Incenter/excenter lemma

Revision as of 16:57, 9 May 2021 by Etmetalakret (talk | contribs) (Created page with "thumb|right|200px|The incenter/excenter lemma. In geometry, the '''incenter/excenter lemma''', sometimes called the '''Trillium theo...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The incenter/excenter lemma.

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 $ABC$ with incenter $I$ and $A$-excenter $I_A$, let $L$ be the midpoint of the arc $BC$ of 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 = \half A + \half B. However, \angle BIL = \angle BAI + \angle ABI = \half A + \half B. Hence, $\triangle BIL$ is isosceles. So $LB = LI$. The rest of the proof proceeds along these lines.