Distance between Incenter and Excenter of Triangle in Terms of Circumradius/Proof

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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

Orthic-Triangle-of-Excenters.png

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

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