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Prove that the area of a triangle $ABC$ is

$$\frac12 (b + c - a)r$$

where $r$ radius of the excircle opposite to $A$ and the rest of the symbols have their usual meaning.

I started off with the following figure:

enter image description here

$BF, CF$ are the angle bisectors of $\angle DBC, \angle BCE$. $GF$ is perpendicular to $BC$. Clearly, $BC = a, GF = r$. Therefore, the conclusion of the problem may be rewritten as follows:

$$\frac12 (b+c)r - \frac12 ar$$

Clearly, the last term is the area of $\Delta BCF$. So the problem reduces to proving that the area of $ACFB$ is the first term of the above. I further tried reducing the given quadrilateral to a triangle of equal area and then proceeding, but it yielded no results. How do I proceed?

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up vote 2 down vote accepted

$\Delta ABC = \Delta ACF + \Delta ABF - \Delta BCF$

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