Difference between revisions of "Incenter/excenter lemma"
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− | [[Image:Incenter_excenter_lemma.png|thumb|right|200px| | + | [[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 | + | 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. | ||
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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. | 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. | ||
− | 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> | + | 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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== See also == | == See also == | ||
* [[Orthic triangle]] | * [[Orthic triangle]] | ||
+ | * [[Geometry/Olympiad | Olympiad geometry]] | ||
[[Category:Geometry]] | [[Category:Geometry]] | ||
− | |||
[[Category:Theorems]] | [[Category:Theorems]] |
Latest revision as of 16:31, 18 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.