Difference between revisions of "Perpendicular bisector"
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− | + | In a [[plane]], the '''perpendicular bisector''' of a [[line segment]] <math>AB</math> is a [[line]] <math>l</math> such that <math>AB</math> and <math>l</math> are [[perpendicular]] and <math>l</math> passes through the [[midpoint]] of <math>AB</math>. | |
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+ | In 3-D space, for each plane containing <math>AB</math> there is a distinct perpendicular bisector in that plane. The [[set]] of lines which are perpendicular bisectors of form a plane which is the plane perpendicularly bisecting <math>AB</math>. | ||
In a [[triangle]], the perpendicular bisectors of all three sides intersect at the [[circumcenter]]. | In a [[triangle]], the perpendicular bisectors of all three sides intersect at the [[circumcenter]]. | ||
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+ | == Locus == | ||
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+ | The perpendicular bisector of <math>AB</math> is also the locus of points [[equidistant]] from <math>A</math> and <math>B</math>. | ||
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+ | To prove this, we must prove that every point on the perpendicular bisector is equidistant from <math>A</math> and <math>B</math>, and also that every point equidistant from <math>A</math> and <math>B</math>. | ||
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+ | The first part we prove as follows: Let <math>P</math> be a point on the perpendicular bisector of <math>AB</math>, and let <math>M</math> be the midpoint of <math>AB</math>. Then we observe that the (possibly [[degenerate]]) [[triangles]] <math>APM</math> and <math>BPM</math> are [[congruent (geometry) | congruent]], by [[SAS congruence]]. Hence the segments <math>PA</math> and <math>PB</math> are congruent, meaning that <math>P</math> is equidistant from <math>A</math> and <math>B</math>. | ||
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+ | To prove the second part, we let <math>P</math> be any point equidistant from <math>A</math> and <math>B</math>, and we let <math>M</math> be the midpoint of the segment <math>AB</math>. If <math>P</math> and <math>M</math> are the same point, then we are done. If <math>P</math> and <math>M</math> are not the same point, then we observe that the triangles <math>PAM</math> and <math>PBM</math> are congruent by [[SSS congruence]], so the angles <math>PAM</math> and <math>PBM</math> are congruent. Since these angles are [[supplementary angles]], each of them must be a [[right angle]]. Hence <math>PM</math> is the perpendicular bisector of <math>AB</math>, and we are done. | ||
[[Category:Definition]] | [[Category:Definition]] |
Latest revision as of 14:20, 1 January 2014
In a plane, the perpendicular bisector of a line segment is a line such that and are perpendicular and passes through the midpoint of .
In 3-D space, for each plane containing there is a distinct perpendicular bisector in that plane. The set of lines which are perpendicular bisectors of form a plane which is the plane perpendicularly bisecting .
In a triangle, the perpendicular bisectors of all three sides intersect at the circumcenter.
Locus
The perpendicular bisector of is also the locus of points equidistant from and .
To prove this, we must prove that every point on the perpendicular bisector is equidistant from and , and also that every point equidistant from and .
The first part we prove as follows: Let be a point on the perpendicular bisector of , and let be the midpoint of . Then we observe that the (possibly degenerate) triangles and are congruent, by SAS congruence. Hence the segments and are congruent, meaning that is equidistant from and .
To prove the second part, we let be any point equidistant from and , and we let be the midpoint of the segment . If and are the same point, then we are done. If and are not the same point, then we observe that the triangles and are congruent by SSS congruence, so the angles and are congruent. Since these angles are supplementary angles, each of them must be a right angle. Hence is the perpendicular bisector of , and we are done.