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.

Triangle Problem

Not a solution, but I hope these may help:

Denote the vertices of an equilateral triangle ABC. The perpendiculars then meet AB at D, BC at E, AC at F.
Let the sides of ABC be of unit length 1.

1. By 1/2bh,the big area A1 is obviously half the unit length times the 3 perpendicular heights PD,PE,PF. Call length PD, i. Call length PE, k. Call length PF, m. A1 = ½(i+k+m)

2. Define as follows:

AD=a
BD=b
BE=c
CE=d
CF=e
AF=f
AP=g
PD=i
BP=j
PE= k
CP=l
PF=m

3. For clarity, let  each of  these expressions stand for its own square: a=a^2 etc.

4. Then we have:
a+i=g b+i=j i=g-a i=j-b
c+k=j d+k=l k=j-c k=l-d
e+m=l f+m=g m=l-e m=g-f

g-a=j-b g=j+a-b  d-c-e=a-b-f
j-c=l-d l=j+d-c  d+b+f=a+c+e
l-e=g-f j+d-c-e=j+a-b-f  
    
Reconverting to line segments: EC^2+BD^2+AF^2 =AD^2+BE^2+CF^2
With BE+EC=AD+BD=AF+CF=1, as stipulated above.
(EC+BE)(EC-BE)+(BD+AD)(BD-AD)=(CF+AF)(CF-AF)
substituting: CE-BE+BD-AD=CF-AF
so that CE+BD+AF=BE+AD+CF =3/2.

Hence the alternate pairs of line segments sum to the same length.