geometry

[DEF]/[ABC] = 5/12

As shown in the diagram,$~BD/DC=2,~CE/EA=3$,  and $AF/FB=4$.
Find $[DEF]/[ABC]$.

$\begin{array}{rcll} \text{ Let } BC&=&a\\ \text{ Let } BD&=&\frac{2}{3}a \\ \text{ Let } DC&=&\frac{1}{3}a \\\\ \text{ Let } CA&=&b \\ \text{ Let } CE&=&\frac{3}{4}b \\ \text{ Let } EA&=&\frac{1}{4}b \\\\ \text{ Let } AB&=&c \\ \text{ Let } AF&=&\frac{4}{5}c \\ \text{ Let } FB&=&\frac{1}{5}c \\ \end{array}$

$\begin{array}{rcll} \text{ Let } \angle CAB&=& A\\ \text{ Let } \angle ABC&=& B\\ \text{ Let } \angle BCA&=& C\\ \end{array}$

$\begin{array}{|rcll|} \hline 2*[ABC] &=& bc\sin(A) \\\\ 2*[AFE] &=& \frac{1}{4}b*\frac{4}{5}c \sin(A) \\ 2*[AFE] &=& \frac{1}{5}bc \sin(A) \\ 2*5*[AFE] &=& bc \sin(A) \\ \hline bc\sin(A) = 2*[ABC] &=& 2*5*[AFE] \\ 2*[ABC] &=& 2*5*[AFE] \\ [ABC] &=& 5*[AFE] \\ \mathbf{[AFE]} &=& \mathbf{\frac{1}{5}[ABC]} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline 2*[ABC] &=& ca\sin(B) \\\\ 2*[FBD] &=& \frac{1}{5}c*\frac{2}{3}a \sin(B) \\ 2*[FBD] &=& \frac{2}{15}ca \sin(B) \\ 2*\frac{15}{2}*[FBD] &=& ca \sin(B) \\ \hline ca \sin(B) = 2*[ABC]&=&2*\frac{15}{2}[FBD] \\ 2*[ABC]&=&2*\frac{15}{2}[FBD] \\ [ABC]&=&\frac{15}{2}[FBD] \\ \mathbf{[FBD]} &=& \mathbf{\frac{2}{15}[ABC]} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline 2*[ABC] &=& ab\sin(C) \\\\ 2*[EDC] &=& \frac{1}{3}a*\frac{3}{4}b \sin(C) \\ 2*[EDC] &=& \frac{1}{4}ab \sin(C) \\ 2*4*[EDC] &=& ab \sin(C) \\ \hline ab \sin(C) = 2*[ABC] &=& 2*4*[EDC] \\ 2*[ABC] &=& 2*4*[EDC] \\ [ABC] &=& 4*[EDC] \\ \mathbf{[EDC]} &=& \mathbf{\frac{1}{4}[ABC]} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline [DEF] + [AFE]+[FBD]+[EDC] &=& [ABC] \\ [DEF] + \frac{1}{5}[ABC]+\frac{2}{15}[ABC]+\frac{1}{4}[ABC] &=& [ABC] \\ [DEF] + [ABC] \left(\frac{1}{5}+\frac{2}{15}+\frac{1}{4} \right) &=& [ABC] \\ [DEF] + \frac{7}{12}[ABC] &=& [ABC] \\ [DEF] &=& [ABC]- \frac{7}{12}[ABC] \\ [DEF] &=& [ABC] \left(1- \frac{7}{12} \right) \\ [DEF] &=& \frac{5}{12}[ABC] \\ \mathbf{\frac{[DEF]}{[ABC]}} &=& \mathbf{\frac{5}{12}} \\ \hline \end{array}$