Area in a triangle divided by a segment that goes through the centroid. In the image, $G$ is the centroid of $\triangle ABC$. If $BN=4NC$, and the area of $ACNM$ is $19$, find the area of $\triangle MNB$.
My try: I draw a segment parallel to $AC$ from $M$ to $BC$. I did the same from $N$ to $AC$. Then I tried to apply some similarity of triangles, drawing the heights of the similar triangles and trapezoids that are created by those segments, but I couldn't proceed further. Any help to do this problem?
P.S: I'm searching a solution that doesn't involve trigonometry, but all solutions are appreciated.

 A: 
Let [.] denote areas and $\frac{AM}{AB} =  x$. Given that $G$ is the centroid, we have $\frac{AD}{DC}=1$ and $\frac{BG}{GD} = 2$.
Then, evaluate the areas below in terms of $I=[ABC]$,
$$[CND] = \frac15 [BDC]=\frac15\cdot \frac12I$$
$$[MBN] = (1-x)[ABN] = (1-x)\cdot \frac45 I$$
$$[MDN] =\frac12 [MBN]= (1-x)\cdot \frac25 I$$
$$[AMD] = x [ABD] =\frac x2 I$$
The sum of above four areas is equal to $I$, which leads to
$$\frac 12 x + \frac65 (1-x)+ \frac1{10}= 1$$
Solve to obtain $x = \frac37$. Then, $[MBN] = (1-x)\frac45 I=\frac{16}{35}I$ and  
$$\frac{[MBN]}{[ACNM]}= \frac{[MBN]}{I-[MBN]} 
= \frac{\frac{16}{35}I}{I-\frac{16}{35}I}=\frac{16}{19}$$
Thus, the area of $MBN$ is
$$[MBN] = \frac{16}{19}[ACNM]= 16$$
A: Let $\vec{BA}=\vec{a},$ $\vec{BC}=\vec{c},$ $\vec{BM}=x\vec{a}$ and $\vec{MG}=y\vec{MN}.$
Thus, $$-\vec{BM}+\vec{BG}=\vec{MG}$$ or
$$-x\vec{a}+\frac{1}{3}\vec{a}+\frac{1}{3}\vec{c}=y\left(-x\vec{a}+\frac{4}{5}\vec{c}\right),$$ which gives the following system:
$$-x+\frac{1}{3}=-xy$$ and
$$\frac{1}{3}=\frac{4}{5}y,$$ which gives
$$y=\frac{5}{12},$$ $$x=\frac{4}{7}$$ and
$$\frac{S_{\Delta BMN}}{S_{\Delta ABC}}=\frac{4}{7}\cdot\frac{4}{5}=\frac{16}{35}.$$
Thus, $$\frac{S_{ACMN}}{S_{\Delta ABC}}=\frac{19}{35},$$ $$S_{\Delta ABC}=35$$ and
$$S_{\Delta BMN}=16.$$
Another way:
Let $BK$ be a median of $\Delta ABC$, $L\in AM$ such that $LK||MN,$ $LK\cap BC=\{P\}$, $Q\in BC$ such that $AQ||MN$ and $BN=4x$.
Thus, $NC=x$ and by the Thales's theorem $$\frac{NP}{4x}=\frac{NP}{BN}=\frac{GK}{BQ}=\frac{1}{2},$$ which gives $$NP=2x$$ and $$PC=x.$$
Now, since $KP||AQ$ and $AK=KC$, we obtain:
$$PQ=PC=x,$$ which gives $$BM:ML:LA=BN:NP:PQ=4:2:1$$ and $$BM=\frac{4}{7}BA.$$
The rest is the same. 
