Difference between revisions of "Incenter/excenter lemma"
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== See also == | == See also == | ||
* [[Orthic triangle]] | * [[Orthic triangle]] | ||
| − | * [[Geometry/Olympiad | Olympiad | + | * [[Geometry/Olympiad | Olympiad geometry]] |
[[Category:Geometry]] | [[Category:Geometry]] | ||
[[Category:Theorems]] | [[Category:Theorems]] | ||
Revision as of 15:29, 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 ![\[\angle LBI = \angle LBC + \angle CBI = \angle LAC + \angle CBI = \angle IAC + \angle CBI = \frac{1}{2} A + \frac{1}{2} B.\]](http://latex.artofproblemsolving.com/e/f/0/ef044ac8f791a66dc6a123cf797af6be15dbb36f.png)
However, ![\[\angle BIL = \angle BAI + \angle ABI = \frac{1}{2} A + \frac{1}{2} B.\]](http://latex.artofproblemsolving.com/6/f/d/6fd54602cd51ee1807548259c2479e05d52fd7b4.png)
Hence, 
is isosceles, so . The rest of the proof proceeds along these lines. 
