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Lösung von Aufgabe 10.4P (WS 12 13) - Versionsgeschichte
2026-08-14T18:36:45Z
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Tutorin Anne am 17. Januar 2013 um 15:11 Uhr
2013-01-17T15:11:43Z
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<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Version vom 17. Januar 2013, 15:11 Uhr</td>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Bei einer Punktspiegelung ist der Schnittpunkt ''S'' der beiden Spiegelgeraden ''a'' und ''b'' Mittelpunkt der Strecke <math>\overline{PP''}</math>, mit <math>P''=a\circ b(P) </math>.<br /></div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Bei einer Punktspiegelung ist der Schnittpunkt ''S'' der beiden Spiegelgeraden ''a'' und ''b'' Mittelpunkt der Strecke <math>\overline{PP''}</math>, mit <math>P''=a\circ b(P) </math>.<br /></div></td></tr>
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Tutorin Anne
http://geometrie.idea-sketch.com/index.php?title=L%C3%B6sung_von_Aufgabe_10.4P_(WS_12_13)&diff=20269&oldid=prev
Schnirch am 17. Januar 2013 um 12:26 Uhr
2013-01-17T12:26:51Z
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<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Version vom 17. Januar 2013, 12:26 Uhr</td>
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<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Beweisen Sie Satz IX.<del style="font-weight: bold; text-decoration: none;">4</del>:</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Beweisen Sie Satz IX.<ins style="font-weight: bold; text-decoration: none;">3</ins>:</div></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Bei einer Punktspiegelung <del style="font-weight: bold; text-decoration: none;">werden Geraden stets auf parallele Bildgeraden abgebildet</del>.<br /></div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Bei einer Punktspiegelung <ins style="font-weight: bold; text-decoration: none;">ist der Schnittpunkt ''S'' der beiden Spiegelgeraden ''a'' und ''b'' Mittelpunkt der Strecke <math>\overline{PP''}</math>, mit <math>P''=a\circ b(P) </math></ins>.<br /></div></td></tr>
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Schnirch
http://geometrie.idea-sketch.com/index.php?title=L%C3%B6sung_von_Aufgabe_10.4P_(WS_12_13)&diff=20211&oldid=prev
Schnirch: Die Seite wurde neu angelegt: „Beweisen Sie Satz IX.4: Bei einer Punktspiegelung werden Geraden stets auf parallele Bildgeraden abgebildet.<br /> Kategorie:Einführung_P“
2013-01-16T13:26:27Z
<p>Die Seite wurde neu angelegt: „Beweisen Sie Satz IX.4: Bei einer Punktspiegelung werden Geraden stets auf parallele Bildgeraden abgebildet.<br /> <a href="/wiki/Kategorie:Einf%C3%BChrung_P" title="Kategorie:Einführung P">Kategorie:Einführung_P</a>“</p>
<p><b>Neue Seite</b></p><div>Beweisen Sie Satz IX.4:<br />
Bei einer Punktspiegelung werden Geraden stets auf parallele Bildgeraden abgebildet.<br /><br />
[[Kategorie:Einführung_P]]</div>
Schnirch