Find the Ratio $BM \colon ME$

Question

In Triangle $\Delta ABC$, the Point $D$ is on $BC$ such that $D$ divides $B$ and $C$ in the Ratio $1 \colon 3$ and there is a point $E$ on $CA$ such that $E$ divides $C$ and $A$ in ratio $1 \colon 3$. if $BE$ and $AD$ meet at Point $M$, Find $BM \colon ME$.

My Try: I assumed $C$ is $(0,0)$ ,$A(x_1,y_1)$ and $B(x_2,y_2)$. Then by Section formulae we have $D(\frac{3x_2}{4},\frac{3y_2}{4})$ and $E(\frac{x_1}{4},\frac{y_1}{4})$. Forming equations of $AD$ and $BE$ and solving thm to find $M$ is laborious. Can i have any other approache

Answer 1

Consider

Area of triangle ABM$=3(x+4y)-3x=12y$

So for triangle ABD and ADC:

$$\frac{12y+y}{4x+3y}=\frac13$$ $$x=9y$$

So $$\frac{BM}{ME}=\frac{3y+y}{x}=\frac{4y}{9y}=\frac4 9$$

Answer 2

Let $F=CM\cap AB$. Ceva's theorem gives:

$$\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$$ hence $\frac{AF}{FB}=9$, then Van Obel's theorem gives: $$\frac{BM}{ME}=\frac{BD}{DC}+\frac{BF}{FA}=\frac{1}{3}+\frac{1}{9}=\color{red}{\frac{4}{9}}.$$