Area of Triangle in Terms of Inradius and Exradii

Theorem
Let $\triangle ABC$ be a triangle whose sides are of lengths $a, b, c$.

Then the area $\AA$ of $\triangle ABC$ is given by:
 * $\AA = \rho_a \paren {s - a} = \rho_b \paren {s - b} = \rho_c \paren {s - c} = \sqrt {\rho_a \rho_b \rho_c \rho}$

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 of the First Part
First, we show that the area is equal to $\rho_a \paren {s - a} = \rho_b \paren {s - b} = \rho_c \paren {s - c}$.

, we pick an excircle $I_a$.



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

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

Also see

 * Heron's Formula, to which this is closely related.