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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.
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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> (proof from Evan Chen)
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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 15:31, 18 May 2021

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 any $\triangle ABC$ with incenter $I$ and $A$-excenter $I_A$, let $L$ be the midpoint of $\overarc{BC}$ on the triangle's circumcenter. Then, the theorem states that $L$ is the center of a circle through $I$, $B$, $I_A$, and $C$.

The incenter/excenter lemma makes frequent appearances in olympiad geometry. Along with the larger lemma, two smaller results follow: first, $A$, $I$, $L$, and $I_A$ are collinear, and second, $I_A$ is the reflection of $I$ across $L$. Both of these follow easily from the main proof.

Proof

Let $A = \angle BAC$, $B = \angle CBA$, $C = \angle ACB$, and note that $A$, $I$, $L$ are collinear (as $L$ is on the angle bisector). We are going to show that $LB = LI$, the other cases being similar. First, notice that \[\angle LBI = \angle LBC + \angle CBI = \angle LAC + \angle CBI = \angle IAC + \angle CBI = \frac{1}{2} A + \frac{1}{2} B.\] However, \[\angle BIL = \angle BAI + \angle ABI = \frac{1}{2} A + \frac{1}{2} B.\] Hence, $\triangle BIL$ is isosceles, so $LB = LI$. The rest of the proof proceeds along these lines. $\square$

See also

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