Lösung von Aufgabe 11.2P (SoSe 12): Unterschied zwischen den Versionen
Aus Geometrie-Wiki
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| − | hier erst mal nen geogebra-bild dazu. <br />das dreieck abc wird an d zu a'b'c', <br />dann an e zu a´´ b´´ c´´ <br />und schließlich an f zu a´´´ | + | hier erst mal nen geogebra-bild dazu. <br />das dreieck abc wird an d zu a'b'c', <br />dann an e zu a´´ b´´ c´´ <br />und schließlich an f zu a´´´ b´´´ c´´´.<br /> |
d, e und f können bewegt werden.<br /> | d, e und f können bewegt werden.<br /> | ||
durch spiegelung von abc an g erhält man ebenfalls a´´´b´´´c´´´<br />--[[Benutzer:Studentin|Studentin]] 23:47, 3. Jul. 2012 (CEST) | durch spiegelung von abc an g erhält man ebenfalls a´´´b´´´c´´´<br />--[[Benutzer:Studentin|Studentin]] 23:47, 3. Jul. 2012 (CEST) | ||
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[[Kategorie:Einführung_P]] | [[Kategorie:Einführung_P]] | ||
Version vom 3. Juli 2012, 22:53 Uhr
Zeigen Sie, dass die Verkettung dreier Punktspiegelungen wieder eine Punktspiegelung ist, wobei das Zentrum der neuen Punktspiegelung auf dem Eckpunkt eines Parallelogramms liegt, dessen drei andere Eckpunkte durch die Zentren der zu ersetzenden drei Punktspiegelungen gebildet werden.
hier erst mal nen geogebra-bild dazu.
das dreieck abc wird an d zu a'b'c',
dann an e zu a´´ b´´ c´´
und schließlich an f zu a´´´ b´´´ c´´´.
d, e und f können bewegt werden.
durch spiegelung von abc an g erhält man ebenfalls a´´´b´´´c´´´
--Studentin 23:47, 3. Jul. 2012 (CEST)
