ABC is a triangle with sides of length BC = a, AC = b, and AB = c. The semiperimeter is
The circle with center O is the inscribed circle of the triangle with points of tangency at D, E, and F.
Point H is constructed on the extension of BC such that BH = AF. Therefore
CH = s.
A perpendicular to CB is constructed at B and a perpendicular to OC is
constructed at O. L is the intersection of these two perpendiculars and
K is the point of intersection of OL with CB.
Click here for a GSP sketch of the figure
which you can manipulate.
The proof can be constructed by considering similar figures (e.g., triangles
AOF and CLB), the area as the sum of the areas of the triangles BOC, AOC,
and AOB, and appropriate substitutions.