Distance between Excenters of Triangle in Terms of Circumradius/Proof

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

Hence: