Tikz: draw a path the same length as another path starting at a specific coordinate and intersecting a given point

Question

I have some tikz code that draws a triangle with labelled verticies.

I would like to draw a line that starts at the origin, intersecting the point labelled P, and has a length equal to the edge labelled A.

I figure that the calc library is what I want for this, but I'm having trouble working out how to feed it the length of A...

Here is my code so far, with my best guess at what I think the edge I want to draw command will look like:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{intersections, calc, through, backgrounds}
\begin{document}
\begin{tikzpicture}[scale=3]
% layout some coordinates for a triangle
\coordinate (P) at ($(0,0) + (rand,rand)$);
\coordinate (Q) at ($(2,-2) + .5(rand, rand)$);
\coordinate (R) at ($(-2, -2) + .5(rand, rand)$);
%convienience for finding the centre
\coordinate (PQR) at ($(P)+(Q)+(R)$);
\coordinate (O) at ($1/3*(PQR)$);
%draw triangle
\draw[name path=A] (P) -- (Q);
\draw[name path=B] (Q) -- (R);
\draw[name path=C] (R) -- (P);
%labels
\node at (O) {$O$};
\node at (P) [above=2pt]{$P$};
\node at (Q) [right=2pt]{$Q$};
\node at (R) [left=2pt]{$R$};
\node at ($(P)!.5!(Q)$) [above right=1em]{$\mathbf{A}$};
%draw a path of length (A) from (O), passing through (P)
%\draw[name path=U] (O) -- ($(O)! !(B)$)
\end{tikzpicture}
\end{document}

Which gives me this output

Answer 1

You can calculate the length of the "vector" (\x,\y) using sqrt(\x1*\x1+\y1*\y1). Then use the construction

($(A)!<len>!(B)-(A)$)

to create a vector from (A) in the direction of (B) – (A) with length <len>:

Here, the blue line has endpoints (A) = (1,0) and (B) = (3,1). The red line is drawn from (C) = (1,1) through (D) = (2,2) with the length of the line from (A) to (B) using the code

\draw[red,thick] let\p1=($(B)-(A)$) in (C)--++($(C)!sqrt(\x1*\x1+\y1*\y1)!(D)-(C)$);

The let command assigns to \p1 the "vector" from (A) to (B), which automatically assigns its x and y components to \x1 and \y1.

If you wish, you could define a command \thru that takes 5 arguments, one optional:

\newcommand{\thru}[5][]{\draw[#1] let\p1=($#5-#4$) in #2--++($#2!sqrt(\x1*\x1+\y1*\y1)!#3-#2$);}

The optional argument would be any tikz attributes you want. The first required argument is your starting point, the second is the through point, the third and fourth are the endpoints of the segment whose length you want.

So \thru[green,thick]{(2,1)}{(3,2)}{(A)}{(B)} would draw the thick green line from (2,1) through (3,2) having the length of segment (A)--(B).

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\newcommand{\thru}[5][]{\draw[#1] let\p1=($#5-#4$) in #2--++($#2!sqrt(\x1\x1+\y1\y1)!#3-#2$);}
\begin{document}
\begin{tikzpicture}
\draw[help lines] (0,0) grid (4,3);
\coordinate(A) at (1,0);
\coordinate(B) at (3,1);
\coordinate(C) at (1,1);
\coordinate(D) at (2,2);
\draw[blue,thick] (A)--(B);
\draw[red,thick] let\p1=($(B)-(A)$) in (C)--++($(C)!sqrt(\x1\x1+\y1\y1)!(D)-(C)$);
\thru[green,thick]{(2,1)}{(3,2)}{(A)}{(B)}
\end{tikzpicture}
\end{document}

Answer 2

Using tkz-euclide, you may use this:

\documentclass[tikz,border=10pt]{standalone}
\usepackage{tkz-euclide}
\begin{document}
\begin{tikzpicture}[scale=3]

    % Coordinates for a triangle
    \coordinate (P) at ($(0,0) + (rand,rand)$);
    \coordinate (Q) at ($(2,-2) + .5*(rand, rand)$);
    \coordinate (R) at ($(-2, -2) + .5*(rand, rand)$);
    \coordinate (O) at (barycentric cs:P=1,Q=1,R=1) ;

    % Draw triangle
    \draw (P) -- (Q) node[midway,above,sloped] {dist $=A$}  -- (R) -- cycle;

    \tkzDrawPoints(P,Q,R,O)

    % Labels
    \node at (O) [left=2pt]{$O$};
    \node at (P) [above right]{$P$};
    \node at (Q) [right=2pt]{$Q$};
    \node at (R) [left=2pt]{$R$};

    % Calculate length of segment PQ and label as dPQ
    \tkzCalcLength(P,Q)\tkzGetLength{dPQ}

    % Create some point on a circle centred on O and with radius = PQ length
    \tkzDefPointOnCircle[angle=90,center=O,radius=\dPQ pt]
    \tkzGetPoint{I}

    % Find intersection between line OP and previous circle
    % Two points, but only one is useful
    \tkzInterLC(O,P)(O,I) \tkzGetPoints{D}{E}

    % Draw the segment from O, passing through P, the length of PQ
    \draw[dashed,blue] (O) -- (E) node[pos=0.7,left] {dist$=A$};

    \tkzDrawPoints(E)
    \tkzLabelPoints[left](E)

\end{tikzpicture}

\end{document}

EDIT
Taking advantage of this other clever option (apart from veclen which I definitely would not use), here comes another shorter version, using tikz-euclide to compute the segment length and a simpler implementation of the drawn segment.

\documentclass[tikz,border=10pt]{standalone}
\usepackage{tkz-euclide}
\begin{document}
\begin{tikzpicture}[scale=3]

    % Coordinates for a triangle
    \coordinate (P) at ($(0,0) + (rand,rand)$);
    \coordinate (Q) at ($(2,-2) + .5*(rand, rand)$);
    \coordinate (R) at ($(-2, -2) + .5*(rand, rand)$);
    \coordinate (O) at (barycentric cs:P=1,Q=1,R=1) ;

    % Draw triangle
    \draw (P) -- (Q) node[midway,above,sloped] {dist $=A$}  -- (R) -- cycle;

    \tkzDrawPoints(P,Q,R,O)

    % Labels
    \node at (O) [left=2pt]{$O$};
    \node at (P) [above right]{$P$};
    \node at (Q) [right=2pt]{$Q$};
    \node at (R) [left=2pt]{$R$};

    % Calculate length of segment PQ
    \tkzCalcLength(P,Q)\tkzGetLength{dPQ}

    % Draw the segment from O, passing through P, the length of PQ
    \draw[orange,thick,dashed] (O)--++ ($(O)!\dPQ pt!(P)-(O)$) coordinate[at end] (E);

    \tkzDrawPoints(E)
    \tkzLabelPoints[left](E)


\end{tikzpicture}

\end{document}