Jtdylktuy

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}

I just wrote such a style in this answer. I slightly changed the syntax, so you need to say

draw[blue,vert={of {(B)--(C)} at (3,0)}];

to draw a vertical line at (3,0) that goes all the way until it hits BC. And I added vert outwards which is just a wrapper of the distance modifiers (see section 13.5.4 The Syntax of Distance Modifiers of the pgfmanual), and can be used as

    draw[blue,vert outwards={from {($(B)!0.3!(C)$)} by 3cm on line to {(C)}}];

MWE

documentclass[tikz,border=3.14mm]{standalone}

usetikzlibrary{calc}

begin{document}

begin{tikzpicture}[vert/.style args={of #1 at #2}{insert path={%

#2 -- (intersection cs:first

  line={#1}, second line={#2--($#2+(0,10)$)}) }},

vert outwards/.style args={from #1 by #2 on line to #3}{insert path={

#1 -- ($#1!#2!90:#3$)

}}]

    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) {};

    draw[blue,vert={of {(B)--(C)} at (3,0)}];

    draw[blue,vert outwards={from {($(B)!0.3!(C)$)} by 3cm on line to {(C)}}];

end{tikzpicture}

end{document}

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}

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}

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}

Another PSTricks solution just for comparison purposes.

I provide some possible tricks but you can remove the parts that you don't want.

documentclass[pstricks,border=12pt]{standalone}

usepackage{pst-eucl}

begin{document}

foreach i in {1,2,3}{%

begin{pspicture}(8,5)

pstTriangle(1,1){A}(7,1){B}(3,4){C}

psline(C)(C|A)

pnode([nodesep=i]{B}C){P}

psline(P)(P|A)

pnode([nodesep=i,offset=i]{B}C){Q}

psline[linecolor=red](P)(Q)

end{pspicture}}

end{document}