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Difference between revisions of "Incenter/excenter lemma"
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[[Image:Incenter_excenter_lemma.png|thumb|right|200px|Diagram as shown.]]
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[[Image:Incenter_excenter_lemma.png|thumb|right|200px|Diagram of the configuration.]]
  
 
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 any <math>\triangle ABC</math> with incenter <math>I</math> and <math>A</math>-excenter <math>I_A</math>, let <math>L</math> be the midpoint of <math>\overarc{BC}</math> on the triangle's circumcenter. Then, the theorem states that <math>L</math> is the center of a circle through <math>I</math>, <math>B</math>, <math>I_A</math>, and <math>C</math>.
 
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 any <math>\triangle ABC</math> with incenter <math>I</math> and <math>A</math>-excenter <math>I_A</math>, let <math>L</math> be the midpoint of <math>\overarc{BC}</math> on the triangle's circumcenter. Then, the theorem states that <math>L</math> is the center of a circle through <math>I</math>, <math>B</math>, <math>I_A</math>, and <math>C</math>.

Revision as of 16:34, 9 May 2021

Diagram of the configuration.

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 any $\triangle ABC$ with incenter $I$ and $A$-excenter $I_A$, let $L$ be the midpoint of $\overarc{BC}$ on 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. $\square$


See also

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