area of triangle In $\triangle ABC$ points $D,E,F$ are on the sides $AB,BC,CA$, respectively, with $AD=DB$, $CE=3BE$ and $AF=2CF$. If the area of $\triangle ABC$ is $480 cm^2$, how do we find the area of $\triangle DEF$?
 A: Hint:
Drop the perpendicular from $B$ to $AC$ and $D$ to $AC$, and use that to find the area of $\triangle DAF$.
Use same idea for others.
A: Denote 
$$
\vec a = \vec{AB}, \qquad \vec b = \vec{AC}
$$
Then
$$
\vec c = -\vec a + \vec b
$$
and
$$
\vec{DE} = \frac12 \vec a + \frac14 (-\vec a + \vec b) = \frac14 \vec a + \frac14 \vec b
$$
$$
\vec{DF} = -\frac12 a + \frac23 \vec b
$$
Area of the $\triangle DEF$ can be calculated as 
$$
\frac12 |\vec{DE} \times \vec{DF}| = \frac12 |(\frac14 \vec a + \frac14 \vec b) \times (-\frac12 a + \frac23 \vec b)| = \frac{7}{48} |\vec a \times \vec b|
$$
since $\vec a \times \vec a = \vec b \times \vec b = 0$ and $\vec a \times \vec b = \vec b \times \vec a$.
Area of $\triangle ABC$ is also
$$
\frac12 |\vec{a} \times \vec{b}| = 480
$$
implying
$$
|\vec{a} \times \vec{b}| = 960
$$
Area of $\triangle DEF$ is therefore
$$
\frac{7}{48} \cdot 960 = 140
$$
A: Hint: Calculate the area of $ADF, BDE, ECF$. Subtract their sum from $ABC$ to get $DEF$.

Further Hint: $\frac{\mbox{Area } ADF } { \mbox {Area } ABC } = \frac{ AD } { AB} \times \frac { AF} { AC} $.
