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

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


Then:

$r = 4 R \sin \dfrac A 2 \sin \dfrac B 2 \sin \dfrac C 2$


Proof

IncenterLengthProof.png

Let $D$, $E$ and $F$ be the points where the incircle is tangent to the sides $AC$, $AB$ and $CB$ respectively.

Let $s$ denote the semiperimeter of $\triangle ABC$.


From Tangent Points of Incircle in Terms of Semiperimeter:

\(\ds AD\) \(=\) \(\ds s - a\)
\(\ds BE\) \(=\) \(\ds s - b\)
\(\ds CF\) \(=\) \(\ds s - c\)


Then:

\(\ds r\) \(=\) \(\ds AI \sin \dfrac A 2\) Definition of Sine of Angle
\(\ds \dfrac {AI} {\sin \frac B 2}\) \(=\) \(\ds \dfrac c {\sin \angle AIB}\) Law of Sines applied to $\triangle AIB$
\(\ds \) \(=\) \(\ds \dfrac c {\map \sin {180 \degrees - \dfrac {A + B} 2} }\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds \dfrac c {\sin \dfrac {A + B} 2}\) Sine of Supplementary Angle
\(\ds \) \(=\) \(\ds \dfrac c {\map \sin {\dfrac {180 \degrees - C} 2} }\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds \dfrac c {\cos \frac C 2}\) Sine of Complement equals Cosine
\(\ds \) \(=\) \(\ds \dfrac {2 R \sin C} {\cos \frac C 2}\) Law of Sines applied to $\triangle ABC$
\(\ds \) \(=\) \(\ds \dfrac {2 R \cdot 2 \sin \frac C 2 \cos \frac C 2} {\cos \frac C 2}\) Double Angle Formula for Sine
\(\ds \) \(=\) \(\ds 4 R \sin \frac C 2\) simplifying
\(\ds \leadsto \ \ \) \(\ds AI\) \(=\) \(\ds 4 R \sin \dfrac B 2 \sin \frac C 2\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds 4 R \sin \dfrac A 2 \sin \dfrac B 2 \sin \frac C 2\)

$\blacksquare$


Sources