Area sum of inner triangles is half of the equilateral triangle Given an equilateral triangle $\triangle ABC$ and let $P$ be an arbitrary point inside this triangle. Moreover let points $V,W,T$ be the orthogonal projections of the point $P$ on to the sides $\overline{AB}, \overline{BC}$, and $\overline{CA}$ respectively. Prove that $$S(AVP) +S(BWP) +S(CTP) =\frac{1}{2} S(ABC), $$ where $S(XYZ)$ denotes the surface area of $\triangle XYZ$.

 A: Here is a very short solution using a powerful technique that can effectively dismantle many problems in euclidean geometry if used together with synthetic geometry. Basically it is using polynomial interpolation to prove geometric facts including both identities and properties like collinearity.
If $P$ is on a line of symmetry of $\triangle ABC$:
  The theorem is true by symmetry
If $P$ is not on a line of symmetry of $\triangle ABC$:
  Let $Q(r)$ be on $AP$ such that $\overline{AQ} = r$ (using directed lengths) for any $r \in \mathbb{R}$
  Then $Area(\triangle QAV)$,$Area(\triangle QBW)$,$Area(\triangle QCT)$ are all quadratic in $r$
  Thus $T(r) = Area(\triangle QAV)+Area(\triangle QBW)+Area(\triangle QCT)$ is a quadratic in $r$
  Also $AP$ intersects the 3 lines of symmetry of $\triangle ABC$ at distinct points
  If $Q(r)$ is at any of these 3 points:
    The theorem is true for $Q(r)$
    Thus $T(r) = \frac{1}{2} Area(\triangle ABC)$
  Therefore $T(r) = \frac{1}{2} Area(\triangle ABC)$ for any $r \in \mathbb{R}$
  Thus the theorem is true for $P$
Note
I've seen this problem a long time ago but don't know how to find it. Any references to its source are welcome!
A: Here is a purely synthetic proof:
Draw lines through $P$ parallel to the 3 sides of $\triangle ABC$. Then there are 6 pairs of triangles of equal area, 3 of them by reflection across the perpendiculars to the sides, and the other 3 of them by rotation within parallelograms. The theorem follows.
