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Difference between revisions of "Incenter"
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{{stub}}
 
{{stub}}
  
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[[Image:Incenter.PNG|left|thumb|300px|Triangle ''ABC'' with incenter ''I'', with [[angle bisector]]s (red), [[incircle]] (blue), and [[incircle]] (green)]]
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[[Image:Incenter.PNG|left|thumb|300px|Triangle ''ABC'' with [[incenter]] ''I'', with [[angle bisector]]s (red), [[incircle]] (blue), and [[inradius|inradii]] (green)]]  
  
 
The '''incenter''' of a [[triangle]] is the intersection of its (interior) [[angle bisector]]s.  The incenter is the center of the [[incircle]].  Every [[nondegenerate]] triangle has a unique incenter.
 
The '''incenter''' of a [[triangle]] is the intersection of its (interior) [[angle bisector]]s.  The incenter is the center of the [[incircle]].  Every [[nondegenerate]] triangle has a unique incenter.
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== Proof of Existence ==
 
== Proof of Existence ==
  
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Consider a triangle <math>\displaystyle ABC</math>.  Let <math>\displaystyle I</math> be the intersection of the respective interior angle bisectors of the [[angle]]s <math>\displaystyle BAC</math> and <math>\displaystyle CBA</math>.  We observe that since <math>\displaystyle I</math> lies on an angle bisector of <math>\displaystyle BAC</math>, is equidistant from <math>\displaystyle AB</math> and <math>\displaystyle CA</math>; likewise, it is equidistant from <math>\displaystyle BC</math> and <math>\displaystyle AB</math>; hence it is equidistant from <math>\displaystyle BC</math> and <math>\displaystyle BC</math> and <math>\displaystyle CA</math> and therefore lies on an angle bisector of <math>\displaystyle ACB</math>.  Since it lies within the triangle <math>\displaystyle ABC</math>, this is the interior angle bisector of <math>\displaystyle ACB</math>.  Since <math>\displaystyle I</math> is equidistant from all three sides of the triangle, it is the incenter.  
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Consider a triangle <math>ABC</math>.  Let <math>I</math> be the intersection of the respective interior angle bisectors of the [[angle]]s <math>BAC</math> and <math>CBA</math>.  We observe that since <math>I</math> lies on an angle bisector of <math>BAC</math>, is equidistant from <math>AB</math> and <math>CA</math>; likewise, it is equidistant from <math>BC</math> and <math>AB</math>; hence it is equidistant from <math>BC</math> and <math>BC</math> and <math>CA</math> and therefore lies on an angle bisector of <math>ACB</math>.  Since it lies within the triangle <math>ABC</math>, this is the interior angle bisector of <math>ACB</math>.  Since <math>I</math> is equidistant from all three sides of the triangle, it is the incenter.  
  
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It should be noted that this proof parallels that for the existance of the [[circumcenter]].
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It should be noted that this proof parallels that for the existence of the [[circumcenter]].
  
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The proofs of existance for the [[excenter]]s is the same, except that certain angle bisectors are exterior.
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The proofs of existence for the [[excenter]]s is the same, except that certain angle bisectors are exterior.
  
 
== Properties of the Incenter ==
 
== Properties of the Incenter ==
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<math>\bullet</math> The unnormalised [[areal coordinates]] of the incenter are <math>(a,b,c)</math>
 
<math>\bullet</math> The unnormalised [[areal coordinates]] of the incenter are <math>(a,b,c)</math>
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<math>\bullet</math> Let <math>D</math> be a point on the [[circumcircle]] of <math>\triangle ABC</math> such that <math>AD</math> bisects <math>\angle BAC</math>. Then points <math>B</math>, <math>C</math>, and <math>I</math> lie on a circle centered at <math>D</math>.
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<center><asy>defaultpen(fontsize(8));
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pair A=(7,10), B=(0,0), C=(10,0), D, I;
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I=incenter(A,B,C);
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draw(A--B--C--A);draw(circumcircle(A,B,C));draw(incircle(A,B,C));
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D=intersectionpoint(A+0.1*expi((angle(B-A)+angle(C-A))/2)--A+20*expi((angle(B-A)+angle(C-A))/2), circumcircle(A,B,C));
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draw(A--D);draw(circumcircle(B,C,I));
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dot(A^^B^^C^^D^^I);label("A",A,(0,1));label("B",B,(-1,0));label("C",C,(1,0));
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label("D",D,(0,-1));label("I",I,(-1,1));</asy></center>
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[[Category:Geometry]]

Latest revision as of 18:38, 24 November 2011

This article is a stub. Help us out by expanding it.

Triangle ABC with incenter I, with angle bisectors (red), incircle (blue), and inradii (green)

The incenter of a triangle is the intersection of its (interior) angle bisectors. The incenter is the center of the incircle. Every nondegenerate triangle has a unique incenter.

Proof of Existence

Consider a triangle $ABC$. Let $I$ be the intersection of the respective interior angle bisectors of the angles $BAC$ and $CBA$. We observe that since $I$ lies on an angle bisector of $BAC$, is equidistant from $AB$ and $CA$; likewise, it is equidistant from $BC$ and $AB$; hence it is equidistant from $BC$ and $BC$ and $CA$ and therefore lies on an angle bisector of $ACB$. Since it lies within the triangle $ABC$, this is the interior angle bisector of $ACB$. Since $I$ is equidistant from all three sides of the triangle, it is the incenter.

It should be noted that this proof parallels that for the existence of the circumcenter.

The proofs of existence for the excenters is the same, except that certain angle bisectors are exterior.

Properties of the Incenter

$\bullet$ The incenter of any triangle lies within the orthocentroidal circle.

$\bullet$ The unnormalised areal coordinates of the incenter are $(a,b,c)$

$\bullet$ Let $D$ be a point on the circumcircle of $\triangle ABC$ such that $AD$ bisects $\angle BAC$. Then points $B$, $C$, and $I$ lie on a circle centered at $D$.

[asy]defaultpen(fontsize(8)); pair A=(7,10), B=(0,0), C=(10,0), D, I; I=incenter(A,B,C); draw(A--B--C--A);draw(circumcircle(A,B,C));draw(incircle(A,B,C)); D=intersectionpoint(A+0.1*expi((angle(B-A)+angle(C-A))/2)--A+20*expi((angle(B-A)+angle(C-A))/2), circumcircle(A,B,C)); draw(A--D);draw(circumcircle(B,C,I)); dot(A^^B^^C^^D^^I);label("A",A,(0,1));label("B",B,(-1,0));label("C",C,(1,0)); label("D",D,(0,-1));label("I",I,(-1,1));[/asy]
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