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:

$BF, CF$ 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?
$\Delta ABC = \Delta ACF + \Delta ABF - \Delta BCF$