Distance between Excenters 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 $I_b$ and $I_c$ be the excenters of $\triangle ABC$ with respect to $b$ and $c$ respectively.

Let $R$ be the circumradius of $\triangle ABC$.


Then:

$I_b I_c = 4 R \cos \dfrac A 2$


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 $\triangle I_a I_b I_c$.

Hence $I_b B$ is an altitude of $\triangle I_a I_b I_c$.

Thus $\angle I_b B I_a$ is a right angle.

From Altitudes of Triangle Bisect Angles of Orthic Triangle:

$\angle CBI = \dfrac B 2$

So:

$\angle I_a B C = 90 \degrees - \dfrac B 2$


By a similar argument, mutatis mutandis:

$\angle I_a C B = 90 \degrees - \dfrac C 2$


Hence:

\(\ds \angle B I_a C\) \(=\) \(\ds 180 \degrees - \paren {\angle I_a B C + \angle I_a C B}\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds 180 \degrees - \paren {\paren {90 \degrees - \dfrac B 2} + \paren {90 \degrees - \dfrac C 2} }\)
\(\ds \) \(=\) \(\ds \dfrac {B + C} 2\)
\(\ds \) \(=\) \(\ds \dfrac {180 \degrees - A} 2\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds 90 \degrees - \dfrac A 2\)
\(\ds \leadsto \ \ \) \(\ds BC\) \(=\) \(\ds I_b I_c \map \cos {90 \degrees - \dfrac A 2}\) Sides of Orthic Triangle of Acute Triangle
\(\ds \leadsto \ \ \) \(\ds I_b I_c\) \(=\) \(\ds \dfrac a {\sin \frac A 2}\) Cosine of Complement equals Sine
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {2 R \sin A} {\sin \frac A 2}\) Law of Sines
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {2 R \cdot 2 \sin \frac A 2 \cos \frac A 2} {\sin \frac A 2}\) Double Angle Formula for Sine
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds 4 R \cos \frac A 2\) Double Angle Formula for Sine

$\blacksquare$


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