Inradius of Triangle in Terms of Exradii
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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.
Then we have the following relation:
- $\dfrac 1 r = \dfrac 1 {\rho_a} + \dfrac 1 {\rho_b} + \dfrac 1 {\rho_c}$
where:
- $r$ is the inradius
- $\rho_a$, $\rho_b$ and $\rho_c$ are the exradii of $\triangle ABC$ with respect to $a$, $b$ and $c$ respectively.
Proof
The area $\AA$ of $\triangle ABC$ is given by:
\(\ds \AA\) | \(=\) | \(\ds \rho_a \paren {s - a}\) | Area of Triangle in Terms of Exradius | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 {\rho_a}\) | \(=\) | \(\ds \frac {s - a} \AA\) | rearranging |
Similarly:
\(\ds \quad \quad \frac 1 {\rho_b}\) | \(=\) | \(\ds \frac {s - b} \AA\) | ||||||||||||
\(\ds \frac 1 {\rho_c}\) | \(=\) | \(\ds \frac {s - c} \AA\) |
Hence:
\(\ds \) | \(\) | \(\ds \frac 1 {\rho_a} + \frac 1 {\rho_b} + \frac 1 {\rho_c}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {s - a} \AA + \frac {s - b} \AA + \frac {s - c} \AA\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {3 s - a - b - c} \AA\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {3 s - 2 s} \AA\) | Definition of Semiperimeter | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac s {r s}\) | Area of Triangle in Terms of Inradius | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 r\) |
$\blacksquare$