https://artofproblemsolving.com/wiki/index.php?title=Equidistant&feed=atom&action=history
Equidistant - Revision history
2024-03-29T14:36:36Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=Equidistant&diff=19175&oldid=prev
1=2 at 14:35, 6 November 2007
2007-11-06T14:35:03Z
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1=2
https://artofproblemsolving.com/wiki/index.php?title=Equidistant&diff=12206&oldid=prev
JBL at 22:32, 16 January 2007
2007-01-16T22:32:00Z
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<p><b>New page</b></p><div>'''Equidistant''' means "at the same distance."<br />
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Thus, for instance, if [[triangle]] <math>\triangle ABC</math> is [[isosceles triangle | isosceles]] with base <math>BC</math>, [[point]]s <math>B</math> and <math>C</math> are equidistant from point <math>A</math>.<br />
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Similarly, the [[perpendicular bisector]] of a [[line segment]] is the [[set]] of points equidistant from the [[endpoint]]s. So, given segment <math>\overline{AB}</math> and a point <math>C</math> such that <math>AC = BC</math>, we know (by definition) that <math>C</math> is on the perpendicular bisector of <math>\overline{AB}</math>. Also, given <math>C</math> on the perpendicular bisector of <math>\overline{AB}</math>, we know that <math>AC = BC</math>.<br />
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JBL