summaryrefslogtreecommitdiff
path: root/Master/texmf-dist/doc/latex/tkz-euclide/latex/TKZdoc-euclide-points.tex
blob: 9cea12304c818538199916929b5b5a51236cddd5 (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
\section{Definition of a point}

 Points can be specified in any of the following ways:
\begin{itemize}
\item Cartesian coordinates 
\item Polar coordinates
\item Named points
\item Relative points
\end{itemize}

Even if it's possible, I think it's a bad idea to work directly with coordinates. Preferable is to use named points.
A point is defined if it has a name linked to a unique pair of decimal numbers. 
 Let $(x,y)$ or $(a:d)$  i.e. ( $x$ abscissa, $y$ ordinate) or  ($a$ angle : $d$ distance ).
 This is possible because the plan has been provided with an orthonormed Cartesian coordinate system.   The working axes are supposed to be (ortho)normed with unity equal to $1cm $ or something equivalent like $0.39370~in$.
 Now by default if you use a grid or axes, the rectangle used is defined by the coordinate points~: $(0,0)$ et $(10,10)$. It's the macro \tkzcname{tkzInit} of the package \tkzNamePack{tkz-base} that creates this rectangle. Look at the following two codes and the result of their compilation:
 
\begin{tkzexample}[latex=10cm,small]
\begin{tikzpicture}
 \tkzGrid
 \tkzDefPoint(0,0){O}
 \tkzDrawPoint[red](O)
 \tkzShowBB[line width=2pt,
                   orange]
\end{tikzpicture}
\end{tkzexample}


\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}
 \tkzDefPoint(0,0){O}
 \tkzDefPoint(5,5){A}
 \tkzDrawSegment[blue](O,A)
 \tkzDrawPoints[red](O,A)
 \tkzShowBB[line width=2pt,orange]
\end{tikzpicture}
\end{tkzexample}
 
 The Cartesian coordinate $(a,b)$ refers to the
 point $a$ centimeters in the $x$-direction and $b$ centimeters in the
 $y$-direction.

 A point in polar coordinates requires an angle $\alpha$, in degrees,
 and distance from the origin, $d$.  Unlike Cartesian coordinates, the
 distance does not have a default dimensional unit, so one must be
 supplied.  The \tikz{} syntax for a point specified in polar
 coordinates is $(\alpha:r\:dim)$, where {\em dim} is a dimensional
 unit such as \texttt{cm}, \texttt{pt}, \texttt{in}, or any other
 \TeX-based unit.  Other than syntax and the required dimensional unit,
 this follows usual mathematical usage. 
 

\begin{minipage}[b]{0.5\textwidth}
 Cartesian coordinates 
\begin{tkzexample}[vbox,small]
\begin{tikzpicture}[scale=1] 
  \tkzInit[xmax=5,ymax=5] 
  \tkzDefPoints{0/0/O,1/0/I,0/1/J}
  \tkzDrawXY[noticks,>=latex]
  \tkzDefPoint(3,4){A} 
  \tkzDrawPoints(O,A) 
  \tkzLabelPoint(A){$A_1 (x_1,y_1)$} 
  \tkzShowPointCoord[xlabel=$x_1$,ylabel=$y_1$](A) 
  \tkzLabelPoints(O,I)
  \tkzLabelPoints[left](J)
  \tkzDrawPoints[shape=cross](I,J) 
\end{tikzpicture}
\end{tkzexample}%
\end{minipage}
\begin{minipage}[b]{0.5\textwidth}
 Polar coordinates
\begin{tkzexample}[vbox,small]
\begin{tikzpicture}[,scale=1]
  \tkzInit[xmax=5,ymax=5]
  \tkzDefPoints{0/0/O,1/0/I,0/1/J}
  \tkzDefPoint(40:4){P}
  \tkzDrawXY[noticks,>=triangle 45]    
  \tkzDrawSegment[dim={$r$,
                 16pt,above=6pt}](O,P)
  \tkzDrawPoints(O,P) 
  \tkzMarkAngle[mark=none,->](I,O,P) 
  \tkzFillAngle[fill=blue!20,
                opacity=.5](I,O,P) 
  \tkzLabelAngle[pos=1.25](I,O,P){$\alpha$}  
  \tkzLabelPoint(P){$P  (\alpha : r )$} 
  \tkzDrawPoints[shape=cross](I,J) 
  \tkzLabelPoints(O,I)
  \tkzLabelPoints[left](J) 
\end{tikzpicture}
\end{tkzexample}
\end{minipage}%

The \tkzNameMacro{tkzDefPoint} macro is used to define a point by assigning coordinates to it. This macro is based on \tkzNameMacro{coordinate}, a macro of \TIKZ\ . It can use \TIKZ-specific options such as \IoptName{TikZ}{shift}. If calculations are required then the \tkzNamePack{xfp} package is chosen. We can use Cartesian or polar coordinates.

\subsection{Defining a named point  \tkzcname{tkzDefPoint}}

\begin{NewMacroBox}{tkzDefPoint}{\oarg{local options}\parg{x,y}\marg{name} ou \parg{a:r}\marg{name}}

\begin{tabular}{lll}
\toprule
arguments &  défaut  & définition  \\ 
\midrule
\TAline{(x,y)}{no default}{x et y sont deux dimensions, par défaut en cm.}
\TAline{(a:d)}{no default}{a est un angle en degré, d une dimension}
\TAline{\{name\}}{no default}{Nom attribué au point : $A$, $T_a$ ,$P1$ etc ...}
\bottomrule
\end{tabular}

\medskip
{Les arguments obligatoires de cette macro sont  deux dimensions exprimées avec des décimaux, dans le premier cas ce sont deux mesures de longueur, dans le second ce sont une mesure de longueur et la mesure d'un angle en degré}

\medskip
\begin{tabular}{lll}
\toprule
options             & default & definition  \\ 
\midrule
\TOline{label} {no default} {permet de placer un label à une distance prédéfinie}
\TOline{shift} {no default} {Ajoute (x,y) ou (a:d) à toutes les coordonnées}
 \bottomrule
\end{tabular}

\end{NewMacroBox}

 \subsubsection{Cartesian coordinates }
 
 \begin{tkzexample}[latex=7cm,small]
   \begin{tikzpicture}
   \tkzInit[xmax=5,ymax=5]
   \tkzDefPoint(0,0){A}
   \tkzDefPoint(4,0){B}
   \tkzDefPoint(0,3){C} 
   \tkzDrawPolygon(A,B,C)
   \tkzDrawPoints(A,B,C)
   \end{tikzpicture}
 \end{tkzexample}

 \subsubsection{Calculations with \tkzNamePack{xfp}}

 \begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
  \tkzInit[xmax=4,ymax=4]
  \tkzGrid
  \tkzDefPoint(-1+2,sqrt(4)){O}
  \tkzDefPoint({3*ln(exp(1))},{exp(1)}){A}
  \tkzDefPoint({4*sin(pi/6)},{4*cos(pi/6)}){B}
  \tkzDrawPoints[color=blue](O,B,A)
\end{tikzpicture}
\end{tkzexample}


\subsubsection{Polar coordinates }

\begin{tkzexample}[latex=7cm,small]
  \begin{tikzpicture}
  \foreach \an [count=\i] in {0,60,...,300}
   { \tkzDefPoint(\an:3){A_\i}}
  \tkzDrawPolygon(A_1,A_...,A_6)
  \tkzDrawPoints(A_1,A_...,A_6)
  \end{tikzpicture}
\end{tkzexample}

\subsubsection{Calculations and coordinates}
You must follow the syntax of \tkzNamePack{fxp} here. It is always possible to go through \tkzNamePack{pgfmath} but in this case, the coordinates must be calculated before using the macro \tkzcname{tkzDefPoint}.

\begin{tkzexample}[latex=6cm,small]
  \begin{tikzpicture}[scale=.5]
  \foreach \an [count=\i] in {0,2,...,358}
   { \tkzDefPoint(\an:sqrt(sqrt(\an mm))){A_\i}}
   \tkzDrawPoints(A_1,A_...,A_180)
  \end{tikzpicture}
\end{tkzexample}


\subsubsection{Relative points}

First, we can use the \tkzNameEnv{scope} environment from \TIKZ\ ..
In the following example, we have a way to define an equilateral triangle.

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
  \tkzSetUpLine[color=blue!60]
 \begin{scope}[rotate=30]
  \tkzDefPoint(2,3){A}
  \begin{scope}[shift=(A)]
     \tkzDefPoint(90:5){B}
     \tkzDefPoint(30:5){C}
  \end{scope}
 \end{scope}
 \tkzDrawPolygon(A,B,C)
\tkzLabelPoints[above](B,C)
\tkzLabelPoints[below](A)
\tkzDrawPoints(A,B,C)
\end{tikzpicture}
\end{tkzexample}

%<--------------------------------------------------------------------------->
\subsection{Point relative to another : \tkzcname{tkzDefShiftPoint}}
\begin{NewMacroBox}{tkzDefShiftPoint}{\oarg{Point}\parg{x,y}\marg{name} ou \parg{a:d}\marg{name}}
\begin{tabular}{lll}
arguments &  default & definition \\
\midrule
\TAline{(x,y)}{no default}{x and y are two dimensions, by default in cm.}
\TAline{(a:d)}{no default}{a is an angle in degrees, d is a dimension}

\midrule
options &  default & definition \\

\midrule
\TOline{[pt]} {no default} {\tkzcname{tkzDefShiftPoint}[A](0:4)\{B\}}
\bottomrule
\end{tabular}

\end{NewMacroBox}

\subsubsection{Isosceles triangle with  \tkzcname{tkzDefShiftPoint}}
This macro allows you to place one point relative to another. This is equivalent to a translation. Here is how to construct an isosceles triangle with main vertex A and angle at vertex of $30^{\circ} $.

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[rotate=-30]
 \tkzDefPoint(2,3){A}
 \tkzDefShiftPoint[A](0:4){B}
 \tkzDefShiftPoint[A](30:4){C}
 \tkzDrawSegments(A,B B,C C,A)
 \tkzMarkSegments[mark=|,
           color=red](A,B A,C)
 \tkzDrawPoints(A,B,C)
 \tkzLabelPoints(B,C)
 \tkzLabelPoints[above left](A)
\end{tikzpicture}
\end{tkzexample}



\subsubsection{Equilateral triangle}
Let's see how to get an equilateral triangle (there is much simpler)


\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
 \tkzDefPoint(2,3){A}
 \tkzDefShiftPoint[A](30:4){B}
 \tkzDefShiftPoint[A](-30:4){C}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawPoints(A,B,C)
 \tkzLabelPoints(B,C)
 \tkzLabelPoints[above left](A)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Parallelogram}
There's a simpler way
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}
 \tkzDefPoint(0,0){A}
 \tkzDefPoint(60:3){B}
 \tkzDefShiftPointCoord[B](30:4){C}
 \tkzDefShiftPointCoord[A](30:4){D}
 \tkzDrawPolygon(A,...,D)
 \tkzDrawPoints(A,...,D)
\end{tikzpicture}
\end{tkzexample}

%<--------------------------------------------------------------------------->

\subsection{Definition of multiple points  : \tkzcname{tkzDefPoints}}

\begin{NewMacroBox}{tkzDefPoints}{\oarg{local options}\marg{$x_1/y_1/n_1,x_2/y_2/n_2$, ...}}
$x_i$ et $y_i$ are the coordinates of a referenced point $n_i$

\begin{tabular}{lll}
\toprule
arguments &  default  & example  \\
\midrule
\TAline{$x_i/y_i/n_i$}{}{\tkzcname{tkzDefPoints\{0/0/O,2/2/A\}}}
\end{tabular}

\medskip
\begin{tabular}{lll}
\toprule
options             & default & definition   \\ 
\midrule
\TOline{label} {no default} {allows you to place a label at a predefined distance}
\TOline{shift} {no default} {Adds (x,y) or (a:d) to all coordinates}
 \bottomrule
\end{tabular}

\end{NewMacroBox}

\subsection{Create a triangle}

\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=1]
 \tkzDefPoints{0/0/A,4/0/B,4/3/C}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawPoints(A,B,C)
\end{tikzpicture}
\end{tkzexample}

\subsection{Create a square}
Note here the syntax for drawing the polygon.
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=1]
 \tkzDefPoints{0/0/A,2/0/B,2/2/C,0/2/D}
 \tkzDrawPolygon(A,...,D)
 \tkzDrawPoints(A,B,C,D)
\end{tikzpicture}
\end{tkzexample}

\newpage
\section{Special points}
The introduction of the dots was done in \tkzname{tkz-base}, the most important macro being \tkzcname{tkzDefPoint}. Here are some special points.

%<--------------------------------------------------------------------------->
\subsection{Middle of a segment \tkzcname{tkzDefMidPoint}}
It is a question of determining the middle of a segment.

\begin{NewMacroBox}{tkzDefMidPoint}{\parg{pt1,pt2}}
The result is in \tkzname{tkzPointResult}. We can access it with \tkzcname{tkzGetPoint}.

 \medskip
\begin{tabular}{lll}
\toprule
arguments & default & definition \\
\midrule
\TAline{(pt1,pt2)}{no default}{pt1 and pt2 are two points}
\end{tabular}
\end{NewMacroBox}

\subsubsection{Use of \tkzcname{tkzDefMidPoint}}
Review the use of \tkzcname{tkzDefPoint} in \NamePack{tkz-base}.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
 \tkzDefPoint(2,3){A}
 \tkzDefPoint(4,0){B}
 \tkzDefMidPoint(A,B) \tkzGetPoint{C}
 \tkzDrawSegment(A,B)
 \tkzDrawPoints(A,B,C)
 \tkzLabelPoints[right](A,B,C)
\end{tikzpicture}
\end{tkzexample}

\subsection{Barycentric coordinates }

$pt_1$, $pt_2$, \dots, $pt_n$ being $n$ points, they define $n$ vectors $\overrightarrow{v_1}$, $\overrightarrow{v_2}$, \dots, $\overrightarrow{v_n}$ with the origin of the referential as the common endpoint. $\alpha_1$, $\alpha_2$,
\dots $\alpha_n$ is $n$ numbers, the vector obtained by :
\begin{align*}
  \frac{\alpha_1 \overrightarrow{v_1} + \alpha_2 \overrightarrow{v_2} + \cdots + \alpha_n \overrightarrow{v_n}}{\alpha_1
    + \alpha_2 + \cdots + \alpha_n}
\end{align*}
defines a single point.

\begin{NewMacroBox}{tkzDefBarycentricPoint}{\parg{pt1=$\alpha_1$,pt2=$\alpha_2$,\ldots}}
\begin{tabular}{lll}
arguments & default & definition \\
\midrule
\TAline{(pt1=$\alpha_1$,pt2=$\alpha_2$,\ldots)}{no default}{Each point has a assigned weight}
\bottomrule
\end{tabular}

\medskip
You need at least two points.
\end{NewMacroBox}


\subsubsection{Using \tkzcname{tkzDefBarycentricPoint} with two points}
In the following example, we obtain the barycentre of points A and B with coefficients 1 and 2, in other words:
\[
  \overrightarrow{AI}= \frac{2}{3}\overrightarrow{AB}
\]

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}
  \tkzDefPoint(2,3){A}
  \tkzDefShiftPointCoord[2,3](30:4){B}
  \tkzDefBarycentricPoint(A=1,B=2)
  \tkzGetPoint{I}
  \tkzDrawPoints(A,B,I)
  \tkzDrawLine(A,B)
  \tkzLabelPoints(A,B,I)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Using \tkzcname{tkzDefBarycentricPoint} with three points}

This time M is simply the centre of gravity of the triangle. For reasons of simplification and homogeneity, there is also \tkzcname{tkzCentroid}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.8]
  \tkzDefPoint(2,1){A}
  \tkzDefPoint(5,3){B}
  \tkzDefPoint(0,6){C}
  \tkzDefBarycentricPoint(A=1,B=1,C=1)
  \tkzGetPoint{M}
  \tkzDefMidPoint(A,B)  \tkzGetPoint{C'}
  \tkzDefMidPoint(A,C)  \tkzGetPoint{B'}
  \tkzDefMidPoint(C,B)  \tkzGetPoint{A'}
  \tkzDrawPolygon(A,B,C)
  \tkzDrawPoints(A',B',C')
  \tkzDrawPoints(A,B,C,M)
  \tkzDrawLines[add=0 and 1](A,M B,M C,M)
  \tkzLabelPoint(M){$M$}
  \tkzAutoLabelPoints[center=M](A,B,C)
  \tkzAutoLabelPoints[center=M,above right](A',B',C')
\end{tikzpicture}
\end{tkzexample}

\subsection{Internal Similitude Center}
The centres of the two homotheties in which two circles correspond are called external and internal centres of similitude.

\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.75,rotate=-30]
 \tkzDefPoint(0,0){O}
 \tkzDefPoint(4,-5){A}
 \tkzDefIntSimilitudeCenter(O,3)(A,1) 
 \tkzGetPoint{I}
 \tkzExtSimilitudeCenter(O,3)(A,1) 
 \tkzGetPoint{J}
 \tkzDefTangent[from with R= I](O,3 cm)  
 \tkzGetPoints{D}{E}
 \tkzDefTangent[from with R= I](A,1 cm)  
 \tkzGetPoints{D'}{E'}
 \tkzDefTangent[from  with R= J](O,3 cm) 
 \tkzGetPoints{F}{G}
 \tkzDefTangent[from with R= J](A,1 cm)   
 \tkzGetPoints{F'}{G'}
 \tkzDrawCircle[R,fill=red!50,opacity=.3](O,3 cm)
 \tkzDrawCircle[R,fill=blue!50,opacity=.3](A,1 cm)
 \tkzDrawSegments[add = .5 and .5,color=red](D,D' E,E')
 \tkzDrawSegments[add= 0 and 0.25,color=blue](J,F J,G)
 \tkzDrawPoints(O,A,I,J,D,E,F,G,D',E',F',G')
 \tkzLabelPoints[font=\scriptsize](O,A,I,J,D,E,F,G,D',E',F',G')
\end{tikzpicture}
\end{tkzexample}

\clearpage \newpage
\section{Special points relating to a triangle}

\subsection{Triangle center : \tkzcname{tkzDefTriangleCenter}}

This macro allows you to define the center of a triangle.


\begin{NewMacroBox}{tkzDefTriangleCenter}{\oarg{local options}\parg{A,B,C}}
\tkzHandBomb\ Be careful, the arguments are lists of three points. This macro is used in conjunction with \tkzcname{tkzGetPoint} to get the center you are looking for. You can use \tkzname{tkzPointResult} if it is not necessary to keep the results.

\medskip
\begin{tabular}{lll}
\toprule
arguments & default & definition \\

\midrule
\TAline{(pt1,pt2,pt3)}{no default}{three points}
\midrule
options             & default & definition                         \\
\midrule
\TOline{ortho}  {circum}{Intersection of the altitudes of a triangle}
\TOline{centroid} {circum}{centre of gravity. Intersection of the medians }
\TOline{circum}{circum}{circle center circumscribed}
\TOline{in}    {circum}{centre du cercle inscrit dans à un triangle }
\TOline{ex}    {circum}{center of a circle exinscribed to a triangle }
\TOline{euler}{circum}{centre of Euler's circle }
\TOline{symmedian} {circum}{Lemoine's point or symmedian centre or Grebe's point }
\TOline{spieker} {circum}{Spieker Circle Center}
\TOline{nagel}{circum}{Nagel Centre}
\TOline{mittenpunkt} {circum}{or else MiddlePoint center}
\TOline{feuerbach}{circum}{Feuerbach Point}

\end{tabular}
\end{NewMacroBox}

\subsubsection{\IoptName{tkzDefTriangleCenter}{ortho}}
 The intersection H of the three altitudes  of a triangle is called the orthocenter.


\begin{tkzexample}[latex=5cm,small]
\begin{tikzpicture}
  \tkzDefPoint(0,0){A}
  \tkzDefPoint(5,1){B}
  \tkzDefPoint(1,4){C}
  \tkzClipPolygon(A,B,C)
  \tkzDefTriangleCenter[ortho](B,C,A)
    \tkzGetPoint{H}
  \tkzDefSpcTriangle[orthic,name=H](A,B,C){a,b,c}
  \tkzDrawPolygon[color=blue](A,B,C)
  \tkzDrawPoints(A,B,C,H)
  \tkzDrawLines[add=0 and 1](A,Ha B,Hb C,Hc)
  \tkzLabelPoint(H){$H$}
  \tkzAutoLabelPoints[center=H](A,B,C)
  \tkzMarkRightAngles(A,Ha,B B,Hb,C C,Hc,A)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\IoptName{tkzDefTriangleCenter}{centroid}}

\begin{tkzexample}[latex=5cm,small]
\begin{tikzpicture}[scale=.75]
  \tkzDefPoints{-1/1/A,5/1/B}
  \tkzDefEquilateral(A,B)
  \tkzGetPoint{C}
  \tkzDefTriangleCenter[centroid](A,B,C)
      \tkzGetPoint{G}
  \tkzDrawPolygon[color=brown](A,B,C)
  \tkzDrawPoints(A,B,C,G)
  \tkzDrawLines[add = 0 and 2/3](A,G B,G C,G)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\IoptName{tkzDefTriangleCenter}{circum}}

\begin{tkzexample}[latex=6cm,small]
 \begin{tikzpicture}
  \tkzDefPoints{0/1/A,3/2/B,1/4/C}
  \tkzDefTriangleCenter[circum](A,B,C)
  \tkzGetPoint{G}
  \tkzDrawPolygon[color=brown](A,B,C)
  \tkzDrawCircle(G,A)
  \tkzDrawPoints(A,B,C,G)
 \end{tikzpicture}
\end{tkzexample}


\subsubsection{\IoptName{tkzDefTriangleCenter}{in}}
 In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is a triangle center called the triangle's incenter.
 The center of the incircle, called the incenter, can be found as the intersection of the three internal angle bisectors. The center of an excircle is the intersection of the internal bisector of one angle (at vertex A, for example) and the external bisectors of the other two. The center of this excircle is called the excenter relative to the vertex A, or the excenter of A.[3] Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system.(\url{https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle})
 
 \medskip
 We get the centre of the inscribed circle of the triangle. The result is of course in \tkzname{tkzPointResult}. We can retrieve it with \tkzcname{tkzGetPoint}.

\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}
  \tkzDefPoints{0/1/A,3/2/B,1/4/C}
  \tkzDefTriangleCenter[in](A,B,C)\tkzGetPoint{I}
  \tkzDefPointBy[projection=onto A--C](I)
  \tkzGetPoint{Ib}
  \tkzDrawPolygon[color=blue](A,B,C)
  \tkzDrawPoints(A,B,C,I)
  \tkzDrawLines[add = 0 and 2/3](A,I B,I C,I)
  \tkzDrawCircle(I,Ib)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\IoptName{tkzDefTriangleCenter}{ex}}


An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle's sides.
(\url{https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle})


 We get the centre of an inscribed circle of the triangle. The result is of course in \tkzname{tkzPointResult}. We can retrieve it with \tkzcname{tkzGetPoint}.

\begin{tkzexample}[latex=8cm,small]
\begin{tikzpicture}[scale=.5]
  \tkzDefPoints{0/1/A,3/2/B,1/4/C}
  \tkzDefCircle[ex](B,C,A)
  \tkzGetFirstPoint{J_c}
  \tkzGetSecondPoint{Tc}
  \tkzDrawPolygon[color=blue](A,B,C)
  \tkzDrawPoints(A,B,C,J_c)
  \tkzDrawCircle[red](J_c,Tc)
  \tkzDrawLines[add=1.5 and 0](A,C B,C)
  \tkzLabelPoints(J_c)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Utilisation de \IoptName{tkzDefTriangleCenter}{euler} }
This macro allows to obtain the center of the circle of the nine points or euler's circle or Feuerbach's circle.
The nine-point circle, also called Euler's circle or the Feuerbach circle, is the circle that passes through the perpendicular feet $H_A$, $H_B$, and $H_C$ dropped from the vertices of any reference triangle ABC on the sides opposite them. Euler showed in 1765 that it also passes through the midpoints $M_A$, $M_B$, $M_C$ of the sides of ABC. By Feuerbach's theorem, the nine-point circle also passes through the midpoints $E_A$, $E_B$, and $E_C$ of the segments that join the vertices and the orthocenter H. These points are commonly referred to as the Euler points. (\url{http://mathworld.wolfram.com/Nine-PointCircle.html})

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
 \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[color=blue](A,B,C)
 \tkzDrawCircle(N,E_A)
 \tkzDrawSegments[blue](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)
\end{tikzpicture}
\end{tkzexample}


\subsubsection{Using option \IoptName{tkzDefTriangleCenter}{symmedian}}

\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}
  \tkzDefPoint(0,0){A}
  \tkzDefPoint(5,0){B}
  \tkzDefPoint(1,4){C}
  \tkzDefTriangleCenter[symmedian](A,B,C)\tkzGetPoint{K}
  \tkzDefTriangleCenter[median](A,B,C)\tkzGetPoint{G}
  \tkzDefTriangleCenter[in](A,B,C)\tkzGetPoint{I}
  \tkzDefSpcTriangle[centroid,name=M](A,B,C){a,b,c}
  \tkzDefSpcTriangle[incentral,name=I](A,B,C){a,b,c}
  \tkzDrawPolygon[color=blue](A,B,C)
  \tkzDrawPoints(A,B,C,K)
  \tkzDrawLines[add = 0 and 2/3,blue](A,K B,K C,K)
  \tkzDrawSegments[red,dashed](A,Ma B,Mb C,Mc)
  \tkzDrawSegments[orange,dashed](A,Ia B,Ib C,Ic)
  \tkzDrawLine(G,I)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Using option \IoptName{tkzDefTriangleCenter}{nagel}}

Let$ Ta$ be the point at which the $Ja$ excircle meets the side BC of a triangle $ABC$, and define Tband Tc similarly. Then the lines $ATa$, $BTb$, and $CTc$ concur in the Nagel point $Na$.
\href{http://mathworld.wolfram.com/NagelPoint.html}{Weisstein, Eric W. "Nagel point." From MathWorld--A Wolfram Web Resource. }


\begin{tkzexample}[latex=8cm,small]
  \begin{tikzpicture}[scale=.5]
  \tkzDefPoints{0/0/A,6/0/B,4/6/C}
  \tkzDefSpcTriangle[ex](A,B,C){Ja,Jb,Jc}
  \tkzDefSpcTriangle[extouch](A,B,C){Ta,Tb,Tc}
  \tkzDrawPoints(Ja,Jb,Jc,Ta,Tb,Tc)
  \tkzLabelPoints(Ja,Jb,Jc,Ta,Tb,Tc)
  \tkzDrawPolygon[blue](A,B,C)
  \tkzDefTriangleCenter[nagel](A,B,C) \tkzGetPoint{Na}
  \tkzDrawPoints[blue](B,C,A)
  \tkzDrawPoints[red](Na)
  \tkzLabelPoints[blue](B,C,A)
  \tkzLabelPoints[red](Na)
  \tkzDrawLines[add=0 and 1](A,Ta B,Tb C,Tc)
  \tkzShowBB\tkzClipBB
  \tkzDrawLines[add=1 and 1,dashed](A,B B,C C,A)
  \tkzDrawCircles[ex,gray](A,B,C C,A,B B,C,A)
  \tkzDrawSegments[dashed](Ja,Ta Jb,Tb Jc,Tc)
  \tkzMarkRightAngles[fill=gray!20](Ja,Ta,C Jb,Tb,A Jc,Tc,B)
  \end{tikzpicture}
\end{tkzexample}


\subsubsection{Option Triangle  "mittenpunkt"} 
\begin{tkzexample}[latex=8cm,small]
\begin{tikzpicture}[scale=.5]
 \tkzDefPoints{0/0/A,6/0/B,4/6/C}
 \tkzDefSpcTriangle[centroid](A,B,C){Ma,Mb,Mc}
 \tkzDefSpcTriangle[ex](A,B,C){Ja,Jb,Jc}
 \tkzDefSpcTriangle[extouch](A,B,C){Ta,Tb,Tc}
 \tkzDefTriangleCenter[mittenpunkt](A,B,C) 
 \tkzGetPoint{Mi}
 \tkzDrawPoints(Ma,Mb,Mc,Ja,Jb,Jc)
 \tkzClipBB
 \tkzDrawPolygon[blue](A,B,C)
 \tkzDrawLines[add=0 and 1](Ja,Ma 
               Jb,Mb Jc,Mc)
 \tkzDrawLines[add=1 and 1](A,B A,C B,C)
 \tkzDrawCircles[gray](Ja,Ta Jb,Tb Jc,Tc)
 \tkzDrawPoints[blue](B,C,A)
 \tkzDrawPoints[red](Mi)
 \tkzLabelPoints[red](Mi)
 \tkzLabelPoints[left](Mb)
 \tkzLabelPoints(Ma,Mc,Jb,Jc)
 \tkzLabelPoints[above left](Ja,Jc)
 \tkzShowBB
\end{tikzpicture}
\end{tkzexample}
%<--------------------------------------------------------------------------->
%<--------------------------------------------------------------------------->
\clearpage \newpage
\section{Draw a point}
\subsubsection{Drawing points \tkzcname{tkzDrawPoint}} \hypertarget{tdrp}{}

\begin{NewMacroBox}{tkzDrawPoint}{\oarg{local options}\parg{name}}
\begin{tabular}{lll}
arguments &  default & definition                 \\
\midrule
\TAline{name of point} {no default}  {Only one point name is accepted}
\bottomrule
\end{tabular}

\medskip
The argument is required. The disc takes the color of the circle, but  lighter. It is possible to change everything. The point is a node and therefore it is invariant if the drawing is modified by scaling.

\medskip
\begin{tabular}{lll}
\toprule
options             & default & definition \\
\midrule
\TOline{shape}  {circle}{Possible \tkzname{cross} ou \tkzname{cross out}}
\TOline{size}  {6}{$6 \times$ \tkzcname{pgflinewidth}}
\TOline{color}  {black}{the default color can be changed }
\bottomrule
\end{tabular}

\medskip
{We can create other forms such as \tkzname{cross}}
\end{NewMacroBox}

\subsubsection{Example of point drawings}
Note that \tkzname{scale} does not affect the shape of the dots. Which is normal.  Most of the time, we are satisfied with a single point shape that we can define from the beginning, either with a macro or by modifying a configuration file.


\begin{tkzexample}[latex=5cm,small]
  \begin{tikzpicture}[scale=.5]
   \tkzDefPoint(1,3){A}
   \tkzDefPoint(4,1){B}
   \tkzDefPoint(0,0){O}
   \tkzDrawPoint[color=red](A)
   \tkzDrawPoint[fill=blue!20,draw=blue](B)
   \tkzDrawPoint[color=green](O)
  \end{tikzpicture}
\end{tkzexample}

It is possible to draw several points at once but this macro is a little slower than the previous one. Moreover, we have to make do with the same options for all the points.

\hypertarget{tdrps}{}
\begin{NewMacroBox}{tkzDrawPoints}{\oarg{local options}\parg{liste}}
\begin{tabular}{lll}
arguments &  default  & definition \\
\midrule
\TAline{points list}{no default}{example \tkzcname{tkzDrawPoints(A,B,C)}}
\bottomrule
\end{tabular}

\medskip
\begin{tabular}{lll}
\toprule
options             & default & definition \\
\midrule
\TOline{shape}  {circle}{Possible \tkzname{cross} ou \tkzname{cross out}}
\TOline{size}  {6}{$6 \times$ \tkzcname{pgflinewidth}}
\TOline{color}  {black}{the default color can be changed }
\bottomrule
\end{tabular}

\medskip
\tkzHandBomb\ Beware of the final "s", an oversight leads to cascading errors if you try to draw multiple points. The options are the same as for the previous macro.
\end{NewMacroBox}

\subsubsection{First example}

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}
  \tkzDefPoint(1,3){A} 
  \tkzDefPoint(4,1){B} 
  \tkzDefPoint(0,0){C} 
  \tkzDrawPoints[size=6,color=red,
               fill=red!50](A,B,C)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Second example}

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.5]
 \tkzDefPoint(2,3){A}  \tkzDefPoint(5,-1){B}
 \tkzDefPoint[label=below:$\mathcal{C}$,
               shift={(2,3)}](-30:5.5){E}
 \begin{scope}[shift=(A)]
    \tkzDefPoint(30:5){C}
 \end{scope}
 \tkzCalcLength[cm](A,B)\tkzGetLength{rAB}
 \tkzDrawCircle[R](A,\rAB cm)
 \tkzDrawSegment(A,B)
 \tkzDrawPoints(A,B,C)
 \tkzLabelPoints(B,C)
 \tkzLabelPoints[above](A)
\end{tikzpicture}
\end{tkzexample}

\section{Point on line or circle}
\subsection{Point on a line}

\begin{NewMacroBox}{tkzDefPointOnLine}{\oarg{local options}\parg{A,B}}
\begin{tabular}{lll}
arguments &  default & definition                 \\
\midrule
\TAline{pt1,pt2} {no default}  {Two points to define a line}
\bottomrule
\end{tabular}

\medskip
\begin{tabular}{lll}
\toprule
options       & default & definition \\
\midrule
\TOline{pos=nb}  {}{nb is a decimal  }
\bottomrule
\end{tabular}

\medskip

\end{NewMacroBox}

\subsubsection{Use of option \tkzname{pos} 1}
\begin{tkzexample}[latex=9cm,small]
  \begin{tikzpicture}
  \tkzDefPoints{0/0/A,4/0/B}
  \tkzDrawLine[red](A,B)
  \tkzDefPointOnLine[pos=1.2](A,B) 
  \tkzGetPoint{P}
  \tkzDefPointOnLine[pos=-0.2](A,B) 
  \tkzGetPoint{R}
  \tkzDefPointOnLine[pos=0.5](A,B) 
  \tkzGetPoint{S}
  \tkzDrawPoints(A,B,P)
  \tkzLabelPoints(A,B)
  \tkzLabelPoint[above](P){pos=$1.2$}
  \tkzLabelPoint[above](R){pos=$-.2$}
  \tkzLabelPoint[above](S){pos=$.5$}
  \tkzDrawPoints(A,B,P,R,S)
  \tkzLabelPoints(A,B)
  \end{tikzpicture}
\end{tkzexample}

\subsection{Point on a circle}

\begin{NewMacroBox}{tkzDefPointOnCircle}{\oarg{local options}\parg{A,B}}
\begin{tabular}{lll}
arguments &  default & definition                 \\
\midrule
\TAline{pt1,pt2} {no default}  {Two points to define a line}
\bottomrule
\end{tabular}

\medskip
\begin{tabular}{lll}
\toprule
options       & default & definition \\
\midrule
\TOline{angle}  {0}{angle formed with the abscissa axis}
\TOline{center}  {tkzPointResult}{circle center}
\TOline{radius}  {|\tkzLengthResult pt|}{radius circle}
\bottomrule
\end{tabular}


\end{NewMacroBox}

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}
  \tkzDefPoints{0/0/A,4/0/B,0.8/3/C}  
  \tkzDefPointOnCircle[angle=90,center=B,
      radius=1 cm]
  \tkzGetPoint{I} 
  \tkzDrawCircle[R,teal](B,1cm) 
   \tkzDrawPoint[teal](I)       
  \tkzDefCircle[circum](A,B,C)
  \tkzGetPoint{G} \tkzGetLength{rG}
  \tkzDefPointOnCircle[angle=30,center=G,
  radius=\rG pt]
 \tkzGetPoint{J}
  \tkzDrawPoints(A,B,C)
  \tkzDrawCircle(G,J)
  \tkzDrawPoint(G)
  \tkzDrawPoint[red](J)
\end{tikzpicture}
\end{tkzexample}


\endinput