Proving the area of a triangle within a triangle Consider a triangle with vertices ABC, we pick a point C' on the line segment AB in such a way that |BC'|=2|AC'|. Similarly, we pick a point B' on the line segment AC and a point A' on the line segment BC, with |AB'|=2|CB'| and |CA'|=2|BA'|.
Draw the lines AA', BB', CC'. Prove the area of the little triangle that appears is 1/7 of the area of triangle ABC.
I've drawn the diagrm, however I still have got no idea how to prove this
 A: Let $D,E,F$ be the intersection point between $AA'$ and $BB'$, $BB'$ and $CC'$, $CC'$ and $AA'$ respectively.
Then, note that
$$\frac{[\triangle{DEF}]}{[\triangle{ABC}]}=\frac{[\triangle{ACA'}]}{[\triangle{ABC}]}\times\frac{[AB'A']}{[\triangle{ACA'}]}\times\frac{[\triangle{DFB'}]}{[\triangle{AB'A'}]}\times\frac{[\triangle{DEF}]}{[\triangle{DFB'}]}$$$$=\frac{A'C}{BC}\times\frac{AB'}{AC}\times\frac{DF}{AA'}\times \frac{DE}{DB'}=\frac 49\times\frac{DF}{AA'}\times \frac{DE}{DB'}\tag1$$
So, all you need is to find $\frac{DF}{AA'}$ and $\frac{DE}{DB'}$.
You can use vector here (can you?), but using Menelaus' theorem should be easier :
For $\triangle{ABA'}$ and $CC'$, 
$$\frac{BC}{CA'}\cdot\frac{A'F}{FA}\cdot\frac{AC'}{C'B}=1\Rightarrow \frac{A'F}{FA}=\frac 43.$$
For $\triangle{A'AC}$ and $BB'$,
$$\frac{CB}{BA'}\cdot\frac{A'D}{DA}\cdot\frac{AB'}{B'C}=1\Rightarrow \frac{A'D}{DA}=\frac 16.$$
So, we have $\frac{DF}{AA'}=\frac 37$. Similarly, we have $\frac{DE}{DB'}=\frac 34$.　
Hence, the result follows from these and $(1)$.
