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Let AB=c, BC=a and AC=b. Draw altitude from A, mark its foot as H and it's intersection with DE as G. Let the ratio of areas is:
$\frac{ADE}{BCDE}=k$
We have:
$\frac{ADE}{BCDE}+1=\frac{ADE+BCDE=ABC}{BCDE}=k+1$
$S=ABC=\frac{AH\times BC}2$
$ADE=\frac{DE\times AG}2$
$$\frac{\frac{AH\times BC}2}{\frac{AH\times BC}2-\frac{DE\times AG}2}=k+1$$
$\Rightarrow \frac S{S-DE\times AG}=k+1 \Rightarrow DE\times AG=S\big(1-\frac 1{k+1}\big)=\frac{S\cdot k}{k+1}$
$S=ABC=\sqrt{p(p-a)(p-b)(p-c)}$
$\Rightarrow AH=\frac {2S} a$
Also:
$\frac{AG}{AH}=\frac{DE}{BC}\Rightarrow \frac{DE}{AG}=\frac {BC}{AH}$
Now we have following system of equations:
$\begin{cases}\frac{DE}{AG}=\frac {BC}{AH}=\frac{a^2}{2S}\\DE\times AG=\frac{S\cdot k}{k+1}\end{cases}$
Solving this system gives DE and AG.Now we have:
$$\frac {AD}{c}=\frac{AG}{AH}$$
$$AD=\frac{c\cdot AG}{AH}$$
I did it this way:
Calling that if a triangle (and any 2D polygon) is scaled by a factor $\alpha$ then its surface is scaled by a factor $\alpha^2$ then we have that \begin{equation} ADE=\left(\dfrac{AD}{AB}\right)^2\cdot ABC. \end{equation} Also, we can write the surface $ABC$ using the ratio $ADE/BDEC$ given in the following way: \begin{equation} ABC=ADE+ADE\cdot\left(\dfrac{BDEC}{ADE}\right)=ADE\cdot\left(1+\dfrac{BDEC}{ADE}\right). \end{equation} Joinng both equations we have that \begin{equation} ADE=\left(\dfrac{AD}{AB}\right)^2ABC=ADE\left(\dfrac{AD}{AB}\right)^2\left(1+\dfrac{BDEC}{ADE}\right). \end{equation} Now, since $ADE\not=0$ (otherwise, the exercise is trivial!) we get that \begin{equation} 1=\left(\dfrac{AD}{AB}\right)^2\cdot\left(1+\dfrac{BDEC}{ADE}\right). \end{equation} Solving for $AD$ we end up with \begin{equation} AD=\dfrac{AB}{\sqrt{1+\dfrac{BDEC}{ADE}}}. \end{equation} Note: Notice that although we divided by $ADE$ in some equations, the final formula does not depend on its value (but in its ratio with $BDEC$, which is given).