Exradius 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 $\rho_a$ be the exradius of $\triangle ABC$ with respect to $a$.

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

Then:

$\rho_a = 4 R \sin \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2$

Proof 1

By construction:

$\ds AF$

$=$

$\ds AD$

$\ds BF$

$=$

$\ds BE$

$\ds CD$

$=$

$\ds CE$

$\ds \leadsto \ \ $

$\ds AB + BE$

$=$

$\ds AC + CE$

$\ds $

$=$

$\ds s$

where $s$ is the semiperimeter

$\ds \leadsto \ \ $

$\ds BE$

$=$

$\ds s - c$

$\ds EC$

$=$

$\ds s - b$

We have:

$\text {(1)}: \quad$

$\ds \rho_a$

$=$

$\ds A I_a \sin \frac A 2$

Definition of Sine of Angle using $\triangle A I_a F$

Then:

$\ds \dfrac {A I_a} {\sin \angle A B I_a}$

$=$

$\ds \dfrac c {\sin \angle A I_a B}$

Law of Sines applied to $\triangle A B I_a$

$\ds \leadsto \ \ $

$\ds \dfrac {A I_a} {\map \sin {90 \degrees + \frac B 2} }$

$=$

$\ds \dfrac c {\map \sin {90 \degrees - \frac B 2 - \frac A 2} }$

$\ds \leadsto \ \ $

$\ds \dfrac {A I_a} {\cos \frac B 2}$

$=$

$\ds \dfrac c {\sin \frac C 2}$

$\ds $

$=$

$\ds \dfrac {2 R \sin C} {\sin \frac C 2}$

$\ds $

$=$

$\ds \dfrac {2 R \cdot 2 \sin \frac C 2 \cos \frac C 2} {\sin \frac C 2}$

Double Angle Formula for Sine

$\ds $

$=$

$\ds 4 R \cos \frac C 2$

simplifying

$\ds \leadsto \ \ $

$\ds A I_a$

$=$

$\ds 4 R \cos \frac B 2 \cos \frac C 2$

$\ds \leadsto \ \ $

$\ds \rho_a$

$=$

$\ds 4 R \sin \frac A 2 \cos \frac B 2 \cos \frac C 2$

from $(1)$

$\blacksquare$

Proof 2

Let $r$ denote the inradius of $\triangle ABC$.

We have:

$\ds r$

$=$

$\ds 4 R \sin \dfrac A 2 \sin \dfrac B 2 \sin \dfrac C 2$

Inradius in Terms of Circumradius

$\ds \leadsto \ \ $

$\ds \rho_a$

$=$

$\ds 4 R \sin \dfrac A 2 \map \sin {\dfrac {180 \degrees - B} 2} \map \sin {\dfrac {180 \degrees - C} 2}$

$\ds $

$=$

$\ds 4 R \sin \dfrac A 2 \map \sin {90 \degrees - \dfrac B 2} \map \sin {90 \degrees - \dfrac C 2}$

$\ds $

$=$

$\ds 4 R \sin \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2$

Sine of Complement equals Cosine

This article, or a section of it, needs explaining.
In particular: Why the second line? From the book it says "It should be noted that this formula can be deduced from Inradius in Terms of Circumradius by writing $\paren {180 \degrees - B}$ for $B$ and $\paren {180 \degrees - C}$ for $C$" but plenty of work is needed to show why this is valid.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
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$\blacksquare$

Sources

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(56)$