How to draw 75-60-45 triangle?

Question

I am preparing thesis about Euler Lines and Schiffler Point; at the end of thesis I need to give example and I chose 75-60-45 triangle. First I need to draw triangle and I must put incentre(I)and draw of course. Then draw Euler lines for each triangles ABI, BIC, CIA and ABC. They are gonne concur at the point which we call Schiffler Point. How can I do that? Thanks for help.

It must looks like;

Note: 1=Euler(CIB) etc

Answer 1

A pstricks solution (you can compile it with pdf-latex if it is launched with the --enable-write18 switch (MiKTeX) or -shell-escape (TeX Live, MacTeX):

\documentclass[12pt, x11names]{article}%

\usepackage{pstricks-add}
\usepackage{pst-eucl}
\usepackage{auto-pst-pdf}

\begin{document}

\psset{unit=10,dotsize = 3pt}
\begin{pspicture}(-3,-1.5)(3,2.5)
    \pstGeonode[PosAngle=180,linecolor=LightSteelBlue3](0,0){A}%(1;45){R}
    \pstGeonode[linecolor=LightSteelBlue3](1,0){B}
    {\psset{PointSymbol=none, PointName=none}
        \pstRotation[RotAngle=45]{A}{B}[R]
        \pstRotation[RotAngle=-60]{B}{A}[T] }
    \pstInterLL[PosAngle=90, linecolor=LightSteelBlue3]{A}{R}{B}{T}{C}
    \pspolygon[linecolor=LightSteelBlue3](A)(B)(C)
    %% Bissectors
    {\psset{PointSymbol=none, PointName=none, linestyle=none}
        \pstBissectBAC{C}{B}{A}{B1}
        \pstBissectBAC{A}{C}{B}{C1}}
    \pstInterLL{B}{B1}{C}{C1}{I}
    \psline(A)(I)(B)\psline(I)(C)
    %%%Centres
    \psset{linestyle=none}
    \pstCircleABC{A}{B}{C}{O}\pstCGravABC{A}{B}{C}{G}
    \pstCircleABC{B}{I}{C}{O1}\pstCGravABC{B}{I}{C}{G1}
    \pstCircleABC{C}{I}{A}{O2}\pstCGravABC{I}{C}{A}{G2}
    \pstCircleABC{A}{I}{B}{O3}\pstCGravABC{A}{I}{B}{G3}
    %%%Euler lines
    \psset{linestyle=solid, linecolor=red, nodesep=0.5,nodesepA=1}
    \pstLineAB[,nodesepA=0.5]{O}{G}
    \pstLineAB{O1}{G1}
    \pstLineAB{O2}{G2}
    \pstLineAB{O3}{G3}
    \color{DarkOrange3}\boldmath\pstInterLL[PointSymbol=o, PosAngle=-30]{O}{G}{O1}{G1}{S}
\end{pspicture}

\end{document} 

Answer 2

Here is a plain Metapost approach, featuring the wonderful whatever keyword in as many places as I could think of. Most of the geometric techniques used here are developed at the end of Ch.9 of The METAFONT Book.

I thought it looked better when I made angle ACB slightly more than a right angle, so that the Euler line for triangle ABC does not pass through point C. The manoeuvres at the end are just to clip the picture to a nice rectangle.

prologues:=3;
outputtemplate:="%j%c.eps";

vardef centroid(expr a, b, c) =
  save m; pair m; m = whatever [a, 1/2[b,c]] = whatever [b, 1/2[a,c]]; m
enddef;

vardef circumcenter(expr a, b, c) =
  save m; pair m;
  m = whatever [a,b] rotatedaround(1/2[a,b],90) 
    = whatever [b,c] rotatedaround(1/2[b,c],90); m
enddef;

vardef orthocenter(expr a, b, c) =
  save m,p,q,r; pair m, p, q, r; 
  p = whatever[b,c]; p-a = whatever * (c-b) rotated 90;
  q = whatever[c,a]; q-b = whatever * (a-c) rotated 90;
  r = whatever[a,b]; r-c = whatever * (b-a) rotated 90;
  m = whatever[a,p] = whatever[b,q]; m
enddef;

vardef incenter(expr a,b,c) = 
  save m; pair m;
  m = whatever [a,b] rotatedaround(a, 1/2 (angle (c-a) - angle (b-a) ) )
    = whatever [b,c] rotatedaround(b, 1/2 (angle (a-b) - angle (c-b) ) ); m
enddef;

% how high is c given a--b as a base?
vardef altitude(expr a,b,c) = 
  save p; pair p; 
  p = whatever[a,b]; p-c = whatever * (b-a) rotated 90;
  length(p-c)
enddef;

newinternal lt_overshoot; lt_overshoot := 10;
vardef line_through (expr a, b) = 
  pair t; t = unitvector(b-a) scaled lt_overshoot;
  -t shifted a -- t shifted b
enddef;   

% Allow en_GB spelling of centre
def circumcentre = circumcenter enddef;
def orthocentre  = orthocenter  enddef;
def incentre     = incenter     enddef;

vardef euler_line(expr a,b,c) = 
  line_through(circumcenter(a,b,c),orthocenter(a,b,c))
enddef;

beginfig(1);

pair A, B, C;

A = origin;
B = 264 right;
C = quartercircle scaled 400 
    intersectionpoint 
    quartercircle rotated 90 scaled 300 shifted B;

pair I; I = incenter(A,B,C);

draw fullcircle scaled 2 altitude(A,B,I) shifted I dashed withdots scaled 1/3 withcolor .7 white;
draw A--B--C--cycle; 
draw A--I;
draw B--I;
draw C--I;

path e[];
e0 = euler_line(A,B,C);
e1 = euler_line(A,B,I);
e2 = euler_line(B,C,I);
e3 = euler_line(C,A,I);

pair S;
S = e0 intersectionpoint e1;

for i=0 upto 3:
  draw e[i] withcolor .67 red;
endfor

draw fullcircle scaled 4 shifted S;
label(btex $S$ etex, S + 12 right);

dotlabel.lft(btex $A$ etex, A);
dotlabel.rt (btex $B$ etex, B);
dotlabel.urt(btex $C$ etex, C);
dotlabel.urt(btex $I$ etex, I);

currentpicture := currentpicture shifted -S;
setbounds currentpicture to unitsquare shifted -(1/2,2/5) xscaled 377 yscaled 233;

endfig;
end

Answer 3

As Percusse told you in the comments, tkz-euclide has French-only documentation, but the examples are very straightforward, especially because the code is always present for each of them and the commands are in English. Performing a search in the document for the commands will usually bring you to the solution.

In order to load the package, you need to add this to your preamble (the second command is loading libraries):

\usepackage{tkz-euclide}
\usetkzobj{all}

Since it's based on TikZ, you need to enclose it in a tikzpicture environment. And actually, it can work together with TikZ, but calculating points/lines is usually easier with this package, whereas TikZ might require some more manual work (it depends on the case though).

My code basically sets up the triangle ABC. From it, you can obtain the triangle incenter and then draw the relative segments. This is provided in the example below to give you an idea on how to get points. As far as the Euler lines are concerned, according to the definition, they are lines that pass through several points relative to a triangle, some of which are:

• the orthocenter
• the circumcenter
• the centroid
• the Exeter point
• the center of the nine-point circle

To form a line, any line, you only need two points. I managed to get the first three for each triangle. Although in the graph, I only use the circumcenter and the orthocenter.

Output

Code

\documentclass[margin=10pt]{standalone}
\usepackage{tkz-euclide}
\usetkzobj{all}


\begin{document}
\begin{tikzpicture}[label style/.style={font=\scriptsize}]
\clip (-1,-1) rectangle (6,4);
\tkzDefPoints{%
    0/0/A,
    5/0/B,
    3.5/3/C}
\tkzDrawPolygon[blue](A,B,C)

\tkzInCenter(A,B,C)\tkzGetPoint{I}
\tkzDrawSegments[blue](A,I B,I C,I)

 For triangle ABC
\tkzCentroid(A,B,C)\tkzGetPoint{ABCc}
\tkzCircumCenter(A,B,C)\tkzGetPoint{ABCcc}
\tkzOrthoCenter(A,B,C)\tkzGetPoint{ABCo}

% For triangle ACI
\tkzCentroid(A,C,I)\tkzGetPoint{ACIc}
\tkzCircumCenter(A,C,I)\tkzGetPoint{ACIcc}
\tkzOrthoCenter(A,C,I)\tkzGetPoint{ACIo}

% For triangle ABI
\tkzCentroid(A,B,I)\tkzGetPoint{ABIc}
\tkzCircumCenter(A,B,I)\tkzGetPoint{ABIcc}
\tkzOrthoCenter(A,B,I)\tkzGetPoint{ABIo}

% For triangle CBI
\tkzCentroid(C,B,I)\tkzGetPoint{CBIc}
\tkzCircumCenter(C,B,I)\tkzGetPoint{CBIcc}
\tkzOrthoCenter(C,B,I)\tkzGetPoint{CBIo}

\tkzDrawSegments[red](ACIcc,ACIo ABIcc,ABIo)
\tkzDrawLines[color=red,add = .7 and .5](CBIo,CBIcc ABCcc,ABCo)
\tkzInterLL(ACIcc,ACIo)(ABIcc,ABIo) \tkzGetPoint{S}

\tkzDrawPoints[color=red,fill=white](S)

\tkzDrawPoints[color=blue, fill=white](A,B,C,I)
\tkzLabelPoints[above left](S)
\tkzLabelPoints[below left](A)
\tkzLabelPoints[above](C)
\tkzLabelPoints[below right](B)
\tkzLabelPoints[above right](I)
\end{tikzpicture}
\end{document}