is the computation of the point in the middle of the line $AB$:
%
\begin{codeexample}[preamble={\usetikzlibrary{calc}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (1.25,0.25);
  \draw (A) -- (B);
  \node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($ (A)!.5!(B) $) {};
\end{tikzpicture}
\end{codeexample}

The computation of the point $D$ in Euclid's second proposition is a bit more
complicated. It can be expressed as follows: Consider the line from $X$ to $B$.
Suppose we rotate this line around $X$ for 90$^\circ$ and then stretch it by a
factor of $\sin(60^\circ) \cdot 2$. This yields the desired point~$D$. We can
do the stretching using the partway modifier above, for the rotation we need a
new modifier: the rotation modifier. The idea is that the second coordinate in
a partway computation can be prefixed by an angle. Then the partway point is
computed normally (as if no angle were given), but the resulting point is
rotated by this angle around the first point.
%
\begin{codeexample}[preamble={\usetikzlibrary{calc}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (1.25,0.25);
  \draw (A) -- (B);
  \node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($ (A)!.5!(B) $) {};
  \node [fill=red,inner sep=1pt,label=above:$D$] (D) at
    ($ (X) ! {sin(60)*2} ! 90:(B) $) {};
  \draw (A) -- (D) -- (B);
\end{tikzpicture}
\end{codeexample}

Finally, it is not necessary to explicitly name the point $X$. Rather, again
like in the |xcolor| package, it is possible to chain partway modifiers:
%
\begin{codeexample}[preamble={\usetikzlibrary{calc}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (1.25,0.25);
  \draw (A) -- (B);
  \node [fill=red,inner sep=1pt,label=above:$D$] (D) at
    ($ (A) ! .5 ! (B) ! {sin(60)*2} ! 90:(B) $) {};
  \draw (A) -- (D) -- (B);
\end{tikzpicture}
\end{codeexample}


\subsubsection{Intersecting a Line and a Circle}

The next step in the construction is to draw a circle around $B$ through $C$,
which is easy enough to do using the |circle through| option. Extending the
lines $DA$ and $DB$ can be done using partway calculations, but this time with
a part value outside the range $[0,1]$:
%
\begin{codeexample}[preamble={\usetikzlibrary{calc,through}}]
\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (0.75,0.25);
  \coordinate [label=above:$C$] (C) at (1,1.5);
  \draw (A) -- (B) -- (C);
  \coordinate [label=above:$D$] (D) at
    ($ (A) ! .5 ! (B) ! {sin(60)*2} ! 90:(B) $) {};
  \node (H) [label=135:$H$,draw,circle through=(C)] at (B) {};
  \draw (D) -- ($ (D) ! 3.5 ! (B) $) coordinate [label=below:$F$] (F);
  \draw (D) -- ($ (D) ! 2.5 ! (A) $) coordinate [label=below:$E$] (E);
\end{tikzpicture}
\end{codeexample}

We now face the problem of finding the point $G$, which is the intersection of
the line $BF$ and the circle $H$. One way is to use yet another variant of the
partway computation: Normally, a partway computation has the form
\meta{p}|!|\meta{factor}|!|\meta{q}, resulting in the point
$(1-\meta{factor})\meta{p} + \meta{factor}\meta{q}$. Alternatively, instead of
\meta{factor} you can also use a \meta{dimension} between the points. In this
case, you get the point that is \meta{dimension} away from \meta{p} on the
straight line to \meta{q}.

We know that the point $G$ is on the way from $B$ to $F$. The distance is given
by the radius of the circle~$H$. Here is the code for computing $H$:
%
{\ifpgfmanualexternalize\tikzexternaldisable\fi
\begin{codeexample}[
    preamble={\usetikzlibrary{calc,through}},
    pre={\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (0.75,0.25);
  \coordinate [label=above:$C$] (C) at (1,1.5);
  \draw (A) -- (B) -- (C);
  \coordinate [label=above:$D$] (D) at
    ($ (A) ! .5 ! (B) ! {sin(60)*2} ! 90:(B) $) {};
  \draw (D) -- ($ (D) ! 3.5 ! (B) $) coordinate [label=below:$F$] (F);
  \draw (D) -- ($ (D) ! 2.5 ! (A) $) coordinate [label=below:$E$] (E);},
    post={\end{tikzpicture}},
]
  \node (H) [label=135:$H$,draw,circle through=(C)] at (B) {};
  \path let \p1 = ($ (B) - (C) $) in
    coordinate [label=left:$G$] (G) at ($ (B) ! veclen(\x1,\y1) ! (F) $);
  \fill[red,opacity=.5] (G) circle [radius=2pt];
\end{codeexample}

However, there is a simpler way: We can simply name the path of the circle and
of the line in question and then use |name intersections| to compute the
intersections.
%
\begin{codeexample}[
    preamble={\usetikzlibrary{calc,intersections,through}},
    pre={\begin{tikzpicture}
  \coordinate [label=left:$A$]  (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (0.75,0.25);
  \coordinate [label=above:$C$] (C) at (1,1.5);
  \draw (A) -- (B) -- (C);
  \coordinate [label=above:$D$] (D) at
    ($ (A) ! .5 ! (B) ! {sin(60)*2} ! 90:(B) $) {};
  \draw (D) -- ($ (D) ! 3.5 ! (B) $) coordinate [label=below:$F$] (F);
  \draw (D) -- ($ (D) ! 2.5 ! (A) $) coordinate [label=below:$E$] (E);},
    post={\end{tikzpicture}},
]
  \node (H) [name path=H,label=135:$H$,draw,circle through=(C)] at (B) {};
  \path [name path=B--F] (B) -- (F);
  \path [name intersections={of=H and B--F,by={[label=left:$G$]G}}];
  \fill[red,opacity=.5] (G) circle [radius=2pt];
\end{codeexample}
}%


\subsubsection{The Complete Code}

\begin{codeexample}[pre={\pgfmathsetseed{1}},preamble={\usetikzlibrary{calc,intersections,through}}]
\begin{tikzpicture}[thick,help lines/.style={thin,draw=black!50}]
  \def\A{\textcolor{orange}{$A$}}   \def\B{\textcolor{input}{$B$}}
  \def\C{\textcolor{input}{$C$}}    \def\D{$D$}
  \def\E{$E$}                       \def\F{$F$}
  \def\G{$G$}                       \def\H{$H$}
  \def\K{$K$}                       \def\L{\textcolor{output}{$L$}}

  \colorlet{input}{blue!80!black}    \colorlet{output}{red!70!black}

  \coordinate [label=left:\A]  (A) at ($ (0,0) + .1*(rand,rand) $);
  \coordinate [label=right:\B] (B) at ($ (1,0.2) + .1*(rand,rand) $);
  \coordinate [label=above:\C] (C) at ($ (1,2) + .1*(rand,rand) $);

  \draw [input] (B) -- (C);
  \draw [help lines] (A) -- (B);

  \coordinate [label=above:\D] (D) at ($ (A)!.5!(B) ! {sin(60)*2} ! 90:(B) $);

  \draw [help lines] (D) -- ($ (D)!3.75!(A) $) coordinate [label=-135:\E] (E);
  \draw [help lines] (D) -- ($ (D)!3.75!(B) $) coordinate [label=-45:\F] (F);

  \node (H) at (B) [name path=H,help lines,circle through=(C),draw,label=135:\H] {};
  \path [name path=B--F] (B) -- (F);
  \path [name intersections={of=H and B--F,by={[label=right:\G]G}}];

  \node (K) at (D) [name path=K,help lines,circle through=(G),draw,label=135:\K] {};
  \path [name path=A--E] (A) -- (E);
  \path [name intersections={of=K and A--E,by={[label=below:\L]L}}];

  \draw [output] (A) -- (L);

  \foreach \point in {A,B,C,D,G,L}
    \fill [black,opacity=.5] (\point) circle [radius=2pt];

  % \node ...
\end{tikzpicture}
\end{codeexample}
