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>. | 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>. | ||
− | In 3-D space, for each plane passing through <math>AB</math> there is a distinct perpendicular bisector. The [[set]] of lines which are perpendicular bisectors of form a plane which is the plane perpendicularly bisecting <math>AB</math>. | + | In 3-D space, for each plane passing through <math>AB</math> that is not [[normal]] to <math>AB</math> there is a distinct perpendicular bisector. 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>\displaystyle AB</math> is also the locus of points [[equidistant]] from <math>\displaystyle A</math> and <math>\displaystyle B</math>. | ||
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+ | To prove this, we must prove that every point on the perpendicular bisector is equidistant from <math>\displaystyle A</math> and <math>\displaystyle B</math>, and also that every point equidistant from <math>\displaystyle A</math> and <math>\displaystyle B</math>. | ||
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+ | The first part we prove as follows: Let <math>\displaystyle P</math> be a point on the perpendicular bisector of <math>\displaystyle AB</math>, and let <math>\displaystyle M</math> be the midpoint of <math>\displaystyle AB</math>. Then we observe that the (possibly [[degenerate]]) [[triangles]] <math>\displaystyle APM</math> and <math>\displaystyle BPM</math> are [[congruent (geometry) | congruent]], by [[side-angle-side congruence]]. Hence the segments <math>\displaystyle PA</math> and <math>\displaystyle PB</math> are congruent, meaning that <math>\displaystyle P</math> is equidistant from <math>\displaystyle A</math> and <math>\displaystyle B</math>. | ||
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+ | To prove the second part, we let <math>\displaystyle P</math> be any point equidistant from <math>\displaystyle A</math> and <math>\displaystyle B</math>, and we let <math>\displaystyle M</math> be the midpoint of the segment <math>\displaystyle AB</math>. If <math>\displaystyle P</math> and <math>\displaystyle M</math> are the same point, then we are done. If <math>\displaystyle P</math> and <math>\displaystyle M</math> are not the same point, then we observe that the triangles <math>\displaystyle PAM</math> and <math>\displaystyle PBM</math> are congruent by [[side-side-side congruence]], so the angles <math>\displaystyle PAM</math> and <math>\displaystyle PBM</math> are congruent. Since these angles are [[supplementary angles]], each of them must be a [[right angle]]. Hence <math>\displaystyle PM</math> is the perpendicular bisector of <math>\displaystyle AB</math>, and we are done. | ||
[[Category:Definition]] | [[Category:Definition]] |
Revision as of 17:13, 25 August 2006
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 passing through that is not normal to there is a distinct perpendicular bisector. 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 side-angle-side 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 side-side-side 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.