Jim Wilson
Heron's formula is a geometric idea and Heron's development of it would have used geometric arguments. One such geometric approach is outlined here. It is the approach usually found in history of mathematics references.
Basically, the idea to to build a demonstration of the area using the perimeter of the triangle and the length r of the radius of the incircle.
Consider the figure at the right.
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 (incircle) of the triangle with points of tangency at D, E, and F. Triangles AOB, BOC, COA all have an altitude equal to the radius of the incircle (r = OD = OE = OF) so
Point H is constructed on the extension of BC such that BH = AF. Therefore CH = s. The area of the triangle is Area = (OD)(CH) = rs.
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.
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. A key observation, or intermediate step, is that COBL is a cyclyic quadrilateral and so opposite angles are supplementary.This is a rather convoluted and involved approach and in the interest of time, I will not completed it here. We have a geometric approach that is much simpler.
