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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>.
  
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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>.
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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 ==
 
== 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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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>.
  
 
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>.
 
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>.

Revision as of 11:13, 7 September 2006

In a plane, the perpendicular bisector of a line segment $AB$ is a line $l$ such that $AB$ and $l$ are perpendicular and $l$ passes through the midpoint of $AB$.

In 3-D space, for each plane containing $AB$ 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 $AB$.

In a triangle, the perpendicular bisectors of all three sides intersect at the circumcenter.

Locus

The perpendicular bisector of $\displaystyle AB$ is also the locus of points equidistant from $\displaystyle A$ and $\displaystyle B$.

To prove this, we must prove that every point on the perpendicular bisector is equidistant from $\displaystyle A$ and $\displaystyle B$, and also that every point equidistant from $\displaystyle A$ and $\displaystyle B$.

The first part we prove as follows: Let $\displaystyle P$ be a point on the perpendicular bisector of $\displaystyle AB$, and let $\displaystyle M$ be the midpoint of $\displaystyle AB$. Then we observe that the (possibly degenerate) triangles $\displaystyle APM$ and $\displaystyle BPM$ are congruent, by side-angle-side congruence. Hence the segments $\displaystyle PA$ and $\displaystyle PB$ are congruent, meaning that $\displaystyle P$ is equidistant from $\displaystyle A$ and $\displaystyle B$.

To prove the second part, we let $\displaystyle P$ be any point equidistant from $\displaystyle A$ and $\displaystyle B$, and we let $\displaystyle M$ be the midpoint of the segment $\displaystyle AB$. If $\displaystyle P$ and $\displaystyle M$ are the same point, then we are done. If $\displaystyle P$ and $\displaystyle M$ are not the same point, then we observe that the triangles $\displaystyle PAM$ and $\displaystyle PBM$ are congruent by side-side-side congruence, so the angles $\displaystyle PAM$ and $\displaystyle PBM$ are congruent. Since these angles are supplementary angles, each of them must be a right angle. Hence $\displaystyle PM$ is the perpendicular bisector of $\displaystyle AB$, and we are done.

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