Question 17
Let $P$ be a point in the interior of the triangle $ABC$. The lines $AP$, $BP$, $CP$ intersect $BC$, $AC$, $AB$ at $A_1$, $B_1$, $C_1$, respectively. Given that
$$
s(PBA_1) + s(PCB_1) + s(PAC_1) = \frac{1}{2}s(ABC),
$$
(where $s(XYZ)$ denotes the area of $XYZ$) prove that $P$ lies on one of the medians of triangle $ABC$.
$$
s(PBA_1) + s(PCB_1) + s(PAC_1) = \frac{1}{2}s(ABC),
$$
(where $s(XYZ)$ denotes the area of $XYZ$) prove that $P$ lies on one of the medians of triangle $ABC$.