LAST EDITED ON Jul-17-03 AT 07:43 PM (EST)
 
I am having trouble reading your construction, so I checked only the part when the area bisector point P is on the triangle. That's just me, but you did not offer any proof of your construction, a serius defficiency, otherwise I like it. So let's prove the first part of the construction now:

Construct the triangle area bisector through a given point P on the triangle circumference: Connect the point P with the opposite vertex (B in our case). Draw a line parallel to this line (BP) through the median heel (Mb) on the same triangle side as the point P. The line intersects the triangle at another point S. PS is the triangle area bisector. See the attached drawing.

http://www.cut-the-knot.org/htdocs/dcforum/User_files/3f0e673c6fff5696.gif

Proof:

Denote x = SB/AB, y = PC/AC. Since the lines BP and SMb are parallel, DABP and DASMb are similar.
Consequently,

(1 - y)/(1/2) = 1/(1 - x) or y = 1 - 1/{2(1 - x)}

Consider yellow DASP and magenta DSBP. Their altitudes to the line AS or SB are the same. Therefore, their areas A(Y) and A(M) (for yellow and magenta) are in the ratio of their bases AS and SB:

A(Y)/A(M) = (1 - x)/x

Consider yellow & magenta DPAB and cyan DCPB. Again, their altitudes to the line PA or CP are the same. Therefore, their areas A(Y) + A(M) and A(C) (for yellow + magenta and cyan) are in the ratio of their bases PA and CP:

(A(Y) + A(M))/A(C) = (1 - y)/y

Let's calculate the area A(M) + A(C):

A(M) + A(C) = A(Y).x/(1 - x) + (A(Y) + A(M)).y/(1 - y) =

= A(Y).x/(1 - x) + {A(Y) + A(Y).x/(1 - x)}.y/(1 - y) =

= A(Y).(x + y - xy)/(1 - x)/(1 - y)

Substituting y = 1 - 1/{2(1-x)} we get A(M) + A(C) = A(Y) and the line PS is indeed the triangle area bisector.

Q.E.D.