How to create a point on a line with an angle constraint?
How to create a point on a line with an angle constraint?
I want to define a point (node) D on line AC such that angle ABC equals to angle CDE. How to do this by using the easiest trick of PSTricks?
D
AC
ABC
CDE
documentclass[pstricks,border=1cm]standalone
usepackagepst-eucl
begindocument
beginpspicture(8,-6)
pstTriangle(0,-6)B(8,-6)A(2,0)C
pstMarkAngleABC
pstGeonode[PosAngle=180]([nodesep=4]BC)E
endpspicture
enddocument
5 Answers
5
The angle between the line ED and the horizontal line is beta-alpha, the reason why we know the slope of the line.
documentclass[pstricks,border=1cm]standalone
usepackagepst-eucl
begindocument
beginpspicture(8,-6)
pstTriangle(0,-6)B(8,-6)A(2,0)C
pstMarkAngleABC
pstGeonode[PosAngle=180]([nodesep=4]BC)E%
(! psGetNodeCenterA psGetNodeCenterB
psGetNodeCenterC psGetNodeCenterE
C.y A.y sub A.x C.x sub atan /Alpha ED
C.y B.y sub C.x B.x sub atan /Beta ED
Beta Alpha sub abs Tan E.y add E.x 1 add exch )D'
pstInterLLCAED'D
pstLineAB[linecolor=red]DE
pstMarkAngle[linecolor=red]CDE
endpspicture
enddocument
Not sure if this is the easiest. But it works.
documentclass[pstricks,border=1cm]standalone
usepackagepst-eucl
begindocument
beginpspicture(8,-6)
pstTriangle(0,-6)B(8,-6)A(2,0)C
pstMarkAngleABC
pstGeonode[PosAngle=180]([nodesep=4]BC)E
pstInterLC[PointSymbol=none,PointName=none]CACEGF
pstTranslation[PointSymbol=none,PointName=none]ABF
pstInterLL[PointSymbol=none,PointName=none]CBFF'D'
pstInterLCCACD'G'D
pstLineABDE
pstMarkAngleCDE
%pstArcOAB[linecolor=blue]CEA
%pstLineABFF'
%pstArcOAB[linecolor=blue]CD'A
endpspicture
enddocument
To see the construction, simply remove the three [PointSymbol=none,PointName=none]’s and uncomment the last three lines within the pspicture.
[PointSymbol=none,PointName=none]
pspicture
Just for comparison, anyone wrestling with the pst-eucl syntax and documentation, might like to try this type of thing in Metapost, using the elegant implicit definition of linear variables.
pst-eucl
documentclass[border=5mm]standalone
usepackageluatex85
usepackageluamplib
begindocument
mplibtextextlabelenable
beginmplibcode
vardef angle_mark(expr a, b, c, r) =
fullcircle scaled 2r
rotated angle (a-b)
shifted b
cutafter (b--c)
enddef;
beginfig(1);
pair A, B, C, D, E;
A = 6 right scaled 1cm;
B = 2 left scaled 1cm;
C = 6 up scaled 1cm;
E = 1/5[B,C]; % or wherever you like along B--C....
numeric a, b, d, e;
a = abs(B-C);
b = abs(C-A);
d = abs(C-E);
a/b = e/d; % implicitly define "e"
D = (e/b)[C,A]; % D is then e/b along C--A...
label.ulft("$a$", 1/2[B,C]) withcolor 2/3 blue;
label.urt ("$b$", 1/2[A,C]) withcolor 2/3 blue;
label.lrt ("$d$", 1/2[C,E]) withcolor 2/3 blue;
label.llft("$e$", 1/2[C,D]) withcolor 2/3 blue;
draw angle_mark(A, B, C, 12) withcolor 2/3 red;
draw angle_mark(C, D, E, 12) withcolor 2/3 red;
draw A--B--C--cycle;
draw D--E;
dotlabel.lrt ("$A$", A);
dotlabel.llft("$B$", B);
dotlabel.top ("$C$", C);
dotlabel.urt ("$D$", D);
dotlabel.ulft("$E$", E);
endfig;
endmplibcode
enddocument
My own solution.
documentclass[pstricks,border=15pt]standalone
usepackagepst-eucl
begindocument
beginpspicture(6,-4)
pstTriangle(0,-4)B(6,-4)A(2,0)C
pstMarkAngleABC
pstGeonode[PosAngle=180]([nodesep=3]BC)E
pstRotation[RotAngle=pstAngleAOBBAC,PointName=none,PointSymbol=none]EC[C']
pstInterLL[PosAngle=30]ACEC'D
pstMarkAngleCDE
nclineED
endpspicture
enddocument
Just for fun: a TikZ solution. Notice that there is the tkz-euclide package which offers a very similar syntax as in these pstricks codes. The point of this answer, however, is just to say that in TikZ there is the calc syntax, which is, admittedly, a bit strange when one sees it for the first time. However, I would like to argue that, once one gets a bit familiar with it, it is much more powerful and universal than the other helpers that are on the market. There is no need to define a new complicated macro for every purpose, calc allows one to deal with all these things in a universal way.
calc
documentclass[tikz,border=3.14mm]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
draw (0,-6) coordinate (B) -- (8,-6) coordinate (A) -- (2,0) coordinate (C)
-- cycle;
draw let p1=($(C)-(B)$),p2=($(A)-(B)$),n1=(atan2(y1,x1)+atan2(y2,x2))/2
in (B) ++ (n1:4) coordinate (D)
(B) -- (intersection cs:first line=(A)--(C), second line=(B)--(D));
endtikzpicture
enddocument
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