Distance between Incenter and Excenter of Triangle in Terms of Circumradius
Theorem
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $I$ be the incenters of $\triangle ABC$.
Let $I_a$ be the excenters of $\triangle ABC$ with respect to $a$.
Let $R$ be the circumradius of $\triangle ABC$.
Then:
- $I I_a = 4 R \sin \dfrac A 2$
Proof
From Triangle is Orthic Triangle of Triangle formed from Excenters, we establish that $\triangle ABC$ is the orthic triangle of $\triangle I_a I_b I_c$.
By the Nine Point Circle Theorem, the Feuerbach circle of $\triangle I_a I_b I_c$ passes through each of $A$, $B$ and $C$.
Therefore the Feuerbach circle of $\triangle I_a I_b I_c$ is the circumcircle of $\triangle ABC$.
Hence the radius of the Feuerbach circle is $R$.
From Radius of Circumcircle is Twice Radius of Feuerbach Circle, the radius of the circumcircle of $\triangle I_a I_b I_c$ is $2 R$.
Hence:
\(\ds I I_a\) | \(=\) | \(\ds 2 \paren {2 R} \map \cos {90 \degrees - \dfrac A 2}\) | Distance from Vertex of Triangle to Orthocenter | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds 4 R \sin \frac A 2\) | Cosine of Complement equals Sine |
Work In Progress In particular: Extract a lemma from Distance between Excenters of Triangle in Terms of Circumradius that gives $\angle B I_a C = 90 \degrees - \dfrac A 2$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{WIP}} from the code. |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(58)$