Given $\triangle ABC$, locate points $A\_1$, $B\_1$, $C\_1$ on respective sides $BC$, $CA$, $AB$ such that
$$\frac{BA\_1}{A\_1C} =\frac{CB\_1}{B\_1A} = \frac{AC\_1}{C\_1B} = 2$$

> 
> How can I show that the area of the triangle formed by the intersections of $AA\_1$, $BB\_1$, and $CC\_1$ is $1/7$ of the area of $\triangle ABC$?
> 
> 
>