Inradius of Triangle in Terms of Exradii

This article has been identified as a candidate for Featured Proof status. If you do not believe that this proof is worthy of being a Featured Proof, please state your reasons on the talk page. To discuss this page in more detail, feel free to use the talk page.

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

\(\ds \)

\(=\)

\(\ds \frac 1 r\)

$\blacksquare$