Wie kann ich ein Parallelogramm, welches mit Koordinaten gezeichnet wird, um eine sie beinhaltende Koordinate drehen?

Hallo zusammen,

ich bin auf ein neues Problem gestoßen und finde bisher leider keine passende Hilfe über div. Suchen. Daher hoffe ich, dass ihr mir weiterhelfen könnt.

Ich möchte mich ein wenig weiter in Beamer, tikz und Animationen üben und daher folgendes bauen: Der Satz des Pythagoras als Scherungsbeweis. Daher möchte ich eines der Kathetenquadrate scheren (diese Animation funktioniert wunderbar auf Folie 1) und anschließend das entstandene Parallelogramm um den Dreieckspunkt, den es beinhaltet (in diesem Falle B), um 90 Grad gegen den Uhrzeigersinn drehen.

Meine Versuche mit der Option rotate around funktionieren absolut nicht, das Gebilde wird nicht im Ansatz irgendwie gedreht. Es verharrt leider total in seiner Ausgangslage! =(

Hier mein aktueller Code:

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Code, hier editierbar zum Übersetzen:

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\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{intersections,angles,quotes,calc,through,babel}
\usepackage{animate}
\usepackage{ifthen}
%%%%%%%%%%%% Für Folie 1%%%%%%%%%%%%%%%%
\newcommand{\PytA}[1]{%
\gdef\setA{#1}
\begin{tikzpicture}[scale=0.5]
\coordinate (A) at (0,0);
\coordinate (B) at (5,0);
\coordinate (C) at ($(B)+(126.8698976:3)$);
\coordinate (AB) at ($(A)!1!270:(B)$);
\coordinate (BA) at ($(B)!1!90:(A)$);
\coordinate (BC) at ($(B)!1!270:(C)$);
\coordinate (CB) at ($(C)!1!90:(B)$);
\coordinate (CA) at ($(C)!1!270:(A)$);
\coordinate (AC) at ($(A)!1!90:(C)$);
\coordinate (HA) at ($(B)!(A)!(C)$);
\coordinate (HB) at ($(A)!(B)!(C)$);
\coordinate (HC) at ($(A)!(C)!(B)$);
\path[name path=A--HA](A) -- ($(A)!12cm!(HA)$);
\path[name path=B--HB](B) -- ($(B)!12cm!(HB)$);
\path[name path=C--HC](C) -- ($(C)!12cm!(HC)$);
\path[name path=BC--CB](BC) -- (CB);
\coordinate (H) at (intersection of B--HB and A--HA);
\coordinate (HA2) at (intersection of BC--CB and A--HA);
\coordinate (HB2) at (intersection of CA--AC and B--HB);
\coordinate (HC2) at (intersection of BA--AB and C--HC);
\pic[draw,fill=orange, opacity=0.5,angle radius=.15cm]{angle=A--C--B};
\draw (A)--(B)--(C)--cycle;
\draw (A)--(B)--(BA)--(AB)--cycle;
\draw (B)--(C)--(CB)--(BC)--cycle;
\draw (C)--(A)--(AC)--(CA)--cycle;
\draw (C)--(HC2);
\coordinate (WanderCB) at ($(C)!0.05*\setA!(A)+(36.86989765:3)$);
\fill [green!20!white, opacity=.8](BC)--(B)--($(C)!0.05*\setA!(A)$)--(WanderCB)-- cycle;
\end{tikzpicture}
}
%%%%%%%%%%%% Für Folie 2 %%%%%%%%%%%%%%%%%%%%%%%
 
 
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Ich hoffe, dass ihr mir weiterhelfen könnt! :)

gefragt 11 Mai '16, 16:41

Wienie2401
140●5●9
Akzeptiert-Rate: 0%

bearbeitet 11 Mai '16, 16:42

Eine Antwort:

active answers älteste Antworten neueste Antworten Beliebteste Antworten

Du musst die Rotation als Option der einzelnen Koordinaten angeben:

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Code, hier editierbar zum Übersetzen:

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\fill [green!20!white, opacity=.8]
  ([crot]BC)--([crot]B)--([crot]A)--([crot]WanderCB)-- cycle;
 
 
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
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und vorher

Open in Online-Editor

Code, hier editierbar zum Übersetzen:

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\gdef\setA{#1}%
\tikzset{crot/.style={{rotate around={\setA:(5,0)}}}}%
 
 
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
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alt text

Code für die Abbildung:

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Code, hier editierbar zum Übersetzen:

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\documentclass[margin=5pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{intersections,angles,quotes,calc,through,babel}
\newcommand{\PytB}[1]{%
  \gdef\setA{#1}%
  \tikzset{crot/.style={{rotate around={\setA:(5,0)}}}}%
  \begin{tikzpicture}[scale=0.5]
    \coordinate (A) at (0,0);
    \coordinate (B) at (5,0);
    \coordinate (C) at ($(B)+(126.8698976:3)$);
    \coordinate (AB) at ($(A)!1!270:(B)$);
    \coordinate (BA) at ($(B)!1!90:(A)$);
    \coordinate (BC) at ($(B)!1!270:(C)$);
    \coordinate (CB) at ($(C)!1!90:(B)$);
    \coordinate (CA) at ($(C)!1!270:(A)$);
    \coordinate (AC) at ($(A)!1!90:(C)$);
    \coordinate (HA) at ($(B)!(A)!(C)$);
    \coordinate (HB) at ($(A)!(B)!(C)$);
    \coordinate (HC) at ($(A)!(C)!(B)$);
    \path[name path=A--HA](A) -- ($(A)!12cm!(HA)$);
    \path[name path=B--HB](B) -- ($(B)!12cm!(HB)$);
    \path[name path=C--HC](C) -- ($(C)!12cm!(HC)$);
    \path[name path=BC--CB](BC) -- (CB);
    \coordinate (H) at (intersection of B--HB and A--HA);
    \coordinate (HA2) at (intersection of BC--CB and A--HA);
    \coordinate (HB2) at (intersection of CA--AC and B--HB);
    \coordinate (HC2) at (intersection of BA--AB and C--HC);
    \pic[draw,fill=orange, opacity=0.5,angle radius=.15cm]{angle=A--C--B};
    \draw (A)--(B)--(C)--cycle;
    \draw (A)--(B)--(BA)--(AB)--cycle;
    \draw (B)--(C)--(CB)--(BC)--cycle;
    \draw (C)--(A)--(AC)--(CA)--cycle;
    \draw (C)--(HC2);
    \coordinate (WanderCB) at ($(A)+(36.86989765:3)$);
    \fill [green!20!white, opacity=.8]
      ([crot]BC)--([crot]B)--([crot]A)--([crot]WanderCB)-- cycle;
  \end{tikzpicture}%
}
\begin{document}
 
 
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
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Permanenter link

beantwortet 11 Mai '16, 22:12

esdd
17.8k●30●42●57
Akzeptiert-Rate: 62%

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Frage-Themen:

tikz ×731
rotate ×2
around ×1
coordinate ×1

gestellte Frage: 11 Mai '16, 16:41

Frage wurde gesehen: 6,174 Mal

zuletzt geändert: 11 Mai '16, 22:12

Wie kann ich ein Parallelogramm, welches mit Koordinaten gezeichnet wird, um eine sie beinhaltende Koordinate drehen?

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