Area of a triangle inside a triangle 
In $ABC$, let $D$, $E$, and $F$ be points on the sides $BC$, $AC$ and $AB$, respectively, such that $BC = 4CD$, $AC=5AE$, and $AB= 6BF$.
If the area of $ABC$ is $120$, what is the area of $DEF$.

I tried connecting the vertices of the inner triangle to the opposite vertices of the outer triangle, but I think that they wont be collinear because of Ceva's Theorem, I cant think of anything else except for assuming the triangle is right and using coordinate geometry. I am asking for a nice solution
 A: We have
$$[\triangle{CDE}]=\frac{CE}{CA}[\triangle{ADC}]=\frac{CE}{CA}\times \left(\frac{CD}{BC}[\triangle{ABC}]\right)=\frac 15[\triangle{ABC}]$$
Similarly, we get
$$[\triangle{BDF}]=\frac{3}{4}[\triangle{BCF}]=\frac 34\times \frac{1}{6}[\triangle{ABC}]=\frac 18[\triangle{ABC}]$$
and
$$[\triangle{AFE}]=\frac 15[\triangle{ACF}]=\frac 15\times\frac{5}{6}[\triangle{ABC}]=\frac 16[\triangle{ABC}]$$
Now note that
$$[\triangle{DEF}]=[\triangle{ABC}]-[\triangle{CDE}]-[\triangle{BDF}]-[\triangle{AFE}]$$
A: 
well, i am a Chinese and i am worried about i can't explain it well. maybe u can look at the pic. first.
A: I use @Ajar picture and use the formula for triangle's area based on 2 sides and the angle between them.
It is easy to understand that $S_{\triangle{DEF}} = S_{\triangle{ABC}} - S_{\triangle{FBD}} - S_{\triangle{AFE}} - S_{\triangle{EDC}}(0)$
Let me assign angles $\alpha,\beta, \gamma$ for $\triangle{ABC}$
Also 
$CD = a, BC = 4*a$
$AE = b, AC = 5*b$
$BF = c, AB = 6*c$
$ S_{\triangle{ABC}} = 120$
So 
$S_{\triangle{AFE}}=\frac{1}{2} * AF*AE*sin\alpha = \frac{1}{2} * 5*c*b*sin\alpha(1)$
$S_{\triangle{ABC}}=\frac{1}{2} * AB*AC*sin\alpha = \frac{1}{2} * 6*c*5*b*sin\alpha(2)$
From (2) $sin\alpha=\frac{S_{\triangle{ABC}}}{15bc} (3)$
Put (3) to (1) 
$S_{\triangle{AFE}}=\frac{1}{2} * 5*c*b*(\frac{S_{\triangle{ABC}}}{15bc}) = \frac{S_{\triangle{ABC}}}{6}(4)$
Do the same with $\triangle{FBD}, \triangle{EDC}$
$S_{\triangle{FBD}} = \frac{S_{\triangle{ABC}}}{2} (5)$
$S_{\triangle{EDC}} = \frac{S_{\triangle{ABC}}}{5} (6)$
From (0) $S_{\triangle{DEF}} = \frac{S_{\triangle{ABC}}}{15} = \frac{120}{15} = 8 (7)$
Questions, errors, comments, suggestions.
