The result is still true for any P-cevian triangle $\triangle A_0B_0C_0.$ Let $A_0A_1$ cut $B_0C_0$ at $U.$ From the harmonic pencil $B_0(A_0,C_0,P,A),$ we deduce that $A_1$ is midpoint of $\overline{UA_0}.$ Similarly, if $A_0A_2$ cuts $B_0C_0$ at $V,$ then $A_2$ is midpoint of $\overline{VA_0}$ $\Longrightarrow$ $A_1A_2$ is the $A_0$ -midline of $\triangle A_0UV.$ If $X_0 \equiv B_0C_0 \cap BC,$ then $A_3$ is midpoint of $\overline{A_0X_0}.$ Defining $Y_0,Z_0$ on $CA,AB$ similarly as $X_0,$ we get that $B_3,C_3$ are midpoints of $\overline{B_0Y_0},$ $\overline{C_0Z_0}.$ $X_0,Y_0,Z_0$ are collinear on the trilinear polar $\tau$ of $P$ WRT $\triangle ABC,$ hence $A_3,B_3,C_3$ lie on the Newton's line of the complete quadrangle bounded by the sidelines of $\triangle A_0B_0C_0$ and $\tau,$ i.e. $\overline{A_3B_3C_3}$ is the trilinear polar of $P^2.$