In triangle ABC, DE ll BC where D is point on AB and E is a point on AC.  DE divides the area of triangle ABC in two equal parts.  DB : AB is equal to:

(\(\sqrt {2}\) - 1) : \(\sqrt {2}\)

Since DE ll BC ;

D, E are mid:- points on AB, AC respectively (mid - point theorem)

Therefore, ΔADE ∼ ΔABC

∴ \(\frac{area of ΔABC}{area of ΔADE}\) = \(\frac{AB^2}{AD^2}\)

⇒ \(\frac{AB^2}{AD^2}\) = \(\frac{2}{1}\)

⇒ \(\frac{AB}{AD}\) = \(\frac{\sqrt {2}}{1}\)

Here, DB =  (AB - AD) = \(\sqrt {2}\) - 1

Therefore,

DB : AB = (\(\sqrt {2}\) - 1) : \(\sqrt {2}\)