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tkz-euclide

Lines, triangles and polygons

15 Triangles

15.1 Definition of triangles \tkzDefTriangle

The following macros define or construct a triangle from at least two points.

The following triangles can be defined:

  • two angles determines a triangle with two angles;

  • equilateral determines an equilateral triangle;

  • isosceles right determines an isosceles right triangle;

  • half determines a right triangle whose legs (the two sides of the right angle) are in the ratio \(2\);

  • pythagore determines a right triangle whose sides are proportional to 3, 4 and 5;

  • school determines a right triangle whose angles are 30, 60 and 90 degrees;

  • golden (or sublime) determines the golden triangle: an isosceles triangle whose base is \([AB]\) and whose base angles are 72 degrees;

  • euclid determines Euclid’s triangle, which has the same angles but whose side is \([AB]\) (36 degrees at \(A\)); in previous versions the option was "euclide" with an "e";

  • gold determines a right triangle, with the right angle at \(B\), whose legs are in the ratio \(\Phi=1.618034\): \(AB/BC=\Phi\). This triangle comes from the golden rectangle.

  • cheops determines a third point such that the triangle is isosceles, with sides proportional to \(2\), \(\Phi\) and \(\Phi\).

  • \tkzDefTriangle[⟨local options⟩](⟨A,B⟩)

  • The order of the points matters, because the triangle is constructed counterclockwise. The macro is used either with \tkzGetPoint, or directly with tkzPointResult if you don’t need to keep the point.

    optionsdefaultdefinition
    two angles= #1 and #2no defauttriangle knowing two angles
    equilateralequilateralequilateral triangle
    halfequilateralB rectangle \(AB=2BC\) \(AC\) hypothenuse
    isosceles rightequilateralisosceles right triangle
    pythagoreequilateralproportional to the pythagorean triangle 3-4-5
    pythagorasequilateralsame as above
    egyptianequilateralsame as above
    schoolequilateralangles of 30, 60 and 90 degrees
    goldequilateralB rectangle and \(AB/AC = \Phi\)
    euclidequilateralangles of 72, 72 and 36 degrees, \(A\) is the apex
    goldenequilateralangles of 72, 72 and 36 degrees, \(C\) is the apex
    sublimeequilateralangles of 72, 72 and 36 degrees, \(C\) is the apex
    cheopsequilateralAC=BC, AC and BC are proportional to \(2\) and \(\Phi\).
    swapfalsegives the symmetric point with respect to \(AB\)

    \tkzGetPoint stores the point; otherwise tkzPointResult can be used immediately.

15.1.1 Option equilateral

Compiled output

\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(4,0){B}
\tkzDefTriangle[equilateral](A,B)
\tkzGetPoint{C}
\tkzDrawPolygons(A,B,C)
\tkzDefTriangle[equilateral](B,A)
\tkzGetPoint{D}
\tkzDrawPolygon(B,A,D)
\tkzMarkSegments[mark=s|](A,B B,C A,C A,D B,D)
\end{tikzpicture}
15.1.2 Option two angles

Compiled output

\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(5,0){B}
\tkzDefTriangle[two angles = 50 and 70](A,B)
\tkzGetPoint{C}
\tkzDrawSegment(A,B)
\tkzDrawPoints(A,B)
\tkzLabelPoints(A,B)
\tkzDrawSegments[new](A,C B,C)
\tkzDrawPoints[new](C)
\tkzLabelPoints[above,new](C)
\tkzLabelAngle[pos=1.4](B,A,C){$50^\circ$}
\tkzLabelAngle[pos=0.8](C,B,A){$70^\circ$}
\end{tikzpicture}
15.1.3 Option school

The angles are 30, 60 and 90 degrees.

Compiled output

\begin{tikzpicture}
\tkzDefPoints{0/0/A,4/0/B}
\tkzDefTriangle[school](A,B)
\tkzGetPoint{C}
\tkzMarkRightAngles(C,B,A)
\tkzLabelAngle[pos=0.8](B,A,C){$30^\circ$}
\tkzLabelAngle[pos=0.8](C,B,A){$90^\circ$}
\tkzLabelAngle[pos=0.8](A,C,B){$60^\circ$}
\tkzDrawSegments(A,B)
\tkzDrawSegments[new](A,C B,C)
\tkzLabelPoints(A,B)
\tkzLabelPoints[above](C)
\end{tikzpicture}
15.1.4 Option pythagore

This triangle has sides whose lengths are proportional to 3, 4 and 5.

Compiled output

\begin{tikzpicture}
\tkzDefPoints{0/0/A,4/0/B}
\tkzDefTriangle[pythagore](A,B)
\tkzGetPoint{C}
\tkzDrawSegments(A,B)
\tkzDrawSegments[new](A,C B,C)
\tkzMarkRightAngles(A,B,C)
\tkzDrawPoints[new](C)
\tkzDrawPoints(A,B)
\tkzLabelPoints[above](A,B)
\tkzLabelPoints[new](C)
\end{tikzpicture}
15.1.5 Option pythagore and swap

This triangle has sides whose lengths are proportional to 3, 4 and 5.

Compiled output

\begin{tikzpicture}
\tkzDefPoints{0/0/A,4/0/B}
\tkzDefTriangle[pythagore,swap](A,B)
\tkzGetPoint{C}
\tkzDrawSegments(A,B)
\tkzDrawSegments[new](A,C B,C)
\tkzMarkRightAngles(A,B,C)
\tkzLabelPoint[above,new](C){$C$}
\tkzDrawPoints[new](C)
\tkzDrawPoints(A,B)
\tkzLabelPoints(A,B)
\end{tikzpicture}
15.1.6 Option golden

Compiled output

\begin{tikzpicture}[scale=.8]
\tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B}
\tkzDefTriangle[golden](A,B)\tkzGetPoint{C}
\tkzDefSpcTriangle[in,name=M](A,B,C){a,b,c}
\tkzDrawPolygon(A,B,C)
\tkzDrawPoints(A,B)
\tkzDrawSegment(C,Mc)
\tkzDrawPoints[new](C)
\tkzLabelPoints(A,B)
\tkzLabelPoints[above,new](C)
\end{tikzpicture}
15.1.7 Option euclid

The euclid and golden triangles have the same shape, but the segment \([AB]\) is a side in the first one and the base in the second one.

Compiled output

\begin{tikzpicture}[scale=.75]
\tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B}
\tkzDefTriangle[euclid](A,B)\tkzGetPoint{C}
\tkzDrawPolygon(A,B,C)
\tkzDrawPoints(A,B,C)
\tkzLabelPoints(C)
\tkzLabelPoints[above](A,B)
\tkzLabelAngle[pos=0.8](A,B,C){$72^\circ$}
\tkzLabelAngle[pos=0.8](B,C,A){$72^\circ$}
\tkzLabelAngle[pos=0.8](C,A,B){$36^\circ$}
\end{tikzpicture}
15.1.8 Option isosceles right

Compiled output

\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(4,0){B}
\tkzDefTriangle[isosceles right](A,B)
\tkzGetPoint{C}
\tkzDrawPolygons(A,B,C)
\tkzDrawPoints(A,B,C)
\tkzMarkRightAngles(A,C,B)
\tkzLabelPoints(A,B)
\tkzLabelPoints[above](C)
\end{tikzpicture}
15.1.9 Option gold

Compiled output

\begin{tikzpicture}
\tkzDefPoints{0/0/A,4/0/B}
\tkzDefTriangle[gold](A,B)
\tkzGetPoint{C}
\tkzDrawPolygon(A,B,C)
\tkzDrawPoints(A,B,C)
\tkzLabelPoints[above](A,B)
\tkzLabelPoints[below](C)
\tkzMarkRightAngle(A,B,C)
\tkzText(0,-2){$\dfrac{AB}{BC}=\varphi$}
\end{tikzpicture}
15.2 Specific triangles with \tkzDefSpcTriangle

The centers of some triangles have been defined in the section on special points; here we determine the three vertices of specific triangles.

  • \tkzDefSpcTriangle[⟨local options⟩](⟨p1,p2,p3⟩){⟨r1,r2,r3⟩}

  • The order of the points is important! p1p2p3 defines a triangle; the result is a triangle whose vertices have as references name followed by r1, r2 and r3. If name is empty, the references are r1, r2 and r3.

    optionsdefaultdefinition
    orthiccentroiddetermined by endpoints of the altitudes ...
    centroid or medialcentroidintersection of the triangle’s three triangle medians
    in or incentralcentroiddetermined with the angle bisectors
    ex or excentralcentroiddetermined with the excenters
    extouchcentroidformed by the points of tangency with the excircles
    intouch or contactcentroidformed by the points of tangency of the incircle
    each of the vertices
    eulercentroidformed by Euler points on the nine-point circle
    symmedialcentroidintersection points of the symmedians
    tangentialcentroidformed by the lines tangent to the circumcircle
    feuerbachcentroidformed by the points of tangency of the nine-point ...
    circle with the excircles
    nameemptyused to name the vertices
15.2.1 How to name the vertices

With \tkzDefSpcTriangle[medial,name=M](A,B,C){_A,_B,_C} you get three vertices named \(M_A\), \(M_B\) and \(M_C\).

With \tkzDefSpcTriangle[medial](A,B,C){a,b,c} you get three vertices named \(a\), \(b\) and \(c\).

With \tkzDefSpcTriangle[medial,name=M_](A,B,C){A,B,C} you also get three vertices named \(M_A\), \(M_B\) and \(M_C\).

15.2.2 Option medial or centroid

The centroid of a triangle is the point \(G\) (sometimes also denoted \(M\)) where the three medians of the triangle meet; it is therefore sometimes called the median point. The centroid always lies inside the triangle.

Weisstein, Eric W. "Centroid triangle" From MathWorld–A Wolfram Web Resource.

In the following example, we get the Euler circle, which passes through the points defined above.

Compiled output

\begin{tikzpicture}[rotate=90,scale=.75]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
\tkzDefTriangleCenter[centroid](A,B,C)
\tkzGetPoint{M}
\tkzDefSpcTriangle[medial,name=M](A,B,C){_A,_B,_C}
\tkzDrawPolygon(A,B,C)
\tkzDrawSegments[dashed,new](A,M_A B,M_B C,M_C)
\tkzDrawPolygon[new](M_A,M_B,M_C)
\tkzDrawPoints(A,B,C)
\tkzDrawPoints[new](M,M_A,M_B,M_C)
\tkzLabelPoints[above](B)
\tkzLabelPoints[below](A,C,M_B)
\tkzLabelPoints[right](M_C)
\tkzLabelPoints[left](M_A)
\tkzLabelPoints[font=\scriptsize](M)
\end{tikzpicture}
15.2.3 Option in or incentral

The incentral triangle is the triangle whose vertices are determined by the intersections of the reference triangle’s angle bisectors with the respective opposite sides.
Weisstein, Eric W. "Incentral triangle" From MathWorld–A Wolfram Web Resource.

Compiled output

\begin{tikzpicture}[scale=1]
\tkzDefPoints{ 0/0/A,5/0/B,2/3/C}
\tkzDefSpcTriangle[in,name=I](A,B,C){_a,_b,_c}
\tkzDefCircle[in](A,B,C) \tkzGetPoints{I}{a}
\tkzDrawCircle(I,a)
\tkzDrawPolygon(A,B,C)
\tkzDrawPolygon[new](I_a,I_b,I_c)
\tkzDrawSegments[dashed,new](A,I_a B,I_b C,I_c)
\tkzDrawPoints(A,B,C,I,I_a,I_b,I_c)
\tkzLabelPoints[below](A,B,I_c)
\tkzLabelPoints[above left](I_b)
\tkzLabelPoints[above right](C,I_a)
\end{tikzpicture}
15.2.4 Option ex or excentral

The excentral triangle of a triangle \(ABC\) is the triangle \(J_aJ_bJ_c\) with vertices corresponding to the excenters of \(ABC\).

Compiled output

\begin{tikzpicture}[scale=.6]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
\tkzDefSpcTriangle[excentral,name=J](A,B,C){_a,_b,_c}
\tkzDefSpcTriangle[extouch,name=T](A,B,C){_a,_b,_c}
\tkzDrawPolygon(A,B,C)
\tkzDrawPolygon[new](J_a,J_b,J_c)
\tkzClipBB
\tkzDrawPoints(A,B,C)
\tkzDrawPoints[new](J_a,J_b,J_c)
\tkzLabelPoints(A,B,C)
\tkzLabelPoints[new](J_b,J_c)
\tkzLabelPoints[new,above](J_a)
\tkzDrawCircles[gray](J_a,T_a J_b,T_b J_c,T_c)
\end{tikzpicture}
15.2.5 Option intouch or contact

The contact triangle of a triangle \(ABC\), also called the intouch triangle, is the triangle formed by the points of tangency of the incircle of \(ABC\) with its sides.
Weisstein, Eric W. "Contact triangle" From MathWorld–A Wolfram Web Resource.

We get the points where the incircle touches the sides.

Compiled output

\begin{tikzpicture}[scale=.75]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
\tkzDefSpcTriangle[intouch,name=X](A,B,C){_a,_b,_c}
\tkzInCenter(A,B,C)\tkzGetPoint{I}
\tkzDefCircle[in](A,B,C) \tkzGetPoints{I}{i}
\tkzDrawCircle(I,i)
\tkzDrawPolygon(A,B,C)
\tkzDrawPolygon[new](X_a,X_b,X_c)
\tkzDrawPoints(A,B,C)
\tkzDrawPoints[new](X_a,X_b,X_c)
\tkzLabelPoints[right](X_a)
\tkzLabelPoints[left](X_b)
\tkzLabelPoints[above](C)
\tkzLabelPoints[below](A,B,X_c)
\end{tikzpicture}
15.2.6 Option extouch

The extouch triangle \(T_aT_bT_c\) is the triangle formed by the points of tangency of the sides of a triangle \(ABC\) with its excircles, with centers \(J_a\), \(J_b\) and \(J_c\). The points \(T_a\), \(T_b\) and \(T_c\) can also be constructed as the points which bisect the perimeter of \(ABC\) starting at \(A\), \(B\) and \(C\).
Weisstein, Eric W. "Extouch triangle" From MathWorld–A Wolfram Web Resource.

We get the points of contact of the excircles, and the triangle formed by the centers of the excircles.

Compiled output

\begin{tikzpicture}[scale=.7]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
\tkzDefSpcTriangle[excentral,
name=J](A,B,C){_a,_b,_c}
\tkzDefSpcTriangle[extouch,
name=T](A,B,C){_a,_b,_c}
\tkzDefTriangleCenter[nagel](A,B,C)
\tkzGetPoint{N_a}
\tkzDefTriangleCenter[centroid](A,B,C)
\tkzGetPoint{G}
\tkzDrawPoints[new](J_a,J_b,J_c)
\tkzClipBB \tkzShowBB
\tkzDrawCircles[gray](J_a,T_a J_b,T_b J_c,T_c)
\tkzDrawLines[add=1 and 1](A,B B,C C,A)
\tkzDrawSegments[new](A,T_a B,T_b C,T_c)
\tkzDrawSegments[new](J_a,T_a J_b,T_b J_c,T_c)
\tkzDrawPolygon(A,B,C)
\tkzDrawPolygon[new](T_a,T_b,T_c)
\tkzDrawPoints(A,B,C,N_a)
\tkzDrawPoints[new](T_a,T_b,T_c)
\tkzLabelPoints[below left](A)
\tkzLabelPoints[below](N_a,B)
\tkzLabelPoints[above](C)
\tkzLabelPoints[new,below left](T_b)
\tkzLabelPoints[new,below right](T_c)
\tkzLabelPoints[new,right=6pt](T_a)
\tkzMarkRightAngles[fill=gray!15](J_a,T_a,B
J_b,T_b,C J_c,T_c,A)
\end{tikzpicture}
15.2.7 Option orthic

Given a triangle \(ABC\), the triangle \(H_AH_BH_C\) whose vertices are endpoints of the altitudes from each of the vertices of ABC is called the orthic triangle, or sometimes the altitude triangle. The three lines \(AH_A\), \(BH_B\) and \(CH_C\) are concurrent at the orthocenter \(H\) of \(ABC\).

Compiled output

\begin{tikzpicture}[scale=.75]
\tkzDefPoints{1/5/A,0/0/B,7/0/C}
\tkzDefSpcTriangle[orthic](A,B,C){H_A,H_B,H_C}
\tkzDefTriangleCenter[ortho](B,C,A)
\tkzGetPoint{H}
\tkzDefPointWith[orthogonal,normed](H_A,B)
\tkzGetPoint{a}
\tkzDrawSegments[new](A,H_A B,H_B C,H_C)
\tkzMarkRightAngles[fill=gray!20,
opacity=.5](A,H_A,C B,H_B,A C,H_C,A)
\tkzDrawPolygon[fill=teal!20,opacity=.3](A,B,C)
\tkzDrawPoints(A,B,C)
\tkzDrawPoints[new](H_A,H_B,H_C)
\tkzDrawPolygon[new,fill=orange!20,
opacity=.3](H_A,H_B,H_C)
\tkzLabelPoints(C)
\tkzLabelPoints[left](B)
\tkzLabelPoints[above](A)
\tkzLabelPoints[new](H_A)
\tkzLabelPoints[new,above left](H_C)
\tkzLabelPoints[new,above right](H_B,H)
\end{tikzpicture}
15.2.8 Option feuerbach

The Feuerbach triangle is the triangle formed by the three points of tangency of the nine-point circle with the excircles.
Weisstein, Eric W. "Feuerbach triangle" From MathWorld–A Wolfram Web Resource.

The points of tangency define the Feuerbach triangle.

Compiled output

\begin{tikzpicture}[scale=1]
\tkzDefPoint(0,0){A}
\tkzDefPoint(3,0){B}
\tkzDefPoint(0.5,2.5){C}
\tkzDefCircle[euler](A,B,C) \tkzGetPoint{N}
\tkzDefSpcTriangle[feuerbach,
name=F](A,B,C){_a,_b,_c}
\tkzDefSpcTriangle[excentral,
name=J](A,B,C){_a,_b,_c}
\tkzDefSpcTriangle[extouch,
name=T](A,B,C){_a,_b,_c}
\tkzLabelPoints[below left](J_a,J_b,J_c)
\tkzClipBB \tkzShowBB
\tkzDrawCircle[purple](N,F_a)
\tkzDrawPolygon(A,B,C)
\tkzDrawPolygon[new](F_a,F_b,F_c)
\tkzDrawCircles[gray](J_a,F_a J_b,F_b J_c,F_c)
\tkzDrawPoints[blue](J_a,J_b,J_c,%
F_a,F_b,F_c,A,B,C)
\tkzLabelPoints(A,B,F_c)
\tkzLabelPoints[above](C)
\tkzLabelPoints[right](F_a)
\tkzLabelPoints[left](F_b)
\end{tikzpicture}
15.2.9 Option tangential

The tangential triangle is the triangle \(T_aT_bT_c\) formed by the lines tangent to the circumcircle of a given triangle \(ABC\) at its vertices. It is therefore the antipedal triangle of \(ABC\) with respect to the circumcenter \(O\).
Weisstein, Eric W. "Tangential Triangle." From MathWorld–A Wolfram Web Resource.

Compiled output

\begin{tikzpicture}[scale=.5,rotate=80]
\tkzDefPoints{0/0/A,6/0/B,1.8/4/C}
\tkzDefSpcTriangle[tangential,
name=T](A,B,C){_a,_b,_c}
\tkzDrawPolygon(A,B,C)
\tkzDrawPolygon[new](T_a,T_b,T_c)
\tkzDrawPoints(A,B,C)
\tkzDrawPoints[new](T_a,T_b,T_c)
\tkzDefCircle[circum](A,B,C)
\tkzGetPoint{O}
\tkzDrawCircle(O,A)
\tkzLabelPoints(A)
\tkzLabelPoints[above](B)
\tkzLabelPoints[left](C)
\tkzLabelPoints[new](T_b,T_c)
\tkzLabelPoints[new,left](T_a)
\end{tikzpicture}
15.2.10 Option euler

The Euler triangle of a triangle \(ABC\) is the triangle \(E_AE_BE_C\) whose vertices are the midpoints of the segments joining the orthocenter \(H\) with the respective vertices. The vertices of the triangle are known as the Euler points, and lie on the nine-point circle.
Weisstein, Eric W. "Euler Triangle." From MathWorld–A Wolfram Web Resource.

Compiled output

\begin{tikzpicture}[rotate=90,scale=1.25]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
\tkzDefSpcTriangle[medial,
name=M](A,B,C){_A,_B,_C}
\tkzDefTriangleCenter[euler](A,B,C)
\tkzGetPoint{N} % I= N nine points
\tkzDefTriangleCenter[ortho](A,B,C)
\tkzGetPoint{H}
\tkzDefMidPoint(A,H) \tkzGetPoint{E_A}
\tkzDefMidPoint(C,H) \tkzGetPoint{E_C}
\tkzDefMidPoint(B,H) \tkzGetPoint{E_B}
\tkzDefSpcTriangle[ortho,name=H](A,B,C){_A,_B,_C}
\tkzDrawPolygon(A,B,C)
\tkzDrawCircle(N,E_A)
\tkzDrawSegments[new](A,H_A B,H_B C,H_C)
\tkzDrawPoints(A,B,C,N,H)
\tkzDrawPoints[red](M_A,M_B,M_C)
\tkzDrawPoints[blue]( H_A,H_B,H_C)
\tkzDrawPoints[green](E_A,E_B,E_C)
\tkzAutoLabelPoints[center=N,font=\scriptsize]%
(A,B,C,M_A,M_B,M_C,H_A,H_B,H_C,E_A,E_B,E_C)
\tkzLabelPoints[font=\scriptsize](H,N)
\tkzMarkSegments[mark=s|,size=3pt,
color=blue,line width=1pt](B,E_B E_B,H)
\tkzDrawPolygon[color=cyan](M_A,M_B,M_C)
\end{tikzpicture}
15.2.11 Option euler and Option orthic

Compiled output

\begin{tikzpicture}[scale=1.25]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
\tkzDefSpcTriangle[euler,name=E](A,B,C){a,b,c}
\tkzDefSpcTriangle[orthic,name=H](A,B,C){a,b,c}
\tkzDefExCircle(A,B,C) \tkzGetPoints{I}{i}
\tkzDefExCircle(C,A,B) \tkzGetPoints{J}{j}
\tkzDefExCircle(B,C,A) \tkzGetPoints{K}{k}
\tkzDrawPoints[orange](I,J,K)
\tkzLabelPoints[font=\scriptsize](A,B,C,I,J,K)
\tkzClipBB
\tkzInterLC(I,C)(I,i) \tkzGetSecondPoint{Fc}
\tkzInterLC(J,B)(J,j) \tkzGetSecondPoint{Fb}
\tkzInterLC(K,A)(K,k) \tkzGetSecondPoint{Fa}
\tkzDrawLines[add=1.5 and 1.5](A,B A,C B,C)
\tkzDefCircle[euler](A,B,C) \tkzGetPoints{E}{e}
\tkzDrawCircle[orange](E,e)
\tkzDrawSegments[orange](E,I E,J E,K)
\tkzDrawSegments[dashed](A,Ha B,Hb C,Hc)
\tkzDrawCircles(J,j I,i K,k)
\tkzDrawPoints(A,B,C)
\tkzDrawPoints[orange](E,I,J,K,Ha,Hb,Hc,Ea,Eb,Ec,Fa,Fb,Fc)
\tkzLabelPoints[font=\scriptsize](E,Ea,Eb,Ec,Ha,Hb,Hc,Fa,Fb,Fc)
\end{tikzpicture}
15.2.12 Option symmedial

The symmedial triangle \(K_AK_BK_C\) is the triangle whose vertices are the intersection points of the symmedians with the reference triangle \(ABC\).

Compiled output

\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(5,0){B}
\tkzDefPoint(.75,4){C}
\tkzDefTriangleCenter[symmedian](A,B,C)\tkzGetPoint{K}
\tkzDefSpcTriangle[symmedial,name=K_](A,B,C){A,B,C}
\tkzDrawPolygon(A,B,C)
\tkzDrawSegments[new](A,K_A B,K_B C,K_C)
\tkzDrawPoints(A,B,C,K,K_A,K_B,K_C)
\tkzLabelPoints(A,B,K,K_C)
\tkzLabelPoints[above](C)
\tkzLabelPoints[right](K_A)
\tkzLabelPoints[left](K_B)
\end{tikzpicture}
15.3 Permutation of two points of a triangle
  • \tkzPermute(⟨\(pt1\),\(pt2\),\(pt3\)⟩)

  • argumentsexampleexplanation
    (pt1,pt2,pt3)\tkzPermute(A,B,C)\(A\) and \(\widehat{B,A,C}\) unchanged, \(B\) and \(C\) exchanged

    The triangle is unchanged.

15.3.1 Modification of the school triangle

This triangle is constructed from the segment \([AB]\) on the ray \([Ax)\).

If we want the segment \([AC]\) to be on \([Ax)\), we just have to swap \(B\) and \(C\).

Compiled output

\begin{tikzpicture}
\tkzDefPoints{0/0/A,4/0/B,6/0/x}
\tkzDefTriangle[school](A,B)
\tkzGetPoint{C}
\tkzPermute(A,B,C)
\tkzDrawSegments(A,B C,x)
\tkzDrawSegments(A,C B,C)
\tkzDrawPoints(A,B,C)
\tkzLabelPoints(A,C,x)
\tkzLabelPoints[above](B)
\tkzMarkRightAngles(C,B,A)
\end{tikzpicture}

Remark: only the first point is unchanged. The order of the last two arguments does not matter.