mjhjmtu

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
usetikzlibrarycalc
begindocument
begintikzpicture
 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) ;
endtikzpicture
enddocument

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
usetikzlibrarycalc
begindocument
begintikzpicture[vert/.style args=of #1 at #2insert path=%
#2 -- (intersection cs:first
 line=#1, second line=#2--($#2+(0,10)$)) ,
vert outwards/.style args=from #1 by #2 on line to #3insert 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)];
endtikzpicture
enddocument

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

begindocument
begintikzpicture
 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);
endtikzpicture
enddocument

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
usetikzlibrarycalc, intersections
begindocument
begintikzpicture
 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) -- ($(A)!(P)!(C)$);
endtikzpicture
enddocument

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

documentclass[pstricks,border=10pt]standalone
usepackagepst-eucl
begindocument
foreach i in .3,.5,.7
beginpspicture[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]BC[M]
pstProjection[PosAngle=-90]ACB[H]
pstProjection[PosAngle=-90]ACM[M']
pcline(M)(M')
pcline(B)(H)
endpspicture
enddocument

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
usepackagepst-eucl
begindocument
foreach i in 1,2,3%
beginpspicture(8,5)
pstTriangle(1,1)A(7,1)B(3,4)C
psline(C)(C
enddocument