Line through triangle's vertex-centroid midpoint

Question

I've been struggling for a couple of days trying to solve this geometry problem. Here it is:

On this triangle, $G$ is its centroid (meaning that $AM = MC$). Also, $BN = NG$. He wants to know the ratio $\frac{BD}{CD}$.

Now, here are some thoughts I came up with. Since $G$ is the triangle's centroid, $BG = 2\cdot GM$. Thus $BN = NG = GM$. Couldn't use this fact to solve the problem.

I also tried to use Menelaus's theorem, but with no success. Could someone help me with this one? I could solve it using analytic geometry, but the whole point is using synthetic geometry. Thanks in advance!

Answer 1

Why your Menelaus doesn't work?

Observe transversal $A-N-D$ with respect to triangle $BCM$:

$${BD\over DC}\cdot {CA\over AM}\cdot{MN\over NB}=1$$ so

$${BD\over DC}\cdot {2\over 1}\cdot{2\over 1}=1$$

and thus $${BD\over DC}={1\over 4}$$

Answer 2

Let $K\in CD$ such that $MK||AD$ and $DK=2x$.

Thus, $KC=DK=2x$, $BD=\frac{1}{2}DK=x$ and $BD=\frac{1}{4}CD$.