If a line through the centroid $G$ of a triangle ABC meets $AB$ in $M$ and $AC$ on $N$ then prove that $AN. MB+AM. NC=AM. AN$ both in magnitude as well as sign
I tired dividing the equation by $AM. AN$ Thus resultant became $\frac{MB}{AM}+\frac{NC}{AN}=1$ and now I tried to do some construction but couldn't figure out the solution.
On
Let $X$ be the intersection of the lines $MN$ and $BC$. Here is a picture:
Apply the theorem of Menelaus for the triangles $\Delta ABD$, $\Delta ADC$ to get (using sign conventions for proportions of two segments on the same line):
$$ \begin{aligned} 1 &= \frac{MB}{MA}\cdot \underbrace{\frac{GA}{GD}}_{=-2}\cdot \frac{XD}{XB}\ , \\ 1 &= \frac{NC}{NA}\cdot \underbrace{\frac{GA}{GD}}_{=-2}\cdot \frac{XD}{XC}\ , \qquad\text{ and from here} \\[3mm] \frac{MB}{MA} + \frac{NC}{NA} &=-\frac 12\left( \frac{XB}{XD} + \frac{XC}{XD}\right) = -\frac 12\cdot\frac{XB+XC}{XD}= -\frac 12\cdot\frac{2XD}{XD} =\boxed{\ -1\ }\ . \end{aligned} $$ $\square$
Note: By sign convention we have $\displaystyle \frac{MB}{MA} = -\frac{MB}{AM}$, and $\displaystyle \frac{NC}{NA} = -\frac{NB}{AN}$, making the sign a plus one for the sum in the question.
Hint: Draw perpendicular lines to line $MN$ through $A$, $B$, $C$, and the midpoint $D$ of $BC$. Turn the ratios you got into ratios of the length of the perpendicular segments.