Xjyuk

I'm trying to draw the following in TikZ:

Such that a=1/2, b=1/4 and c=1/4.These lines must be at right angles from the triangle sides.

Finally, the triangle has a height of 1.

Here's my MWE:

documentclass[tikz]standalone

usepackagetikz

begindocument
begintikzpicture

coordinate (A) at (0,0) ;
coordinate (B) at (sqrt(4/3, 0) ;
coordinate (C) at ((sqrt(4/3)/2,1) ;

node at (A) [below left] 1;

node at (B) [below right]2;

node at (C) [above]3;

filldraw[opacity=.3, blue] (A) -- (B) -- (C) -- cycle;

endtikzpicture
enddocument 

The calculations is described in following drawing:

For a triangle whose the height is 1, b=c=0.25 and a=0.5.

documentclass[tikz,margin=3mm]standalone
usetikzlibraryintersections,calc
begindocument
begintikzpicture[scale=2]

coordinate (A) at (0,0);
coordinate (B) at (sqrt(4/3), 0) ;
coordinate (C) at ((sqrt(4/3)/2,1) ;
filldraw[opacity=.3, blue] (A) -- (B) -- (C) -- cycle;
node at (A) [below left] 1;
node at (B) [below right]2;
node at (C) [above]3;

%draw (O)--++(0:1)coordinate(A)--++(120:1)coordinate(B)--cycle;
draw ($(A)!0.375!(B)$)coordinate(X)--++(90:0.25)coordinate(P)--++(150:0.25)coordinate(Y);
draw (P)--++(30:0.5)coordinate(Z);

path[] let p1 = ($ (X) - (P) $) in (X) -- (P) node[midway,below=-1mm,sloped]scalebox0.25 pgfmathparseveclen(x1,y1)/28.4pgfmathresult cm;
path[] let p1 = ($ (Y) - (P) $) in (Y) -- (P) node[above=-0.8mm,midway,sloped]scalebox0.25 pgfmathparseveclen(x1,y1)/28.4pgfmathresult cm;
path[] let p1 = ($ (Z) - (P) $) in (Z) -- (P) node[above=-0.8mm,midway,sloped]scalebox0.25 pgfmathparseveclen(x1,y1)/28.4pgfmathresult cm;

endtikzpicture
enddocument

(too long for a comment) So many people here for a question without MWE and incorrect data! I will delete if OP does not provide at least correct data.

documentclass[tikz]standalone
usetikzlibrarycalc,decorations.pathreplacing
begindocument
begintikzpicture[scale=4]
% suppose the altitude is 1
pgfmathsetmacroa2*sqrt(3)/3
draw[teal]
(0,0) coordinate (1) node[below left]1--
(a,0) coordinate (2) node[below right]2--
([turn]120:a) coordinate (3) node[above]3--cycle;
path
(.5*a,0) coordinate (M)
+(90:.5) coordinate (I)
($(1)!(I)!(3)$) coordinate(N)
($(2)!(I)!(3)$) coordinate (P);
draw[red] 
(I)--(M) node[midway,right=1pt,cyan]$a$ 
(I)--(N) node[midway,below left,cyan]$b$
(I)--(P) node[midway,above left,cyan]$c$;
draw[decorate,decoration=brace,raise=1pt,cyan] (I)--(M);
draw[decorate,decoration=brace,raise=1pt,cyan] (I)--(N);
draw[decorate,decoration=brace,raise=1pt,cyan] (I)--(P);
endtikzpicture
enddocument

To me this looks like an XY question. What you really may be after (or what you were really asked to do) is to produce a so-called ternary diagram. Luckily there exists a library for this specifically: usepgfplotslibraryternary. It comes with pgfplots, which is based on TikZ. I added the braces for fun, but also think you'd be better off with just the diagram. Notice that there are already several posts on this site that discuss how you can customize these diagrams, just do a google search for site:tex.stackexchange.com ternary diagram to find them.

documentclass[tikz,border=3mm]standalone
usetikzlibrarycalc,decorations.pathreplacing
usepackagepgfplots
pgfplotssetwidth=7cm,compat=1.16
usepgfplotslibraryternary 
begindocument
begintikzpicture
 beginternaryaxis
 addplot3 coordinates (0.25,0.5,0.25) ;
 path (0.25,0.5,0.25) coordinate (M)
 (1,0,0) coordinate (C) (0,1,0) coordinate (A) (0,0,1) coordinate (B);
 endternaryaxis
 draw (M) -- ($(B)!(M)!(C)$); 
 draw (M) -- ($(A)!(M)!(B)$);
 draw (M) -- ($(C)!(M)!(A)$);
 beginscope[thick,decoration=brace,raise=1pt]
 draw[decorate] (M) -- ($(B)!(M)!(C)$) node[midway,above=2pt,sloped]$0.5$; 
 draw[decorate] (M) -- ($(A)!(M)!(B)$) node[midway,right=2pt]$0.25$; 
 draw[decorate] ($(C)!(M)!(A)$) -- (M) node[midway,above=2pt,sloped]$0.25$; 
 endscope 
endtikzpicture
enddocument

I wrote a macro that builds such a triangle. But its sides and height do not measure 1 unit. Probabilities are the arguments.

For example, we call it proba.5.25.25 or proba.2.3.5

If that's all right with you, I'll explain the construction.

documentclass[tikz,border=5mm]standalone 
usepackagexcolor
usetikzlibrarycalc,angles,decorations.pathreplacing
definecolormygreenRGB63,186,143

newcommandproba[3]
begintikzpicture[auto=left,decoration=brace,amplitude=5pt,raise=5pt]
coordinate(I) at (0,0);
coordinate(c) at (-90:#3*10);
coordinate(b) at (150:#2*10);
coordinate(a) at (30:#1*10);
coordinate(c') at ($(c)!1!-90:(I)$);
coordinate(b') at ($(b)!1!-90:(I)$);
coordinate(a') at ($(a)!1!-90:(I)$);
coordinate[label=left:1](1) at (intersection of c--c' and b--b');
coordinate[label=right:2](2) at (intersection of a--a' and c--c');
coordinate[label=above:3](3) at (intersection of a--a' and b--b');
draw (1)--(2)--(3)--cycle;
foreach p in a,b,c
 draw[red,postaction=draw=mygreen,decorate,
 decoration=brace,amplitude=5pt,raise=5pt] (p)--(I);
 path($(p)!5mm!90:(I)$)--($(I)!5mm!-90:(p)$)node[midway,mygreen,font=bf]p; 
 pic [draw]right angle = I--p--p';

endtikzpicture

begindocument

proba.5.25.25

proba.2.3.5

enddocument

Thanks everybody. Here's what I ended up doing, inspired in part by the other answers. I also used tkz-euclid to draw the lines at right angles. I ended up ditching the exact measures.

documentclass[tikz]standalone

usepackagetkz-euclide 
usetkzobjall 

usetikzlibrarycalc,decorations.pathreplacing

begindocument
begintikzpicture[scale=1.2]

coordinate (A) at (0,0) ;
coordinate (B) at (sqrt(4/3, 0) ;
coordinate (C) at ((sqrt(4/3)/2,1) ;

filldraw[opacity=.3,blue] (A) -- (B) -- (C) -- cycle;

node at (A) [below left] 1;
node at (B) [below right]2;
node at (C) [above]3;

coordinate (x) at ($(A) + (.4,.25)$);

tkzDefPointBy[projection=onto A--C](x) tkzGetPointE
tkzDefPointBy[projection=onto A--B](x) tkzGetPointF
tkzDefPointBy[projection=onto B--C](x) tkzGetPointG


draw (x) -- (E);
draw (x) -- (F);
draw (x) -- (G);



node at ($(x)!0.5!(G)$)[above left=0.5pt]footnotesize a;
node at ($(x)!0.5!(E)$)[below left=0.5pt]footnotesize b;
node at ($(x)!0.5!(F)$)[right=0.5pt]footnotesize c;

draw[decorate,decoration=brace,raise=1pt] (x)--(E);
draw[decorate,decoration=brace,raise=1pt] (x)--(F);
draw[decorate,decoration=brace,raise=1pt] (x)--(G);


endtikzpicture
enddocument