Difference between revisions of "Incenter/excenter lemma"
Etmetalakret (talk | contribs) (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 == | ||
− | Let <math>A = \angle BAC</math>, <math>B = \angle CBA</math>, <math>C = \angle ACB</math>, | + | 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>. 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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Revision as of 17:00, 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 a triangle with incenter
and
-excenter
, let
be the midpoint of the arc
of 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.