In $ABC$ we have $AB = 25$, $BC = 39$, and $AC=42$. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$. What is the [ratio](https://artofproblemsolving.com/wiki/index.php/Ratio) of the area of triangle $ADE$ to the area of the [quadrilateral](https://artofproblemsolving.com/wiki/index.php/Quadrilateral) $BCED$?
$\textbf{(A) } \frac{266}{1521}\qquad \textbf{(B) } \frac{19}{75}\qquad \textbf{(C) }\frac{1}{3}\qquad \textbf{(D) } \frac{19}{56}\qquad \textbf{(E) } 1$
We have
\[\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.\]

But $[BCED] = [ABC] - [ADE]$, so
\begin{align*} \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ &= \frac{1}{\frac{ABC}{ADE} - 1} \\ &= \frac{1}{\frac{75}{19} - 1} \\ &= \boxed{\textbf{(D) }\frac{19}{56}}. \end{align*}

Note: If it is hard to understand why \[\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC}\], you can use the fact that the area of a triangle equals $\frac{1}{2} \cdot ab \cdot \sin(C)$. If angle $DAE = Z$, we have that \[\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}\].