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 | + | 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 == | ||
− | 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>. | + | 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 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 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.