Hard at first sight

Since only areas and ratios within the same line are in play, result obtained for one triangle will apply to all triangles obtainable from it by linear transformation, that is to all triangles. So I picked a right triangle with bottom side 1, as shown above.

Triangles $AED$ and $ACB$ are similar, so $\frac{CB}{1}=\frac{ED}{AD}=\frac{h}{x}$ .

Area of triangle $AED=\frac{1}{2}xh$ . Area of triangle $ABC=\frac{1}{2}\frac{h}{x}$ .

Area of triangle $ACB$ should be twice as big as area of triangle $AED$ .

$\frac{1}{2}xh\times2=\frac{1}{2}\frac{h}{x}$ .

This gives us $x=\frac{1}{\sqrt{2}}$

$\frac{BD}{1}=1-x=1-\frac{1}{\sqrt{2}}\approx0.293$

$DE || BC \\ \Delta ADE \sim \Delta ABC \text{ (AA Similarity)} \\ \dfrac{\text{ar(ABC)}}{\text{ar(ADE)}} = (\dfrac{AB}{AD})^2 \\ \dfrac{\text{ar(ABC)}}{2\text{ar(ADE)}} = (\dfrac{AB}{AD})^2 \text{ (Since, ar(ADE) = ar(DECB) = 2ar(ABC))} \\ \dfrac{1}{2} = (\dfrac{AB}{AD})^2 \\ \dfrac{1}{\sqrt[]{2}} = \dfrac{AB}{AD} \\ \sqrt[]{2} = \dfrac{AD}{AB} \\ -\sqrt[]{2} = -\dfrac{AD}{AB} \\ 1 - \sqrt[]{2} = 1 - \dfrac{AD}{AB} \\ 1 - \dfrac{\sqrt[]{2}}{2} = \dfrac{AB - AD}{AB} \\ \dfrac{DB}{AB} = \dfrac{2 - \sqrt[]{2}}{2} = 0.293\\$

Here is the same question posted by me ealier