Satz des Thales: Unterschied zwischen den Versionen

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(→‎induktive Satzfindung)
(→‎Umkehrung Satz des Thales)
 
(36 dazwischenliegende Versionen von 9 Benutzern werden nicht angezeigt)
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=Test=
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=Ein wenig Didaktik=
 
Hier geben Ihnen die Didaktikspezialisten Tipps zum Satz des Thales
 
Hier geben Ihnen die Didaktikspezialisten Tipps zum Satz des Thales
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==Geogebratest==
 
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=Satzfindung=
  
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==Induktive Satzfindung==
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==Funktionale Betrachtung==
 
==Funktionale Betrachtung==
−
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−
==Beweisführung==
+
===Variante 1===
−
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+
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−
=induktive Satzfindung=
 
−
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−
=Sehnen verschieben=
+
===Variante 2===
−
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+
−
--[[Benutzer:Mirasol|Mirasol]] 13:52, 9. Jul. 2010 (UTC)
+
−
Bewege den Punkt A, um die Sehne zu verschieben!
+
  
−
=induktive Satzfindung einer Umkehrung=
+
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 +
--[[Benutzer:&quot;chris&quot;07|&quot;chris&quot;07]] 21:12, 14. Jul. 2010 (UTC)
  
−
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(vgl. Idee von Frau "komplizierter Doppelname" in einer alten Klausur)<br />
+
===Variante 3===
−
Kann bitte jemand den Satz auf richtige Formulierung kontrollieren?! <br />
+
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−
Welche speziellen Art der Einführung von Sätzen im induktiven Bereich lässt sich dieses Beispiel zuordnen?
+
--[[Benutzer:&quot;chris&quot;07|&quot;chris&quot;07]] 21:12, 14. Jul. 2010 (UTC)
−
=Funktional=
+
 
−
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+
=Beweisfindung=
 +
 
 +
 
 +
==ikonisches/halbikonisches Beweisen==
 +
 
 +
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framePossible = "false" showResetIcon = "true" showAnimationButton = "true" enableRightClick = "false" errorDialogsActive = "true" enableLabelDrags = "false" showMenuBar = "false" showToolBar = "false" showToolBarHelp = "false" showAlgebraInput = "false" allowRescaling = "true" />--[[Benutzer:&quot;chris&quot;07|&quot;chris&quot;07]] 17:07, 15. Jul. 2010 (UTC)
 +
 
 +
==Beweisen am Beispiel==
 
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 +
 +
=induktive Satzfindung der allgemeinen Umkehrung=
 +
 +
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 +
 +
Nimmt man statt eines Rechtecks ein Parallelogramm, so lässt sich die Umkehrung des Periphriewinkelsatzes finden:<br />
 +
Die Scheitel kongruente Winkel, deren Schenkel die Eckpunkte einer Strecke AB enthalten, liegen auf einem Kreis, der AB als Sehne hat.--[[Benutzer:Tja???|Tja???]] 09:39, 23. Jul. 2010 (UTC)
 +
 +
=Beweisführung=
 +
 +
==Satz des Thales==
 +
===Satz des Thales===
 +
Es sei k ein Kreis mit einem Durchmesser <math> \overline {AB} </math>. Jeder Peripheriewinkel von k über <math> \overline {AB} </math> ist ein rechter Winkel.--[[Benutzer:Löwenzahn|Löwenzahn]] 15:07, 23. Jul. 2010 (UTC)
 +
 +
Ein Versuch den Satz des Thales  mit dem EP zu beweisen:
 +
 +
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 +
Vor: Dreieck ABC, A,B und C element von k, Kreis K<br />
 +
Beh: y= 90<br />
 +
 +
1) Konstruieren eine Parallele zu AB durch C  (Nach dem EP)<br />
 +
2)Der Winkel < ACE ist Kongruent zu alpha  (Wechselwinkelsatz)<br />
 +
3)Delta ist kongruent zu Betha  (Wechselwinkelsatz)<br />
 +
4)Winkel < ACM ist Kongruent zu alpha  (Basiswinkelsatz)<br />
 +
5) Winkel < MCB ist kongruent zu Betha  (Basiswinkelsatz)<br />
 +
6) <ACE+<ACM+<MCB+<BCD= 180<br />
 +
7)alpha+alpha+betha+Betha= 180  (einsetzten der Kongruenzen)<br />
 +
8) 2*(alpha+betha)= 180    (rechenen in R)<br />
 +
9) alpha+betha=90            (rechenen in R)<br />
 +
10) Y=90<br />
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q.e.d<br />
 +
 +
==Umkehrung 1: Satz des Thales==
 +
===Umkehrung Satz des Thales===
 +
Ist <math> \overline {ABC} </math> ein Dreieck mit einem rechten WInkel bei <math> C </math>, so liegt der Punkt <math> C </math> auf dem Thaleskreis, wobei <math> \overline {AB} </math> einen Durchmesser des Kreises <math> k </math>bildet.--[[Benutzer:Löwenzahn|Löwenzahn]] 15:07, 23. Jul. 2010 (UTC)<br />
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Anmerkung--[[Benutzer:TimoRR|TimoRR]] 16:13, 26. Jul. 2010 (UTC): Du meinst <math>\overline{AB}</math> ist Durchmesser des Kreises <math>k</math>. !?? Jupp, Danke!--[[Benutzer:Löwenzahn|Löwenzahn]] 16:25, 26. Jul. 2010 (UTC)
 +
 +
==Umkehrung 2: Satz des Thales==
 +
===Umkehrung Satz des Thales===
 +
 +
<br />Ist ein Peripheriewinkel <math>\gamma </math> über einer Sehne <math> s </math> eines Kreises <math> k </math> ein rechter Winkel, so ist die Sehne s ein Durchmesser des Kreises <math> k </math>.--[[Benutzer:Löwenzahn|Löwenzahn]] 15:07, 23. Jul. 2010 (UTC)
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<br />Ein [[Beweise_von_Studenten#Umkehrung_des_Satz_des_Thales | Versuch eines Beweises]] besser: zwei Beweis-Ideen, eine über Winkelkonstruktion, die andere via Zentri-Peripheriewinkelsatz...
 +
<br />--[[Benutzer:Heinzvaneugen|Heinzvaneugen]] 10:29, 26. Jul. 2010 (UTC)
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 +
===Kommentar zu den Umkehrungen des Thalesstzes--[[Benutzer:*m.g.*|*m.g.*]] 20:43, 23. Jul. 2010 (UTC)===
 +
Es sei <math>\ \alpha</math> ein Winkel und <math>\ k</math> ein Kreis.
 +
Der Satz des Thales hat zwei Voraussetzungen:
 +
 +
# <math>\ \alpha</math> ist Peripheriewinkel von <math>\ k</math>
 +
# über einem Durchmesser von <math> \ k</math>.
 +
 +
Die Behauptung des Thalessatzes: <math>\ \alpha</math> ist ein rechter Winkel.
 +
 +
Aus Gründen der Übersicht benenne ich die Voraussetzungen V1 und V2. Für die Behauptung schreibe ich B.
 +
 +
Satz des Thales:
 +
 +
Aus V1 und V2 folgt B.
 +
 +
Die eigentliche Umkehrung:
 +
 +
Aus B folgt V1 und V2.
 +
 +
Gemischte Umkehrung 1:
 +
 +
Aus B und V1 folgt V2.
 +
 +
Gemischte Umkehrung 2:
 +
 +
Aus B und V2 folgt V1.
 +
 +
<br /><br />Also fehlt uns noch die
 +
 +
==== Eigentliche Umkehrung des Satz von Thales ====
 +
Mein Vorschlag: <br />Es sei ein Dreieck <math> \overline {ABC} </math> mit den schulüblichen Bezeichnungen. Ist <math>\ \gamma</math> ein rechter Winkel, so ist <math> \ c</math>  identisch mit einem Durchmesser des Umkreises des Dreiecks.
 +
<br />--[[Benutzer:Barbarossa|Barbarossa]] 08:38, 25. Jul. 2010 (UTC)

Aktuelle Version vom 26. Juli 2010, 17:25 Uhr

Inhaltsverzeichnis

Ein wenig Didaktik

Hier geben Ihnen die Didaktikspezialisten Tipps zum Satz des Thales

Satzfindung

Induktive Satzfindung

--Gubbel 12:10, 21. Jul. 2010 (UTC)

Funktionale Betrachtung

Variante 1

--"chris"07 21:47, 15. Jul. 2010 (UTC)


Variante 2

--"chris"07 21:12, 14. Jul. 2010 (UTC)


Variante 3

--"chris"07 21:12, 14. Jul. 2010 (UTC)

Beweisfindung

ikonisches/halbikonisches Beweisen

--"chris"07 17:07, 15. Jul. 2010 (UTC)

Beweisen am Beispiel

induktive Satzfindung der allgemeinen Umkehrung

Nimmt man statt eines Rechtecks ein Parallelogramm, so lässt sich die Umkehrung des Periphriewinkelsatzes finden:
Die Scheitel kongruente Winkel, deren Schenkel die Eckpunkte einer Strecke AB enthalten, liegen auf einem Kreis, der AB als Sehne hat.--Tja??? 09:39, 23. Jul. 2010 (UTC)

Beweisführung

Satz des Thales

Satz des Thales

Es sei k ein Kreis mit einem Durchmesser \overline {AB} . Jeder Peripheriewinkel von k über \overline {AB} ist ein rechter Winkel.--Löwenzahn 15:07, 23. Jul. 2010 (UTC)

Ein Versuch den Satz des Thales mit dem EP zu beweisen:

Vor: Dreieck ABC, A,B und C element von k, Kreis K
Beh: y= 90

1) Konstruieren eine Parallele zu AB durch C (Nach dem EP)
2)Der Winkel < ACE ist Kongruent zu alpha (Wechselwinkelsatz)
3)Delta ist kongruent zu Betha (Wechselwinkelsatz)
4)Winkel < ACM ist Kongruent zu alpha (Basiswinkelsatz)
5) Winkel < MCB ist kongruent zu Betha (Basiswinkelsatz)
6) <ACE+<ACM+<MCB+<BCD= 180
7)alpha+alpha+betha+Betha= 180 (einsetzten der Kongruenzen)
8) 2*(alpha+betha)= 180 (rechenen in R)
9) alpha+betha=90 (rechenen in R)
10) Y=90
q.e.d

Umkehrung 1: Satz des Thales

Umkehrung Satz des Thales

Ist \overline {ABC} ein Dreieck mit einem rechten WInkel bei C , so liegt der Punkt C auf dem Thaleskreis, wobei \overline {AB} einen Durchmesser des Kreises k bildet.--Löwenzahn 15:07, 23. Jul. 2010 (UTC)
Anmerkung--TimoRR 16:13, 26. Jul. 2010 (UTC): Du meinst \overline{AB} ist Durchmesser des Kreises k. !?? Jupp, Danke!--Löwenzahn 16:25, 26. Jul. 2010 (UTC)

Umkehrung 2: Satz des Thales

Umkehrung Satz des Thales


Ist ein Peripheriewinkel \gamma über einer Sehne s eines Kreises k ein rechter Winkel, so ist die Sehne s ein Durchmesser des Kreises k .--Löwenzahn 15:07, 23. Jul. 2010 (UTC)


Ein Versuch eines Beweises besser: zwei Beweis-Ideen, eine über Winkelkonstruktion, die andere via Zentri-Peripheriewinkelsatz...
--Heinzvaneugen 10:29, 26. Jul. 2010 (UTC)

Kommentar zu den Umkehrungen des Thalesstzes--*m.g.* 20:43, 23. Jul. 2010 (UTC)

Es sei \ \alpha ein Winkel und \ k ein Kreis. Der Satz des Thales hat zwei Voraussetzungen:

  1. \ \alpha ist Peripheriewinkel von \ k
  2. über einem Durchmesser von \ k.

Die Behauptung des Thalessatzes: \ \alpha ist ein rechter Winkel.

Aus Gründen der Übersicht benenne ich die Voraussetzungen V1 und V2. Für die Behauptung schreibe ich B.

Satz des Thales:

Aus V1 und V2 folgt B.

Die eigentliche Umkehrung:

Aus B folgt V1 und V2.

Gemischte Umkehrung 1:

Aus B und V1 folgt V2.

Gemischte Umkehrung 2:

Aus B und V2 folgt V1.



Also fehlt uns noch die

Eigentliche Umkehrung des Satz von Thales

Mein Vorschlag:
Es sei ein Dreieck \overline {ABC} mit den schulüblichen Bezeichnungen. Ist \ \gamma ein rechter Winkel, so ist \ c identisch mit einem Durchmesser des Umkreises des Dreiecks.
--Barbarossa 08:38, 25. Jul. 2010 (UTC)

Von „http://geometrie.zum.de/index.php?title=Satz_des_Thales&oldid=3800“