Exradius of Triangle in Terms of Circumradius/Proof 2

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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 $\rho_a$ 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

Area-of-Triangle-by-Exradius.png

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

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