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Difference between revisions of "Perpendicular bisector"

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m (Congruence Theorems)
 
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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>AB</math> is also the locus of points [[equidistant]] from <math>A</math> and <math>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>A</math> and <math>B</math>, and also that every point equidistant from <math>A</math> and <math>B</math>.
  
−
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>.
+
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>\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.
+
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 $AB$ 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 $AB$ is also the locus of points equidistant from $A$ and $B$.

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

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

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

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