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Tikz: Perpendicular FROM a line

Rotate a node but not its content: the case of the ellipse decorationIntersection of paths with constructed namesHow to define the default vertical distance between nodes?Numerical conditional within tikz keys?Why do I get an extra white page before my TikZ picture?TikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of themHow to draw a square and its diagonals with arrows?Using tikz Calc package to add cordinates

4

0

With tikz-pgf, I can draw a perpendicular from a point to a line. Is there a robust method to draw a perpendicular from a point? For example, in the following code, say I want to raise a perpendicular, from a specific point on line BC outwards.

documentclass[tikz,border=10pt]{standalone}

usetikzlibrary{calc}

begin{document}

begin{tikzpicture}

    coordinate (A) at (0,0);

    coordinate (B) at (2,4);

    coordinate (C) at (8,0);



    draw(A)--(B)--(C)--cycle;

    draw[red] (B) -- ($(A)!(B)!(C)$);



    node[label={below left:$A$}] at (A) {};

    node[label={above:$B$}] at (B) {};

    node[label={below right:$C$}] at (C) {};

end{tikzpicture}

end{document}

tikz-pgf

share|improve this question

edited 1 hour ago

blackened

asked 1 hour ago

blackenedblackened

1,498714

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You changed the position of the starting point in your question : so to draw from a point P on BC you can use (P) -- ++({2/sqrt(13)},{3/sqrt(13)}) (or to for faster compile (P) -- ++(0.5547001962252291,0.8320502943378437)), or you can use as I said calc to draw (P) -- ($(P)!1cm!90:(C)$).
  
  – Kpym
  55 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Kpym I edited it because in the given form the suggested approach was obvious. I am looking for a general solution.
  
  – blackened
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Hmm! I was being a fool, again. Say the point on BC is the midpoint. I can simply define that coordinate as, say, coordinate (P) at ($(B)!0.5!(C)$). Then I can do, draw (P)--($(P)!0.5!90:(C)$).
  
  – blackened
  39 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @blackened or draw[red, thick] (P) -- ($(A)!(P)!(C)$); as you already did for your line from B.
  
  – CarLaTeX
  29 mins ago
  

  
  
  
  

add a comment | 

4

0

With tikz-pgf, I can draw a perpendicular from a point to a line. Is there a robust method to draw a perpendicular from a point? For example, in the following code, say I want to raise a perpendicular, from a specific point on line BC outwards.

documentclass[tikz,border=10pt]{standalone}

usetikzlibrary{calc}

begin{document}

begin{tikzpicture}

    coordinate (A) at (0,0);

    coordinate (B) at (2,4);

    coordinate (C) at (8,0);



    draw(A)--(B)--(C)--cycle;

    draw[red] (B) -- ($(A)!(B)!(C)$);



    node[label={below left:$A$}] at (A) {};

    node[label={above:$B$}] at (B) {};

    node[label={below right:$C$}] at (C) {};

end{tikzpicture}

end{document}

tikz-pgf

share|improve this question

edited 1 hour ago

blackened

asked 1 hour ago

blackenedblackened

1,498714

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You changed the position of the starting point in your question : so to draw from a point P on BC you can use (P) -- ++({2/sqrt(13)},{3/sqrt(13)}) (or to for faster compile (P) -- ++(0.5547001962252291,0.8320502943378437)), or you can use as I said calc to draw (P) -- ($(P)!1cm!90:(C)$).
  
  – Kpym
  55 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Kpym I edited it because in the given form the suggested approach was obvious. I am looking for a general solution.
  
  – blackened
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Hmm! I was being a fool, again. Say the point on BC is the midpoint. I can simply define that coordinate as, say, coordinate (P) at ($(B)!0.5!(C)$). Then I can do, draw (P)--($(P)!0.5!90:(C)$).
  
  – blackened
  39 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @blackened or draw[red, thick] (P) -- ($(A)!(P)!(C)$); as you already did for your line from B.
  
  – CarLaTeX
  29 mins ago
  

  
  
  
  

add a comment | 

With tikz-pgf, I can draw a perpendicular from a point to a line. Is there a robust method to draw a perpendicular from a point? For example, in the following code, say I want to raise a perpendicular, from a specific point on line BC outwards.

documentclass[tikz,border=10pt]{standalone}

usetikzlibrary{calc}

begin{document}

begin{tikzpicture}

    coordinate (A) at (0,0);

    coordinate (B) at (2,4);

    coordinate (C) at (8,0);



    draw(A)--(B)--(C)--cycle;

    draw[red] (B) -- ($(A)!(B)!(C)$);



    node[label={below left:$A$}] at (A) {};

    node[label={above:$B$}] at (B) {};

    node[label={below right:$C$}] at (C) {};

end{tikzpicture}

end{document}

tikz-pgf

share|improve this question

edited 1 hour ago

blackened

asked 1 hour ago

blackenedblackened

1,498714

documentclass[tikz,border=10pt]{standalone}

usetikzlibrary{calc}

begin{document}

begin{tikzpicture}

    coordinate (A) at (0,0);

    coordinate (B) at (2,4);

    coordinate (C) at (8,0);



    draw(A)--(B)--(C)--cycle;

    draw[red] (B) -- ($(A)!(B)!(C)$);



    node[label={below left:$A$}] at (A) {};

    node[label={above:$B$}] at (B) {};

    node[label={below right:$C$}] at (C) {};

end{tikzpicture}

end{document}

share|improve this question

edited 1 hour ago

blackened

asked 1 hour ago

blackenedblackened

1,498714

share|improve this question

edited 1 hour ago

blackened

asked 1 hour ago

blackenedblackened

1,498714

asked 1 hour ago

blackenedblackened

1,498714

asked 1 hour ago

blackenedblackened

1,498714

3 Answers
3

active

oldest

votes

3

if you willing to define specific point on line (B)--(C) with its relative position, than you can write your mwe as following simple solution:

documentclass[tikz,border=10pt]{standalone}



begin{document}

begin{tikzpicture}

    coordinate[label=below left:$A$]   (A) at (0,0);

    coordinate[label=above:$B$]        (B) at (2,4);

    coordinate[label=below right:$C$]  (C) at (8,0);



    draw(A)--(B)-- coordinate[pos=0.3] (aux) % <--- coordinate of the point

                    (C)--cycle;

    draw[red] (aux) -- (aux |- A);

end{tikzpicture}

end{document}

share|improve this answer

answered 39 mins ago

ZarkoZarko

125k867164

add a comment | 

2

Sorry, not tikz. I understand @hpekris's idea.

documentclass[pstricks,border=10pt]{standalone}

usepackage{pst-eucl}

begin{document}

foreach i in {.3,.5,.7}{

begin{pspicture}[PointSymbol=none,linejoin=1](0,-1)(8,4)

pnodes(0,0){A}(2,4){B}(8,0){C}(4,0){I}

psline(A)(B)(C)(A)

pstHomO[HomCoef=i,PosAngle=75]{B}{C}[M]

pstProjection[PosAngle=-90]{A}{C}{B}[H]

pstProjection[PosAngle=-90]{A}{C}{M}[M']

pcline(M)(M')

pcline(B)(H)

end{pspicture}}

end{document}

share|improve this answer

edited 11 secs ago

answered 39 mins ago

chishimutojichishimutoji

7441320

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Good, please could you decrease the velocity of the animation :-)?
  
  – Sebastiano
  23 mins ago
  

  
  
  
  

add a comment | 

0

Answer to the first version of the question

With intersections you can give a name to your triangle path, then build a "fake path" from (4,0) to a point after the intersection with the triangle (you don't need to calculate it exactly, see name path=riga and draw it to understand what I'm saying) and name it, then you can draw a line between the intersections with your triangle and your "fake path".

Answer to the second version of the question

If you want to start from a point on BC, you can use coordinate (P) at ($(B)!.5!(C)$); with any value you want instead of .5 and do like you did for the line from B (see the thick red line in my MWE).

documentclass[tikz,border=10pt]{standalone}

usetikzlibrary{calc, intersections}

begin{document}

begin{tikzpicture}

    coordinate[label={below left:$A$}] (A) at (0,0);

    coordinate[label={above:$B$}] (B) at (2,4);

    coordinate[label={below right:$C$}] (C) at (8,0);



    draw[name path=trian](A)  --(B)--(C)--cycle;

    draw[red] (B) -- ($(A)!(B)!(C)$);

    path [name path=riga] (4,0) -- ++(0,3);

    path [name intersections={of=trian and riga}];

    draw (intersection-1) -- (intersection-2);

    coordinate (P) at ($(B)!.5!(C)$);

    draw[red, thick] (P) -- ($(A)!(P)!(C)$);

end{tikzpicture}

end{document}

share|improve this answer

edited 6 mins ago

answered 1 hour ago

CarLaTeXCarLaTeX

31.8k551133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I understood the question as OP want a line starting from som point on BC and going perpendicular to BC outward some distance. (ok now I see that the question has been edited).
  
  – hpekristiansen
  55 mins ago
  
  
  

  
  
  
  

add a comment | 

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3

if you willing to define specific point on line (B)--(C) with its relative position, than you can write your mwe as following simple solution:

documentclass[tikz,border=10pt]{standalone}



begin{document}

begin{tikzpicture}

    coordinate[label=below left:$A$]   (A) at (0,0);

    coordinate[label=above:$B$]        (B) at (2,4);

    coordinate[label=below right:$C$]  (C) at (8,0);



    draw(A)--(B)-- coordinate[pos=0.3] (aux) % <--- coordinate of the point

                    (C)--cycle;

    draw[red] (aux) -- (aux |- A);

end{tikzpicture}

end{document}

share|improve this answer

answered 39 mins ago

ZarkoZarko

125k867164

add a comment | 

3

if you willing to define specific point on line (B)--(C) with its relative position, than you can write your mwe as following simple solution:

documentclass[tikz,border=10pt]{standalone}



begin{document}

begin{tikzpicture}

    coordinate[label=below left:$A$]   (A) at (0,0);

    coordinate[label=above:$B$]        (B) at (2,4);

    coordinate[label=below right:$C$]  (C) at (8,0);



    draw(A)--(B)-- coordinate[pos=0.3] (aux) % <--- coordinate of the point

                    (C)--cycle;

    draw[red] (aux) -- (aux |- A);

end{tikzpicture}

end{document}

share|improve this answer

answered 39 mins ago

ZarkoZarko

125k867164

add a comment | 

if you willing to define specific point on line (B)--(C) with its relative position, than you can write your mwe as following simple solution:

documentclass[tikz,border=10pt]{standalone}



begin{document}

begin{tikzpicture}

    coordinate[label=below left:$A$]   (A) at (0,0);

    coordinate[label=above:$B$]        (B) at (2,4);

    coordinate[label=below right:$C$]  (C) at (8,0);



    draw(A)--(B)-- coordinate[pos=0.3] (aux) % <--- coordinate of the point

                    (C)--cycle;

    draw[red] (aux) -- (aux |- A);

end{tikzpicture}

end{document}

share|improve this answer

answered 39 mins ago

ZarkoZarko

125k867164

documentclass[tikz,border=10pt]{standalone}



begin{document}

begin{tikzpicture}

    coordinate[label=below left:$A$]   (A) at (0,0);

    coordinate[label=above:$B$]        (B) at (2,4);

    coordinate[label=below right:$C$]  (C) at (8,0);



    draw(A)--(B)-- coordinate[pos=0.3] (aux) % <--- coordinate of the point

                    (C)--cycle;

    draw[red] (aux) -- (aux |- A);

end{tikzpicture}

end{document}

share|improve this answer

answered 39 mins ago

ZarkoZarko

125k867164

answered 39 mins ago

ZarkoZarko

125k867164

answered 39 mins ago

ZarkoZarko

125k867164

2

Sorry, not tikz. I understand @hpekris's idea.

documentclass[pstricks,border=10pt]{standalone}

usepackage{pst-eucl}

begin{document}

foreach i in {.3,.5,.7}{

begin{pspicture}[PointSymbol=none,linejoin=1](0,-1)(8,4)

pnodes(0,0){A}(2,4){B}(8,0){C}(4,0){I}

psline(A)(B)(C)(A)

pstHomO[HomCoef=i,PosAngle=75]{B}{C}[M]

pstProjection[PosAngle=-90]{A}{C}{B}[H]

pstProjection[PosAngle=-90]{A}{C}{M}[M']

pcline(M)(M')

pcline(B)(H)

end{pspicture}}

end{document}

share|improve this answer

edited 11 secs ago

answered 39 mins ago

chishimutojichishimutoji

7441320

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Good, please could you decrease the velocity of the animation :-)?
  
  – Sebastiano
  23 mins ago
  

  
  
  
  

add a comment | 

2

Sorry, not tikz. I understand @hpekris's idea.

documentclass[pstricks,border=10pt]{standalone}

usepackage{pst-eucl}

begin{document}

foreach i in {.3,.5,.7}{

begin{pspicture}[PointSymbol=none,linejoin=1](0,-1)(8,4)

pnodes(0,0){A}(2,4){B}(8,0){C}(4,0){I}

psline(A)(B)(C)(A)

pstHomO[HomCoef=i,PosAngle=75]{B}{C}[M]

pstProjection[PosAngle=-90]{A}{C}{B}[H]

pstProjection[PosAngle=-90]{A}{C}{M}[M']

pcline(M)(M')

pcline(B)(H)

end{pspicture}}

end{document}

share|improve this answer

edited 11 secs ago

answered 39 mins ago

chishimutojichishimutoji

7441320

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Good, please could you decrease the velocity of the animation :-)?
  
  – Sebastiano
  23 mins ago
  

  
  
  
  

add a comment | 

Sorry, not tikz. I understand @hpekris's idea.

documentclass[pstricks,border=10pt]{standalone}

usepackage{pst-eucl}

begin{document}

foreach i in {.3,.5,.7}{

begin{pspicture}[PointSymbol=none,linejoin=1](0,-1)(8,4)

pnodes(0,0){A}(2,4){B}(8,0){C}(4,0){I}

psline(A)(B)(C)(A)

pstHomO[HomCoef=i,PosAngle=75]{B}{C}[M]

pstProjection[PosAngle=-90]{A}{C}{B}[H]

pstProjection[PosAngle=-90]{A}{C}{M}[M']

pcline(M)(M')

pcline(B)(H)

end{pspicture}}

end{document}

share|improve this answer

edited 11 secs ago

answered 39 mins ago

chishimutojichishimutoji

7441320

documentclass[pstricks,border=10pt]{standalone}

usepackage{pst-eucl}

begin{document}

foreach i in {.3,.5,.7}{

begin{pspicture}[PointSymbol=none,linejoin=1](0,-1)(8,4)

pnodes(0,0){A}(2,4){B}(8,0){C}(4,0){I}

psline(A)(B)(C)(A)

pstHomO[HomCoef=i,PosAngle=75]{B}{C}[M]

pstProjection[PosAngle=-90]{A}{C}{B}[H]

pstProjection[PosAngle=-90]{A}{C}{M}[M']

pcline(M)(M')

pcline(B)(H)

end{pspicture}}

end{document}

share|improve this answer

edited 11 secs ago

answered 39 mins ago

chishimutojichishimutoji

7441320

answered 39 mins ago

chishimutojichishimutoji

7441320

answered 39 mins ago

chishimutojichishimutoji

7441320

0

Answer to the first version of the question

With intersections you can give a name to your triangle path, then build a "fake path" from (4,0) to a point after the intersection with the triangle (you don't need to calculate it exactly, see name path=riga and draw it to understand what I'm saying) and name it, then you can draw a line between the intersections with your triangle and your "fake path".

Answer to the second version of the question

If you want to start from a point on BC, you can use coordinate (P) at ($(B)!.5!(C)$); with any value you want instead of .5 and do like you did for the line from B (see the thick red line in my MWE).

documentclass[tikz,border=10pt]{standalone}

usetikzlibrary{calc, intersections}

begin{document}

begin{tikzpicture}

    coordinate[label={below left:$A$}] (A) at (0,0);

    coordinate[label={above:$B$}] (B) at (2,4);

    coordinate[label={below right:$C$}] (C) at (8,0);



    draw[name path=trian](A)  --(B)--(C)--cycle;

    draw[red] (B) -- ($(A)!(B)!(C)$);

    path [name path=riga] (4,0) -- ++(0,3);

    path [name intersections={of=trian and riga}];

    draw (intersection-1) -- (intersection-2);

    coordinate (P) at ($(B)!.5!(C)$);

    draw[red, thick] (P) -- ($(A)!(P)!(C)$);

end{tikzpicture}

end{document}

share|improve this answer

edited 6 mins ago

answered 1 hour ago

CarLaTeXCarLaTeX

31.8k551133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I understood the question as OP want a line starting from som point on BC and going perpendicular to BC outward some distance. (ok now I see that the question has been edited).
  
  – hpekristiansen
  55 mins ago
  
  
  

  
  
  
  

add a comment | 

0

Answer to the first version of the question

With intersections you can give a name to your triangle path, then build a "fake path" from (4,0) to a point after the intersection with the triangle (you don't need to calculate it exactly, see name path=riga and draw it to understand what I'm saying) and name it, then you can draw a line between the intersections with your triangle and your "fake path".

Answer to the second version of the question

If you want to start from a point on BC, you can use coordinate (P) at ($(B)!.5!(C)$); with any value you want instead of .5 and do like you did for the line from B (see the thick red line in my MWE).

documentclass[tikz,border=10pt]{standalone}

usetikzlibrary{calc, intersections}

begin{document}

begin{tikzpicture}

    coordinate[label={below left:$A$}] (A) at (0,0);

    coordinate[label={above:$B$}] (B) at (2,4);

    coordinate[label={below right:$C$}] (C) at (8,0);



    draw[name path=trian](A)  --(B)--(C)--cycle;

    draw[red] (B) -- ($(A)!(B)!(C)$);

    path [name path=riga] (4,0) -- ++(0,3);

    path [name intersections={of=trian and riga}];

    draw (intersection-1) -- (intersection-2);

    coordinate (P) at ($(B)!.5!(C)$);

    draw[red, thick] (P) -- ($(A)!(P)!(C)$);

end{tikzpicture}

end{document}

share|improve this answer

edited 6 mins ago

answered 1 hour ago

CarLaTeXCarLaTeX

31.8k551133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I understood the question as OP want a line starting from som point on BC and going perpendicular to BC outward some distance. (ok now I see that the question has been edited).
  
  – hpekristiansen
  55 mins ago
  
  
  

  
  
  
  

add a comment | 

Answer to the first version of the question

With intersections you can give a name to your triangle path, then build a "fake path" from (4,0) to a point after the intersection with the triangle (you don't need to calculate it exactly, see name path=riga and draw it to understand what I'm saying) and name it, then you can draw a line between the intersections with your triangle and your "fake path".

Answer to the second version of the question

If you want to start from a point on BC, you can use coordinate (P) at ($(B)!.5!(C)$); with any value you want instead of .5 and do like you did for the line from B (see the thick red line in my MWE).

documentclass[tikz,border=10pt]{standalone}

usetikzlibrary{calc, intersections}

begin{document}

begin{tikzpicture}

    coordinate[label={below left:$A$}] (A) at (0,0);

    coordinate[label={above:$B$}] (B) at (2,4);

    coordinate[label={below right:$C$}] (C) at (8,0);



    draw[name path=trian](A)  --(B)--(C)--cycle;

    draw[red] (B) -- ($(A)!(B)!(C)$);

    path [name path=riga] (4,0) -- ++(0,3);

    path [name intersections={of=trian and riga}];

    draw (intersection-1) -- (intersection-2);

    coordinate (P) at ($(B)!.5!(C)$);

    draw[red, thick] (P) -- ($(A)!(P)!(C)$);

end{tikzpicture}

end{document}

share|improve this answer

edited 6 mins ago

answered 1 hour ago

CarLaTeXCarLaTeX

31.8k551133

documentclass[tikz,border=10pt]{standalone}

usetikzlibrary{calc, intersections}

begin{document}

begin{tikzpicture}

    coordinate[label={below left:$A$}] (A) at (0,0);

    coordinate[label={above:$B$}] (B) at (2,4);

    coordinate[label={below right:$C$}] (C) at (8,0);



    draw[name path=trian](A)  --(B)--(C)--cycle;

    draw[red] (B) -- ($(A)!(B)!(C)$);

    path [name path=riga] (4,0) -- ++(0,3);

    path [name intersections={of=trian and riga}];

    draw (intersection-1) -- (intersection-2);

    coordinate (P) at ($(B)!.5!(C)$);

    draw[red, thick] (P) -- ($(A)!(P)!(C)$);

end{tikzpicture}

end{document}

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