Area of Triangle in Terms of Inradius and Exradii

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

In this formula: $$s$$ is the semiperimeter

$$I$$ is the incenter

$$\rho$$ is the inradius

$$I_a$$, $$I_b$$, and $$I_c$$ are the excenters.

$$\rho_a$$, $$\rho_b$$, and $$\rho_c$$ are the exradii from $$I_a$$,$$I_b$$, $$I_c$$, respectively.

Proof
we start with the formula with the exinradius

$$(ABC) = \rho_a(s-a)=\rho_b(s-b)=\rho_c(s-c) \,\!$$

We take the exinradius from $$I_a$$



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

Now With the formula $$(ABC)=\rho s$$

We take inradius $$I$$



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And now for the formula $$\sqrt{\rho_a\rho_b\rho_c\rho}$$

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