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 \(XYZ\).)**

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