Question 139
The point $P$ is in the interior of the triangle $ABC$, and the lines $AP$, $BP$, $CP$ meet the sides $BC$, $AC$, $AB$ at $A_1$, $B_1$, $C_1$, respectively. It is known that
$$
s\left(PBA_1\right) + s\left(PCB_1\right) + s\left(PAC_1\right) = \frac{1}{2} s(ABC)
$$
where $s(XYZ)$ denotes the area of the triangle $XYZ$. Show that $P$ lies on a median of the triangle $ABC$.
$$
s\left(PBA_1\right) + s\left(PCB_1\right) + s\left(PAC_1\right) = \frac{1}{2} s(ABC)
$$
where $s(XYZ)$ denotes the area of the triangle $XYZ$. Show that $P$ lies on a median of the triangle $ABC$.