Reduktionssatz: Jede Bewegung ist die NAF von zwei oder drei Geradenspiegelungen: Unterschied zwischen den Versionen

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Hier nochmal mit Bearbeitungsmöglichkeit (Einfügen von Mittelsenkrechten, Strecken etc.)
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<br />
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--[[Benutzer:HecklF|Flo60]] 14:11, 29. Apr. 2012 (CEST)
 +
=====Fall 4.2=====
 +
Umlaufsinn bleibt {{Schrift_orange|nicht}} erhalten
 +
<br />
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--[[Benutzer:HecklF|Flo60]] 14:11, 29. Apr. 2012 (CEST)
  
 
[[Kategorie:Elementargeometrie]]
 
[[Kategorie:Elementargeometrie]]

Version vom 29. April 2012, 14:11 Uhr

Inhaltsverzeichnis

Drei nicht kollineare Punkte reichen aus

Satz:

Jede Bewegung ist durch drei nicht kollineare Punkte und deren Bilder eindeutig bestimmt.

Der Reduktionssatz

Satz: Reduktionssatz

Jede Bewegung ist die Nacheinanderausführung von zwei oder drei Geradenspiegelungen.

Beweis

Es seien A, B, C drei nicht kollineare Punkte und \varphi eine Bewegung.
A'=\varphi(A), B'=\varphi(B), C'=\varphi(C) seien die Bilder von A, B, C bei \varphi

Fall 1

A=A', B=B', C=C'

Fall 2

o.B.d.A. A=A', B=B'

Wo muss C`liegen

C' muss auf dem Kreis um A durch C liegen.
Begründung: Bewegungen sind abstandserhaltend.

C' muss auf dem Kreis um B durch C liegen.
Begründung: Bewegungen sind abstandserhaltend.

C' liegt damit in der Schnittmenge der beiden Kreise.

Warum wird C durch eine Spiegelung an AB auf C' abgebildet?

Es genügt zu zeigen, dass AB die Mittelsenkrechte von \overline{CC'} ist.

AB ist die Mittelsenkrechte von \overline{CC'} weil

|AC|=|AC'| und |BC|=|BC'|.

Fall 3

o.B.d.A. A=A'

Reduktionssatz fall 2.png


Spiegelung an der Mittelsenkrechten von \overline{BB'} führt auf Fall 2 zurück.

Fall 4

A \not= A', B\not=B', C\not=C'

Fall 4.1

Umlaufsinn bleibt erhalten



Hier nochmal mit Bearbeitungsmöglichkeit (Einfügen von Mittelsenkrechten, Strecken etc.)


--Flo60 14:11, 29. Apr. 2012 (CEST)

Fall 4.2

Umlaufsinn bleibt nicht erhalten


--Flo60 14:11, 29. Apr. 2012 (CEST)