Difference between revisions of "Incenter/excenter lemma"
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[[Image:Incenter_excenter_lemma.png|thumb|right|200px|The incenter/excenter lemma.]] | [[Image:Incenter_excenter_lemma.png|thumb|right|200px|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 | + | 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>. |
The incenter/excenter lemma makes frequent appearances in olympiad geometry. Along with the larger lemma, two smaller results follow: first, <math>A</math>, <math>I</math>, <math>L</math>, and <math>I_A</math> are collinear, and second, <math>I_A</math> is the reflection of <math>I</math> across <math>L</math>. Both of these follow easily from the main proof. | The incenter/excenter lemma makes frequent appearances in olympiad geometry. Along with the larger lemma, two smaller results follow: first, <math>A</math>, <math>I</math>, <math>L</math>, and <math>I_A</math> are collinear, and second, <math>I_A</math> is the reflection of <math>I</math> across <math>L</math>. Both of these follow easily from the main proof. |
Revision as of 16:05, 9 May 2021
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 with incenter and -excenter , let be the midpoint of on the triangle's circumcenter. Then, the theorem states that is the center of a circle through , , , and .
The incenter/excenter lemma makes frequent appearances in olympiad geometry. Along with the larger lemma, two smaller results follow: first, , , , and are collinear, and second, is the reflection of across . Both of these follow easily from the main proof.
Proof
Let , , , and note that , , are collinear (as is on the angle bisector). We are going to show that , the other cases being similar. First, notice that However, Hence, is isosceles, so . The rest of the proof proceeds along these lines.