Areal Mystery

Geometry Level 3

Let A B C ABC be a triangle and let P P be a point in its interior. Lines P A PA , P B PB , P C PC intersect sides B C BC , C A CA , A B AB at D D , E E , F F , respectively. Then

[ P A F ] + [ P B D ] + [ P C E ] = 1 2 [ A B C ] [PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC] if and only if P P lies on

A A : at least one of the medians of triangle A B C ABC .

B B : at least one of the altitudes of triangle A B C ABC .

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

D D : at least one of the perpendicular side bisector of triangle A B C ABC .

(Here [ X Y Z ] [XYZ] denotes the area of triangle X Y Z XYZ .)

A D B C

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