Exradius of Triangle in Terms of Circumradius

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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 1

Area-of-Triangle-by-Exradius.png

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

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




$\blacksquare$


Sources