If a line through the centroid G of triangle ABC meets AB in M and AC in N then prove that AN.MB +AM.NC = AM.AN both in magnitude and sign. 
If a line through the centroid $G$ of $\triangle ABC$ meets $AB$ in $M$ and $AC$ in $N$ then prove that $$AN.MB +AM.NC=AM.AN$$

both in magnitude and sign.
 A: Let $MN$ meet line $BC$ at $K$ and let $D$ be a midpoint for $BC$. Also let $BK = y$ and $BD=CD =x$.

By Menelaus theorem for $\triangle ABD$ and transversal $K-M-G$ we have:
$${BM\over AM}{AG\over GD} {DK\over KB} = 1\implies {BM\over AM} = {y\over 2(x+y)}$$
By Menelaus theorem for $\triangle ACD$ and transversal $K-G-N$ we have:
$${CN\over AN}{AG\over GD} {DK\over KC} = 1\implies {CN\over AN} = {y+2x\over 2(x+y)}$$
Finally:
$${BM\over AM} + {CN\over AN}  = {y\over 2(x+y)}+ {y+2x\over 2(x+y)}=1$$
A: Let $\vec{AB}=\vec{a},$ $\vec{AC}=\vec{b},$ $\vec{AM}=x\vec{a}$ and $\vec{AN}=y\vec{b}.$
Thus, $$\vec{AG}=\frac{2}{3}\left(\frac{1}{2}\vec{a}+\frac{1}{2}\vec{b}\right)=\frac{1}{3}(\vec{a}+\vec{b})$$ and there is $z$ for which $\vec{MN}=z\vec{MG}.$
Hence,
$$-x\vec{a}+y\vec{b}=z\left(-x\vec{a}+\frac{1}{3}(\vec{a}+\vec{b})\right),$$ which gives
$$-x=-zx+\frac{z}{3}$$ and $$y=\frac{z}{3},$$ which gives
$$x+y=3xy$$ or
$$\frac{AM}{AB}+\frac{AN}{AC}=\frac{3AM\cdot AN}{AB\cdot AC}$$ or
$$AM(AN+NC)+AN(AM+MC)=3AM\cdot AN$$ or
$$AM\cdot NC+AN\cdot MB=AM\cdot AN.$$
A: 
So $AM\times NC+AN\times BM=AM\times AN$ is equivalent to $AM\times AC+AN\times AB = 3AM\times AN$. Further equivalently, $S_{ACM}+S_{ABN}=3S_{AMN}$ where $S$ denotes area. Further equivalently, $S_{BMN}+S_{CMN}=S_{AMN}$.
Now let $P$ be the midpoint of $AC$ and mirror $N$ around $P$ so $PN=PN'$. Note $S_{MNN'} = 2S_{MNP} = S_{MNB}$. (Imagine connecting $BP$ and $BG=2GP$ for the latter equality where $G$ is the centroid.) 
So $S_{BMN}+S_{CMN}= S_{MN'C}=S_{AMN}$ because $AN=N'C$ and the result proven.
