# arc that passes from (B) having a center on (A), automatically calculates (AB) distance

I want to create an arc which automatically calculates the distance (AB). Inspired by page 58 at pgf manual. But the code block % \draw (H) ++arc (15:60: $(A) ! - ! (B)$); does not help much.

Namely I want one arc that passes from (B) having a center on (A). Any idea?

These answer actually gives a lot of useful information but I can not figure out. How can I compute the distance between two coordinates in TikZ? and TikZ: Drawing an arc between two coordinates?

 \documentclass[b5paper,11pt]{book}

\usepackage{tikz}
\usetikzlibrary{decorations.markings,calc,intersections,through,backgrounds}

\begin{document}

\begin{center}
\begin{tikzpicture}[scale=1.25]%,cap=round,>=latex]

\draw[step=.5cm,gray,very thin] (-2.0cm,-2.0cm) grid (2.cm,2.0cm);

\coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
\coordinate [label=right:$C$] (C) at (1.5cm,-1.0cm);
\coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
\draw (A) -- node[above] {$g$} (B) -- node[right] {$a$} (C) -- node[below] {$b$} (A);

\filldraw [ color = green!10!white, draw=green!10!black] (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

%     \draw (H)  ++arc (15:60: $(A) ! - ! (B)$);

\draw  (5:2cm) arc(15:60:2cm);

\end{tikzpicture}
\end{center}
\end{document}


Αs a final result I want to create something like this

-

First you can use the calc library to make the dashed hypotenuse a little rotated via using the angle argument of ($(node1)!fraction!angle:(node2)$) syntax. It means that the path is going to be drawn from A to B but then it's going to be rotated angle degrees.

Then, you can use a let operation and compute the distance and the angle. Then you can use it for forming an arc that necessarily passes from (B) and rotated (B).

\documentclass[b5paper,11pt]{book}
\usepackage{tikz}
\usetikzlibrary{decorations.markings,calc,intersections,through,backgrounds}

\begin{document}

\begin{center}
\begin{tikzpicture}[scale=1.25]%,cap=round,>=latex]

\draw[step=.5cm,gray!50,very thin] (-2.0cm,-2.0cm) grid (2.cm,2.0cm);

\coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
\coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
\coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
\draw (A) -- node[above] {$g$} (B) -- node[right] {$a$} (C) -- node[below] {$b$} (A);

\filldraw [fill = green!10!white,
draw=green!10!black] (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

\draw[dashed] (A)--  ($(A)!1!-10:(B)$) |- (A);
\draw let                        % We initiate a let operation inside a path
\p1=($(B)-(A)$),         % Declare a vector from A to B
\n1={veclen(\x1,\y1)},   % Compute it's length, let's call it "l"
\n2={atan2(\x1,\y1)}     % Compute it's angle with respect to A, say theta
in                         % finish off the \p and \n register declaration
(A) ++(\n2-15:\n1)       % Move from A to a point in polar form (theta-15:l)
arc (\n2-15:\n2+15:\n1)  % Draw an arc from that point that covers
% 30 degrees with a start angle of theta-15
(A) ++(5mm,0) arc (0:\n2:5mm) % goes back to A and moves horizontally
% to start an arc theta angle
(A) node at ++(\n2/2:7mm) {$\theta$};  % goes again to (A) and places label
\end{tikzpicture}
\end{center}
\end{document}


-
Extremely simple solution. It's easy to explain the command? \draw let \p1=($(B)-(A)$),\n1={veclen(\x1,\y1)},\n2={atan2(\x1,\y1)} in (A) ++(\n2-15:\n1) arc (\n2-15:\n2+15:\n1); – karathan Mar 12 '13 at 12:22
@karathan Sure, in a bit, I'll add some more info. – percusse Mar 12 '13 at 12:26
@karathan Does it make it a little clearer? – percusse Mar 12 '13 at 12:40
@karathan Try this : \usetikzlibrary{calc} \begin{tikzpicture} \node (a) at (0,0) {A}; \node (b) at (2,0) {B}; \draw (a) -- ($(a)!0.7!(b)$); \end{tikzpicture} and play with 0.7 with different numbers to see how fraction modifies the length. 1 is the full length 0.5 is the halfway – percusse Mar 18 '13 at 22:34
After that example now adding \draw (a) -- ($(a)!0.7!10:(b)$); makes the path rotated 10 degrees after it has been calculated. Start with 0 degrees which is equal to not putting an angle and gradually increase the angle and you'll see what I mean. – percusse Mar 18 '13 at 22:55

A good job for tkz-euclide with several possibilities.

First \draw[dashed] (A)-- ($(A)!1!-10:(B)$) |- (C); is preferable (C instead of A) With tkz-euclide the mark for the right angle is obtained with \tkzMarkRightAngle[fill=green!20](A,C,B)

We can get the arc with \tkzCompasss[color = blue,delta = 15](A,B). We draw an arc with center A through B 15 degrees before and after B

\documentclass[11pt]{article}
\usepackage{tkz-euclide}
\usetkzobj{all}
\usetikzlibrary{decorations.markings,calc,intersections,through,backgrounds}

\begin{document}

\begin{center}
\begin{tikzpicture}[scale=1.25]

\draw[step=.5cm,gray!50,very thin] (-2.0cm,-2.0cm) grid (2.cm,2.0cm);

\coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
\coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
\coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
\draw (A) -- node[above](g) {$g$} (B)
-- node[right] {$a$} (C)
-- node[below] {$b$} (A);

\tkzMarkRightAngle[fill=green!20](A,C,B)
\draw[dashed] (A)--  ($(A)!1!-10:(B)$) |- (C);
% \tkzCompasss[color  = orange,length = 2](A,B)
\tkzCompasss[color  = blue,delta = 15](A,B)
% \tkzDrawArc[delta=15](A,B)(B)

% \tkzCalcLength[cm](A,B)\tkzGetLength{rAB}
% \node [above=0.5cm,font=\small] at (g)  { \rAB cm};
\end{tikzpicture}
\end{center}
\end{document}


With

  \tkzCompasss[color  = orange,length = 2](A,B)


I defined the length of the arc

Another possibility is to use the more general macro \tkzDrawArc

\tkzDrawArc[delta=15](A,B)(B)


this is a special use of the macro because I draw an arc with center A from B to B but I define delta = 15 to extend the arc.

Now each of these macros calculates the length AB. It's possible to know this length with :

     \tkzCalcLength[cm](A,B)\tkzGetLength{rAB}
\node [above=0.5cm,font=\small] at (g)  { \rAB cm};


-
very good job, but now I'm confused .... if you want have a look here ... tex.stackexchange.com/questions/102232/… – karathan Mar 13 '13 at 7:16

With PSTricks:

## Method 1 (with Curved Abscissa)

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}[showgrid](-2,-2)(2,2)
\pstGeonode[PosAngle={180,45,-90}]
(-1.5,-1){A}
(1.5,1){B}
(1.5,-1){C}
\pstRightAngle[RightAngleSize=0.2,fillstyle=solid,fillcolor=green!50]{A}{C}{B}
\pstArcOAB[arcsepB=-1,arcsepA=-2.5]{A}{B}{B}
\pstCurvAbsNode[CurvAbsNeg=true,PointName=none]{A}{B}{D}{\pstDistVal{.75}}
\psline[linestyle=dashed](A)(D)(D|C)(C)
\psset{shortput=nab,labelsep=-3pt}
\ncline{A}{B}^{$g$}
\ncline{B}{C}^{$a$}
\ncline{C}{A}^{$b$}
\end{pspicture}
\end{document}


## Method 2 (with PostScript Pyth2 operator)

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\psset{saveNodeCoors}
\begin{document}
\begin{pspicture}[showgrid](-2,-2)(2,2)
\pstGeonode[PosAngle={180,45,-90}]
(-1.5,-1){A}
(1.5,1){B}
(1.5,-1){C}
\pstRightAngle[RightAngleSize=0.2,fillstyle=solid,fillcolor=green!50]{A}{C}{B}
\pstVerb{/Dist {N-A.x N-A.y N-B.x N-B.y Pyth2} def}%
\pnode[A](!Dist 21 PtoC){D}\psdots(D)
\psarc[origin={A},arcsep=-1](A){!Dist}{(D)}{(B)}
\psline[linestyle=dashed](A)(D)(D|C)(C)
\psset{shortput=nab,labelsep=-3pt}
\ncline{A}{B}^{$g$}
\ncline{B}{C}^{$a$}
\ncline{C}{A}^{$b$}
\end{pspicture}
\end{document}


## Method 3 (with rotation)

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}[showgrid](-2,-2)(2,2)
\pstGeonode[PosAngle={180,45,-90}]
(-1.5,-1){A}
(1.5,1){B}
(1.5,-1){C}
\pstRightAngle[RightAngleSize=0.2,fillstyle=solid,fillcolor=green!50]{A}{C}{B}
\pstRotation[PointName=none,RotAngle=-12]{A}{B}[D]
\pstArcOAB[arcsep=-1]{A}{D}{B}
\psline[linestyle=dashed](A)(D)(D|C)(C)
\psset{shortput=nab,labelsep=-3pt}
\ncline{A}{B}^{$g$}
\ncline{B}{C}^{$a$}
\ncline{C}{A}^{$b$}
\end{pspicture}
\end{document}


## Method 4 (with intersection)

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}[showgrid](-2,-2)(2,2)
\pstGeonode[PosAngle={180,45,-90}]
(-1.5,-1){A}
(1.5,1){B}
(1.5,-1){C}
\pstRightAngle[RightAngleSize=0.2,fillstyle=solid,fillcolor=green!50]{A}{C}{B}
\pnode[A](1;21){D'}
\pstInterLC{A}{D'}{A}{B}{E}{D}% E is not used but we have to provide a name for it!
\pstArcOAB[arcsep=-1]{A}{D}{B}
\psline[linestyle=dashed](A)(D)(D|C)(C)
\psset{shortput=nab,labelsep=-3pt}
\ncline{A}{B}^{$g$}
\ncline{B}{C}^{$a$}
\ncline{C}{A}^{$b$}
\end{pspicture}
\end{document}


Red alert: For the sake of simplicity, I left the nodes D in the given methods slightly different. It is not difficult to make them identical.

-
setting the origin for \psarc is superfluous; it is already done by setting (A) – Herbert Mar 12 '13 at 15:48
@Herbert: I don't think so. Try it by removing arcsep. – kiss my armpit Mar 12 '13 at 15:50
\psarc[arcsepB=-1](A){!Dist}{(D)}{(B)} – Herbert Mar 12 '13 at 15:54
@Herbert: I have tried \psarc[origin={A}](A){!Dist}{(D)}{(B)} != \psarc(A){!Dist}{(D)}{(B)}. – kiss my armpit Mar 12 '13 at 15:56
Note: I will not append % (to remove a trailing white space) to the end of \pstRightAngle in this answer because it will be fixed internally in the next update of pst-eucl. – kiss my armpit Mar 12 '13 at 19:03