Area of Triangle in Terms of Inradius and Exradii

Theorem
The area of a $$\triangle ABC$$ 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 of the First Part
First, we show that the area is equal to $$\rho_a(s-a)=\rho_b(s-b)=\rho_c(s-c) \,\!$$

We pick an excircle, WLOG $$I_a$$.



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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 $$\rho$$



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

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