imc

We have $$\frac{[ADE]}{[ABC]}=\frac{AD}{AB}\cdot \frac{AE}{AC}=\frac{19}{25}\cdot \frac{14}{42}=\frac{19}{75}.$$ (Area of a triangle is base times height, so the area ratio of triangles, that have a common vertex (height) and bases on a common line, is the base length ratio. This is applied twice, using different pairs of bases, and corresponding altitudes for height.). $[BCED]=[ABC]-[ADE]$, so $$\begin{array}{rl}\frac{[ADE]}{[BCED]}& =\frac{[ADE]}{[ABC]-[ADE]}\\ & =\frac{1}{\frac{ABC}{ADE}-1}\\ & =\frac{1}{\frac{75}{19}-1}\\ & =\boxed{\frac{19}{56}}.\end{array}$$ 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 $$\frac{[ADE]}{[ABC]}=\frac{\frac{1}{2}\cdot 19\cdot 14\cdot \sin (Z)}{\frac{1}{2}\cdot 25\cdot 42\cdot \sin (Z)}=\frac{19\cdot 14}{25\cdot 42}=\frac{ab}{cd}$$.