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Let the line through $G$ parallel to $AB$ intersect $BC$ at $X$, and the line through $G$ parallel to $BC$ intersect $AB$ at $Y$. Mark the length of $GX$ as $a$, and the length of $GY$ as $b$. Its easy to prove that $a = AB/3$ and $b = BC/3$ by solving the system of equations of lines AG and CG.

Let $T$ be the intersection of $GY$ and $CD$.

By using the similar triangles $NDM$ and $NTG$ we get:

$frac{ND}{MD} =frac{NT}{GT} =frac{ND + 2a}{2b} = frac{ND}{2b} + frac{a}{b}$

multiply by: $frac{3b}{ND}$

$frac{3b}{MD} = frac{3}{2} + frac{3a}{ND}$ (1)

Rearranging the original equation and using (1) we get:

$frac{CN}{ND} – frac{AM}{MD} = frac{3a + ND}{ND} – frac{3b – MD}{MD} = frac{3a}{ND} – frac{3b}{MD} + 2 = frac{1}{2}$

This was the ‘only if’ part of the ‘if and only if’. For the ‘if’ part, because we only used equalities, and every equation was equivalent to the previous, we can say that from

$frac{CN}{ND} – frac{AM}{MD} = frac{1}{2}$

follows that

$frac{ND}{MD} =frac{NT}{GT}$

This, along with the fact that the corresponding angles are equal due to parallel lines, shows that triangles $NDM$ and $NTG$ are similar.

From the similarity, we get that the angle $angle MND = angle GNT$.

$M$, $D$, and $T$ are collinear, so $angle GND = angle GNT$.

From the last two sentences: $angle MND = angle GND$, which is a proof that $N$, $M$, and $G$ are collinear.

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