Areal Mystery

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. Then

$[PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC]$ if and only if $P$ lies on

$A$ : at least one of the medians of triangle $ABC$ .

$B$ : at least one of the altitudes of triangle $ABC$ .

$C$ : at least one of the angle bisector of triangle $ABC$ .

$D$ : at least one of the perpendicular side bisector of triangle $ABC$ .

(Here $[XYZ]$ denotes the area of triangle $XYZ$ .)