US-TST 2003 Problem 2

Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$ , $PB$ , $PC$ intersect sides $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ , respectively. Prove that $$[PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC]$$ if and only if $P$ lies on at least one of the medians of triangle $ABC$ . (Here $[XYZ]$ denotes the area of triangle $XYZ$ .)