Area of Triangle in Terms of Inradius and Exradii

Theorem
The area of a $\triangle ABC$ is given by the formula:
 * $(ABC) = \rho_a \left({s - a}\right) = \rho_b \left({s - b}\right) = \rho_c \left({s - c}\right) = \rho s = \sqrt {\rho_a \rho_b \rho_c \rho}$

In this formula:


 * $s$ 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 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 $I_a$.



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

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 $\rho$:



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

Comment
This formula for the area of a triangle is closely related to Heron's Formula.