Let P be a point in the interior of an equilateral triangle.

Three line segments connect P with the vertices of the 
triangle and three line segments connect P perpendicularly
to the sides of the triangle.

These six line segments divide the triangle into six smaller
triangles that surround P.

If u, v, w, x, y, and z denote the areas of the triangles 
around P in that order, then prove that

                     u + w + y = v + x + z.

Using 1. the same definitions in my last post (as lengths, not squares of lengths) 2. the short proof  from the last occasion of the result that A1=(i+k+m)/2 (the longer proof is not needed) and 3. working on the rectangles rather than the triangles to get rid of the fractions, we have:

ai+bi+ck+dk+em+fm=i+k+m: which we may write as

ai+ck+em=i+k+m-bi-dk-fm, or =i(1-b)+k(1-d)+m(1-f)

but 1-b=a etc, so ai+ck+em=i-ai+k-ck+m-em, or

2ai+2ck+2em=i+k+m; similarly, 2bi+2dk+2fm=i+k+m:

so ai+ck+em=bi+dk+fm.

QED