Hi Bractals

You are correct - we do not lose any generality by looking only at an equilateral triangle, or an isosceles right angle triangle which I used in my analysis (because it simplifies the analysis, for one family of bisectors at least, even more than the equilateral triangle). You can imagine a projection in 3D space, but you have to be careful what kind of projection you use - it must be the parallel projection. Other projections of a plane into another plane (perspective projection) do not have to preserve area ratios.

But I do not think you have to go into the 3D space. You can just perform a linear transformation using any 2x2 matrix with non-zero determinant and apply it to every 2D vector in a plane - namely the 3 triangle vertices and the point through which the area bisector should pass:

x' = a11x + a12y
y' = a21x + a22y

(I cannot post a matrix equation)

D = a11a22 - a12a21 ¹ 0

Such transformation preserves the area ratios. If two shapes are the same before such transformation, they are also the same after such transformation. You can divide the area of any shape before the transformation into extremaly little squares, the area ratio of two different shapes being the ratio of the number of their little squares. Each little square is transformed into the same little parallelogram, but of course the number of little parallelograms in each transformed shape equals to the number of little squares in each shape before the transformation.

This transformation can be performed using the straightedge and compass. For the triangle itself, it is trivial, because you choose the transformation to get the desired triangle shape and position. So you just draw the desired triangle. You only have to use the straightedge and compass construction to transform the point through which the area bisector must pass at the start and transform back the point at which the area bisector intersects the triangle (or another arbitrary point of the bisector) at the end. I tried that and it only complicated the construction. However, a suitable transformation makes the analysis much simpler.

Regards, Vladimir