2

I have a diagram of a line intersecting triangle ABC on two of its sides. Four points on this line are labeled. I want the labels for these points to be typeset 0.15cm above this line. The green line is 0.15cm above the line that intersects the triangle. The labels should be typeset on the green line. P is typeset too low, Q is typeset too high ... and F_1 and F_2 are typeset just right!

\documentclass{amsart}

\usepackage{mathtools}

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


\begin{document}

\begin{tikzpicture}

%triangle{ABC} is drawn. P is a point on AB, Q is a point on AC, and R is the intersection of the line
%through B and C and the line through P and Q. Line PR is drawn. \triangle{BPQ} and \triangle{CPQ} are
%drawn, and the regions bound by them are shaded.
\path (1,3.5) coordinate (A) (-3,0) coordinate (B) (2,0) coordinate (C);
%
\draw let \p1=($(A)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(A)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1+\n2+180)+180}, inner sep=0, font=\footnotesize] at ($(A) + ({0.5*(\n1+\n2+180)}:0.15)$){\textit{A}};
\path node[anchor=north, inner sep=0, font=\footnotesize] at ($(B) + (0,-0.15)$){\textit{B}};
\path node[anchor=north, inner sep=0, font=\footnotesize] at ($(C) + (0,-0.15)$){\textit{C}};
%
\coordinate (P) at ($(A)!{1/3}!(B)$);
\coordinate (Q) at ($(A)!{3/5}!(C)$);
%
\path[name path=a_path_to_locate_R_on_BC] (C) -- ($(C)!-3.5cm!(B)$);
\path[name path=another_path_to_locate_R_on_BC] (Q) -- ($(Q)!-3.5cm!(P)$);
\coordinate[name intersections={of=a_path_to_locate_R_on_BC and another_path_to_locate_R_on_BC, by=R}];
%
%The left arrowhead of the line through P, Q, and R is as high as A and as far right as the right arrowhead
%of the line through B, C, and R.
\path[name path=a_path_to_locate_left_arrowhead] (A) -- ($(A) +(-3.75,0)$);
\path[name path=another_path_to_locate_left_arrowhead] (P) -- ($(P)!-2.7cm!(Q)$);
\coordinate[name intersections={of=a_path_to_locate_left_arrowhead and another_path_to_locate_left_arrowhead, by=left_arrowhead}];
\path[name path=a_path_to_locate_right_arrowhead] ($(R) +(1.5,0)$) -- ($(R) +(1.5,-0.75)$);
\draw[green,  name path=another_path_to_locate_right_arrowhead] (Q) -- ($(R)!-1.7cm!(Q)$);
\coordinate[name intersections={of=a_path_to_locate_right_arrowhead and another_path_to_locate_right_arrowhead, by=right_arrowhead}];
\draw[latex-latex] ($(B)!-1.5cm!(R)$) -- ($(R)!-1.5cm!(B)$);
\draw[latex-latex] (left_arrowhead) -- (right_arrowhead);
%
\draw[green,  name path=a_path_for_the_label_for_P] ($(A)!0.15cm!-90:(B)$) -- ($(B)!0.15cm!90:(A)$);
\draw[green,  name path=another_path_for_the_label_for_P] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$);
\coordinate[name intersections={of=a_path_for_the_label_for_P and another_path_for_the_label_for_P, by=label_for_P}];
\path let \p1=($(A)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(P)-(R)$), \n2={atan(\y2/\x2)} in node[draw=green,  anchor={0.5*(\n1+(\n2+180))-180}, inner sep=0, font=\footnotesize] at (label_for_P){\textit{P}};
%
\draw[green,  name path=a_path_for_the_label_for_Q] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
\draw[green,  name path=another_path_for_the_label_for_Q] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$);
\coordinate[name intersections={of=a_path_for_the_label_for_Q and another_path_for_the_label_for_Q, by=label_for_Q}];
\path let \p1=($(A)-(C)$), \n1={atan(\y1/\x1)}, \p2=($(P)-(R)$), \n2={atan(\y2/\x2)} in node[draw=green,  anchor={0.5*((\n1+180)+\n2)-180}, inner sep=0, font=\footnotesize] at (label_for_Q){\textit{Q}};
%
\path[name path=a_path_for_the_label_for_R] ($(B) +(0,-0.15)$) -- ($(R) +(0,-0.15)$);
\path[name path=another_path_for_the_label_for_R] let \p1=($(P)-(R)$), \n1={atan(\y1/\x1)} in (R) -- ($(R) +({0.5*(\n1-180)}:0.25)$);
\coordinate[name intersections={of=a_path_for_the_label_for_R and another_path_for_the_label_for_R, by=label_for_R}];
\path let \p1=($(P)-(R)$), \n1={atan(\y1/\x1)} in node[anchor={0.5*(\n1-180)+180}, inner sep=0, font=\footnotesize] at (label_for_R){\textit{R}};
%
%
%The regions bound by \triangle{BPQ} and \triangle{CPQ} are shaded.
\path[name path=a_path_to_delineate_shading] (B) -- (Q);
\path[name path=another_path_to_delineate_shading] (C) -- (P);
\coordinate[name intersections={of=a_path_to_delineate_shading and another_path_to_delineate_shading, by=intersection-1}];
\draw[fill=gray!25] (B) -- (intersection-1) --  (P) -- cycle;
\draw[fill=gray!25] (C) -- (intersection-1) --  (Q) -- cycle;
\draw[fill=gray!75] (P) -- (intersection-1) --  (Q) -- cycle;
\draw[dashed] (B) -- (Q);
\draw[dashed] (C) -- (P);
\draw (A) -- (B) -- (C) -- cycle;


%The foot of the altitude of \triangle{BPQ} from B is located. It is labeled F_2.
\coordinate (F_2) at ($(P)!(B)!(Q)$);
\path let \p1=($(P)-(R)$), \n1={atan(\y1/\x1)} in node[draw=green,  anchor={\n1-90}, inner sep=0, font=\footnotesize] at ($(F_2) +({\n1+90}:0.15)$){$F_{2}$};
\draw[dashed] (F_2) -- (B);

%A right-angle mark is drawn at the foot of the altitude of \triangle{BPQ} from B.
\coordinate (U) at ($(F_2)!3mm!-45:(Q)$);
\draw ($(P)!(U)!(Q)$) -- (U) -- ($(B)!(U)!(F_2)$);


%The foot of the altitude of \triangle{CPQ} from C is located. It is labeled F_3.
\coordinate (F_3) at ($(P)!(C)!(Q)$);
\path let \p1=($(P)-(R)$), \n1={atan(\y1/\x1)} in node[draw=green,  anchor={\n1-90}, inner sep=0, font=\footnotesize] at ($(F_3) +({\n1+90}:0.15)$){$F_{3}$};
\draw[dashed] (F_3) -- (C);

%A right-angle mark is drawn at the foot of the altitude of \triangle{CPQ} from C.
\coordinate (V) at ($(F_3)!3mm!-45:(R)$);
\draw ($(P)!(V)!(R)$) -- (V) -- ($(C)!(V)!(F_3)$);

\end{tikzpicture}

\end{document}
  • 1
    What does "above this line" mean? Which point of the node should be 0.15cm far (perpendicular?) from the green line? You draw F_3 and F_2 using this distance but P and Q not, so it's normal that they don't respect it. Did you try drawing P and Q like F_2 and F_3? – Ignasi Apr 16 '18 at 17:58
  • I have edited the post. – Adelyn Apr 16 '18 at 18:19
  • I suppose that P, F_1, and F_2 should have an anchor of south west to be typeset on the green line. – Adelyn Apr 16 '18 at 18:20
  • What anchor should Q have so that the letter - not the box containing the letter - is on the green line? – Adelyn Apr 16 '18 at 18:21
  • @Adelyn To whom are you making all these comments? You have a very delicate way of computing the positions of these labels. It might be easier if you explain why you have gone that way (rather than, say, just adding labels to the respective coordinates). – user121799 Apr 16 '18 at 18:30
2

I think that the difference comes from the subscripts. So I added "fake subscripts" to P and Q. Is this closer to what you want?

\documentclass{amsart}

\usepackage{mathtools}

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


\begin{document}

\begin{tikzpicture}

%triangle{ABC} is drawn. P is a point on AB, Q is a point on AC, and R is the intersection of the line
%through B and C and the line through P and Q. Line PR is drawn. \triangle{BPQ} and \triangle{CPQ} are
%drawn, and the regions bound by them are shaded.
\path (1,3.5) coordinate (A) (-3,0) coordinate (B) (2,0) coordinate (C);
%
\draw let \p1=($(A)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(A)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1+\n2+180)+180}, inner sep=0, font=\footnotesize] at ($(A) + ({0.5*(\n1+\n2+180)}:0.15)$){\textit{A}};
\path node[anchor=north, inner sep=0, font=\footnotesize] at ($(B) + (0,-0.15)$){\textit{B}};
\path node[anchor=north, inner sep=0, font=\footnotesize] at ($(C) + (0,-0.15)$){\textit{C}};
%
\coordinate (P) at ($(A)!{1/3}!(B)$);
\coordinate (Q) at ($(A)!{3/5}!(C)$);
%
\path[name path=a_path_to_locate_R_on_BC] (C) -- ($(C)!-3.5cm!(B)$);
\path[name path=another_path_to_locate_R_on_BC] (Q) -- ($(Q)!-3.5cm!(P)$);
\coordinate[name intersections={of=a_path_to_locate_R_on_BC and another_path_to_locate_R_on_BC, by=R}];
%
%The left arrowhead of the line through P, Q, and R is as high as A and as far right as the right arrowhead
%of the line through B, C, and R.
\path[name path=a_path_to_locate_left_arrowhead] (A) -- ($(A) +(-3.75,0)$);
\path[name path=another_path_to_locate_left_arrowhead] (P) -- ($(P)!-2.7cm!(Q)$);
\coordinate[name intersections={of=a_path_to_locate_left_arrowhead and another_path_to_locate_left_arrowhead, by=left_arrowhead}];
\path[name path=a_path_to_locate_right_arrowhead] ($(R) +(1.5,0)$) -- ($(R) +(1.5,-0.75)$);
\draw[green,  name path=another_path_to_locate_right_arrowhead] (Q) -- ($(R)!-1.7cm!(Q)$);
\coordinate[name intersections={of=a_path_to_locate_right_arrowhead and another_path_to_locate_right_arrowhead, by=right_arrowhead}];
\draw[latex-latex] ($(B)!-1.5cm!(R)$) -- ($(R)!-1.5cm!(B)$);
\draw[latex-latex] (left_arrowhead) -- (right_arrowhead);
%a_path_for_the_label_for_P
\draw[green,  name path=a_path_for_the_label_for_P] ($(A)!0.15cm!-90:(B)$) -- ($(B)!0.15cm!90:(A)$);
\draw[green,  name path=another_path_for_the_label_for_P] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$);
\coordinate[name intersections={of=a_path_for_the_label_for_P and another_path_for_the_label_for_P, by=label_for_P}];
%\path let \p1=($(A)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(P)-(R)$), \n2={atan(\y2/\x2)} in node[draw=green,  anchor={0.5*(\n1+(\n2+180))-180}, inner sep=0, font=\footnotesize] at (label_for_P){\textit{P}};
%a_path_for_the_label_for_Q
\draw[green,  name path=a_path_for_the_label_for_Q] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
\draw[green,  name path=another_path_for_the_label_for_Q] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$);
\coordinate[name intersections={of=a_path_for_the_label_for_Q and another_path_for_the_label_for_Q, by=label_for_Q}];
\node[draw=green,font=\footnotesize,anchor=south,outer sep=0pt,inner sep=0pt] at
(label_for_Q) {$Q_{\vphantom{1}}$};
%
\coordinate[name intersections={of=a_path_for_the_label_for_P and another_path_for_the_label_for_P, by=label_for_Q}];
\node[draw=green,font=\footnotesize,anchor=south,outer sep=0pt,inner sep=0pt] at (label_for_P) {$P_{\vphantom{1}}$};
%
%\path let \p1=($(A)-(C)$), \n1={atan(\y1/\x1)}, \p2=($(P)-(R)$), \n2={atan(\y2/\x2)} in node[draw=green,  anchor={0.5*((\n1+180)+\n2)-180}, inner sep=0, font=\footnotesize] at (label_for_Q){\textit{Q}};
%
\path[name path=a_path_for_the_label_for_R] ($(B) +(0,-0.15)$) -- ($(R) +(0,-0.15)$);
\path[name path=another_path_for_the_label_for_R] let \p1=($(P)-(R)$), \n1={atan(\y1/\x1)} in (R) -- ($(R) +({0.5*(\n1-180)}:0.25)$);
\coordinate[name intersections={of=a_path_for_the_label_for_R and another_path_for_the_label_for_R, by=label_for_R}];
%\path let \p1=($(P)-(R)$), \n1={atan(\y1/\x1)} in node[anchor={0.5*(\n1-180)+180}, inner sep=0, font=\footnotesize] at (label_for_R){\textit{R}};
%
%
%The regions bound by \triangle{BPQ} and \triangle{CPQ} are shaded.
\path[name path=a_path_to_delineate_shading] (B) -- (Q);
\path[name path=another_path_to_delineate_shading] (C) -- (P);
\coordinate[name intersections={of=a_path_to_delineate_shading and another_path_to_delineate_shading, by=intersection-1}];
\draw[fill=gray!25] (B) -- (intersection-1) --  (P) -- cycle;
\draw[fill=gray!25] (C) -- (intersection-1) --  (Q) -- cycle;
\draw[fill=gray!75] (P) -- (intersection-1) --  (Q) -- cycle;
\draw[dashed] (B) -- (Q);
\draw[dashed] (C) -- (P);
\draw (A) -- (B) -- (C) -- cycle;


%The foot of the altitude of \triangle{BPQ} from B is located. It is labeled F_2.
\coordinate (F_2) at ($(P)!(B)!(Q)$);
\path let \p1=($(P)-(R)$), \n1={atan(\y1/\x1)} in node[draw=green,  anchor={\n1-90}, inner sep=0, font=\footnotesize] at ($(F_2) +({\n1+90}:0.15)$){$F_{2}$};
\draw[dashed] (F_2) -- (B);

%A right-angle mark is drawn at the foot of the altitude of \triangle{BPQ} from B.
\coordinate (U) at ($(F_2)!3mm!-45:(Q)$);
\draw ($(P)!(U)!(Q)$) -- (U) -- ($(B)!(U)!(F_2)$);


%The foot of the altitude of \triangle{CPQ} from C is located. It is labeled F_3.
\coordinate (F_3) at ($(P)!(C)!(Q)$);
\path let \p1=($(P)-(R)$), \n1={atan(\y1/\x1)} in node[draw=green,  anchor={\n1-90}, inner sep=0, font=\footnotesize] at ($(F_3) +({\n1+90}:0.15)$){$F_{3}$};
\draw[dashed] (F_3) -- (C);

%A right-angle mark is drawn at the foot of the altitude of \triangle{CPQ} from C.
\coordinate (V) at ($(F_3)!3mm!-45:(R)$);
\draw ($(P)!(V)!(R)$) -- (V) -- ($(C)!(V)!(F_3)$);

\end{tikzpicture}

\end{document}

enter image description here

  • I am particular with labels. The P that you have typeset is too close to both the triangle and the line, and the Q is much too close to the triangle. – Adelyn Apr 16 '18 at 19:27
  • 1
    Would \makebox[0pt][l]{1} instruct TikZ to not make room for them? – Adelyn Apr 16 '18 at 19:54
  • 1
    @Adelyn Yes, I used \vphantom since it does only make vertical room. (Note also that \textit{Q} and $Q$ are not the same, and here many people are very peculiar ;-) – user121799 Apr 16 '18 at 19:57
  • 1
    @Adelyn I agree with Mico. I also said that it's not the same. – user121799 Apr 16 '18 at 20:35
  • 2
    @Adelyn Well, you could do \phantom{O}\makebox[0pt][r]{Q}. Another possibility is to set the anchors of all the nodes to north and increase the distance to the lines. As long as you do not have superscripts, this might give you what you want. – user121799 Apr 17 '18 at 22:23
2

Here I have an alternative using the tkz-euclide package that has optimized commands, the adjustment can be done using the \tkzLabelAngle command, which positions the label in the bisector line at a distance determined by the pos modifier.

In transparent red the default positions are shown. To position F2 and R auxiliary points Z and X are used.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% By J. Leon V.  coded based on the BSD, MIT, Beerware licences.
\documentclass[border=2mm]{standalone}
\usepackage{xcolor}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}
    \begin{tikzpicture}

    % Set limits.
        \tkzInit[xmax=7,xmin=-5,ymax=5, ymin=-2]
        \tkzGrid[sub,color=blue!10!,subxstep=.5,subystep=.5]
        \tkzClip
    %Calculate points.
        \tkzDefPoint(1,3.5){A} 
        \tkzDefPoint(-3,0){B} 
        \tkzDefPoint(2,0){C}
    %Auxiliar points
        \tkzDefPoint(-3,4){Y}
        \tkzDefPoint(6,0){W1}
        \tkzDefPoint(6,1){W2}
    %Calculated points
        \tkzInCenter(A,B,C) \tkzGetPoint{G}
        \tkzDefBarycentricPoint(A=2,B=1) \tkzGetPoint{P}
        \tkzDefBarycentricPoint(A=2,C=3) \tkzGetPoint{Q}
        \tkzInterLL(B,C)(P,Q) \tkzGetPoint{R}
        \tkzDefPointBy[projection=onto Q--R](C) \tkzGetPoint{F}
        \tkzDefPointBy[projection=onto Q--R](B) \tkzGetPoint{H}
        % Auxiliary Points
            \tkzInterLL(B,Y)(P,Q) \tkzGetPoint{Z}
            \tkzInterLL(W1,W2)(P,Q) \tkzGetPoint{X}

        % For dummy lines
            \coordinate (a) at ($ (A)!-.15cm!90:(B) $); % a is .15cm separated orthogonal from midpoint line A-B
            \tkzDefLine[parallel=through a](A,B) \tkzGetPoint{b}
            \coordinate (c) at ($ (H)!.15cm!90:(F) $); % c is .15cm separated orthogonal from midpoint line H-F
            \tkzDefLine[parallel=through c](H,F) \tkzGetPoint{d}
            \coordinate (e) at ($ (A)!.15cm!90:(C) $); % e is .15cm separated orthogonal from midpoint line A-C
            \tkzDefLine[parallel=through e](A,C) \tkzGetPoint{f}
    % Traces
        \tkzFillPolygon[color=yellow!30](B,P,Q)
        \tkzFillPolygon[color=blue!30,opacity=.5](C,P,Q)
        \tkzDrawPolygon[color=black](A,B,C)
        \tkzDrawSegments[style=dashed](C,F B,H)
        \tkzDrawSegments(B,Q P,C)
        \tkzDrawLine[color=green](a,b)
        \tkzDrawLine[color=green](c,d)
        \tkzDrawLine[color=green](e,f)
        \tkzDrawPoint(G)
        \tkzDrawPoints(P,Q,R)
        {%instruction only afects commands inside {}
        \tikzset{line style/.append style={<->},>=latex} 
        \tkzDrawLine[add=1.5cm and 1.65cm](H,R)
        \tkzDrawLine[add=1cm and 1.5cm](B,R)
        }
        \tkzMarkRightAngle[fill=black](C,F,R)
        \tkzMarkRightAngle[fill=black](B,H,R)

    %create labels
        \tkzLabelPoints[color=red,opacity=.3](A,B,C,P,Q,R,F,H) %reference points.
        \tkzLabelPoints[below](B,C)
    % Use label angle method. you can control the distance in a reference line angle bisector of Incenter.
        \tkzLabelAngle[pos = -.3](B,A,C){$A$}
        \tkzLabelAngle[pos = .3](H,P,A){$P$}
        \tkzLabelAngle[pos = .3](A,Q,F){$Q$}
        \tkzLabelAngle[pos = .3](R,F,Q){$F_3$}
        \tkzLabelAngle[pos = .3](P,H,Z){$F_2$}
        \tkzLabelAngle[pos = -.3](C,R,X){$R$}
        \node[left] at (G) {$I$};
    \end{tikzpicture}
\end{document}

And get the following result:

enter image description here

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.