Ratio of lines intersecting in a triangle

Question

In $ABC$ triangle $M$ is the mid point of $BC$ and $N$ is a point on $AB$ that such that $AN:NB = 2:1$. $AM$ and $CN$ are intersected at the point $D$. What is the ratio of $AD:DM$?

Answer 1

By way of mass points we get the ratio is 4:1.

Place a weight 1 on B and C. Then the weight on M is 2. Since AN:NB is 2:1, and B has weight 1, A has a weight .5. Now M has weight 2 and A has weight .5 so the ratio of AD:DM is 2: .5 = 4:1.

Answer 2

$$area(\triangle DMC):area(\triangle BCD) =MC:BC=1:2$$ and $$area(\triangle BCD):area(\triangle ACD) = BN:AN=1:2.$$ So $$DM:AD = area(\triangle DMC):area(\triangle ACD) = 1:4.$$