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Difference between revisions of "Incenter/excenter lemma"

(Created page with "thumb|right|200px|The incenter/excenter lemma. In geometry, the '''incenter/excenter lemma''', sometimes called the '''Trillium theo...")
 
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== Proof ==
 
== Proof ==
  
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Let <math>A = \angle BAC</math>, <math>B = \angle CBA</math>, <math>C = \angle ACB</math>,
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Let <math>A = \angle BAC</math>, <math>B = \angle CBA</math>, <math>C = \angle ACB</math>, and note that <math>A</math>, <math>I</math>, <math>L</math> are collinear (as <math>L</math> is on the angle bisector). We are going to show that <math>LB = LI</math>, the other cases being similar.
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and note that <math>A</math>, <math>I</math>, <math>L</math> are collinear (as <math>L</math> is on the angle bisector).
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First, notice that <cmath>\angle LBI = \angle LBC + \angle CBI = \angle LAC + \angle CBI = \angle IAC + \angle CBI = \frac{1}{2} A + \frac{1}{2} B.</cmath> However, <cmath>\angle BIL = \angle BAI + \angle ABI = \frac{1}{2} A + \frac{1}{2} B</cmath>. Hence, <math>\triangle BIL</math> is isosceles, so <math>LB = LI</math>. The rest of the proof proceeds along these lines. <math>\square</math>
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We are going to show that <math>LB = LI</math>, the other cases being similar.
 
−
 
 
−
First, notice that
 
−
\angle LBI = \angle LBC + \angle CBI
 
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= \angle LAC + \angle CBI
 
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= \angle IAC + \angle CBI
 
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= \half A + \half B.
 
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However,
 
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\angle BIL = \angle BAI + \angle ABI
 
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= \half A + \half B.  
 
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Hence, <math>\triangle BIL</math> is isosceles.
 
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So <math>LB = LI</math>.
 
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The rest of the proof proceeds along these lines.
 

Revision as of 16:00, 9 May 2021

The incenter/excenter lemma.

In geometry, the incenter/excenter lemma, sometimes called the Trillium theorem, is a result concerning a relationship between the incenter and excenter of a triangle. Given a triangle $ABC$ with incenter $I$ and $A$-excenter $I_A$, let $L$ be the midpoint of the arc $BC$ of the triangle's circumcenter. Then, the theorem states that $L$ is the center of a circle through $I$, $B$, $I_A$, and $C$.

The incenter/excenter lemma makes frequent appearances in olympiad geometry. Along with the larger lemma, two smaller results follow: first, $A$, $I$, $L$, and $I_A$ are collinear, and second, $I_A$ is the reflection of $I$ across $L$. Both of these follow easily from the main proof.

Proof

Let $A = \angle BAC$, $B = \angle CBA$, $C = \angle ACB$, and note that $A$, $I$, $L$ are collinear (as $L$ is on the angle bisector). We are going to show that $LB = LI$, the other cases being similar. First, notice that \[\angle LBI = \angle LBC + \angle CBI = \angle LAC + \angle CBI = \angle IAC + \angle CBI = \frac{1}{2} A + \frac{1}{2} B.\] However, \[\angle BIL = \angle BAI + \angle ABI = \frac{1}{2} A + \frac{1}{2} B\]. Hence, $\triangle BIL$ is isosceles, so $LB = LI$. The rest of the proof proceeds along these lines. $\square$

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