Question 66
The medians $AD$, $BE$ and $CF$ of triangle $ABC$ intersect at $G$. Let $P$ be a point lying in the interior of the triangle, not belonging to any of its medians. The line through $P$ parallel to $AD$ intersects the side $BC$ at $A_1$. Similarly one defines the points $B_1$ and $C_1$. Prove that
$$
\overrightarrow{A_1D} + \overrightarrow{B_1E} + \overrightarrow{C_1F} = \frac{3}{2} \overrightarrow{PG}.
$$
$$
\overrightarrow{A_1D} + \overrightarrow{B_1E} + \overrightarrow{C_1F} = \frac{3}{2} \overrightarrow{PG}.
$$