Incenter/excenter lemma

Diagram as shown.

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. $\square$

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Incenter/excenter lemma

Proof

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