# Areal Mystery

Geometry Level 3

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 $$XYZ$$.)

×