Difference between revisions of "Incenter/excenter lemma"
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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> Hence, <math>\triangle BIL</math> is isosceles, so <math>LB = LI</math>. The rest of the proof proceeds along these lines. <math>\square</math> | 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 == |
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.