Area of Triangle in Terms of Exradii

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Theorem

The area of a $\triangle ABC$ is given by the formula:

$(ABC) = \rho_a \left({s - a}\right) = \rho_b \left({s - b}\right) = \rho_c \left({s - c}\right) = \rho s = \sqrt {\rho_a \rho_b \rho_c \rho}$

where:

$s$ is the semiperimeter
$I$ is the incenter
$\rho$ is the inradius
$I_a, I_b, I_c$ are the excenters
$\rho_a, \rho_b, \rho_c$ are the exradii from $I_a, I_b, I_c$, respectively.


Proof

Proof of the First Part

First, we show that the area is equal to $\rho_a \left({s - a}\right) = \rho_b \left({s - b}\right) = \rho_c \left({s - c}\right)$.

We pick an excircle, WLOG $I_a$.

Area1.PNG

\(\displaystyle (ABC)\) \(=\) \(\displaystyle (ABI_a) + (ACI_a) - (CBI_a)\) (see figure above)
\(\displaystyle \) \(=\) \(\displaystyle \frac {c \rho_a} 2 + \frac {b \rho_a} 2 - \frac {a \rho_a} 2\) Area of Triangle in Terms of Side and Altitude
\(\displaystyle \) \(=\) \(\displaystyle \rho_a \frac {b + c + a} 2 - \rho_a a\)
\(\displaystyle \) \(=\) \(\displaystyle \rho_a s - \rho_a a\)
\(\displaystyle (ABC)\) \(=\) \(\displaystyle \rho_a \left({s - a}\right)\)

A similar argument can be used to show that the statement holds for the other excircles.

$\Box$


Proof of the Second Part

Second, we show the area is equal to $(ABC) = \rho s$.

We take the incircle with incenter at $I$ and inradius $\rho$:

T3.PNG

\(\displaystyle (ABC)\) \(=\) \(\displaystyle (ABI) + (BCI) + (CAI)\) (see figure above)
\(\displaystyle \) \(=\) \(\displaystyle \frac {a \rho} 2 + \frac {b \rho} 2 + \frac {c \rho} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\rho} 2 \left({a + b + c}\right)\)
\(\displaystyle (ABC)\) \(=\) \(\displaystyle \rho s\)

$\Box$


Proof of the Third Part

Finally, we show that the area is equal to $\sqrt{\rho_a \rho_b \rho_c \rho}$:

\(\displaystyle (ABC)^4\) \(=\) \(\displaystyle \rho_a \left({s - a}\right) \rho_b \left({s - b}\right) \rho_c \left({s - c}\right) \rho s\)
\(\displaystyle (ABC)^4\) \(=\) \(\displaystyle s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \rho_a \rho_b \rho_c \rho\)
\(\displaystyle (ABC)^4\) \(=\) \(\displaystyle (ABC)^2 \rho_a \rho_b \rho_c \rho\) Heron's Formula
\(\displaystyle (ABC)^2\) \(=\) \(\displaystyle \rho_a \rho_b \rho_c \rho\)
\(\displaystyle (ABC)\) \(=\) \(\displaystyle \sqrt{\rho_a \rho_b \rho_c \rho}\)

$\blacksquare$


Also see


Sources