Since ABCD is a square, $ AD \parallel EC $. Hence $ \triangle ECF \cong \triangle ADF $, by AA Congruence, and $ \triangle ECF \sim \triangle ADF $, by AA Similarity. Now, let $ 2 EC = FC = x $. Since $ [ECF] = 8 $, Now, $ [ECF] = x \cdot \frac{x}{2} \cdot \frac{1}{2} = 8 $. Solving this Eq, we get $ x = 4\sqrt{2} $. Now, since $ AB \parallel DF $ , $ \triangle AGF \sim \triangle FED $, and $ \frac{BG}{GD} = \frac{AB}{DF} = \frac{1}{2} $. Let $ BG = y $ , then $ GD = 2y $ and $ BD = 3y $. By Pythagoras Theorem, we get $ (AB)^2 + (AD)^2 = 32 + 32 = 64 = (BD)^2 = 9y^2 $. Solving this Eq, we get, $ BG = y = \frac{\sqrt{8}}{3} $. We know, $ BG = \frac {\sqrt{8}}{3} $, $ BE = \sqrt{8} $ and $ \angle DBC = 45\deg $ (Since BD is a Diagonal). The Area of a triangle is $ \Delta = \frac{1}{2} \cdot a \cdot b \cdot sin C $, we will use this to $ \triangle GBE $ and Now, $ [GBE] = \frac{1}{2} \cdot GB \cdot BE \cdot sin(\angle GBE ) = \frac{4}{3} $ (

Lot of Right Angles mean Lot of Similar and Congruent Triangles Hint2Use Pythagoras and find as many Lengths as possible, which are are near to $ \triangle GBE $. Hint3Unable to Find $ [GBE] $ ? Think about $\frac{1}{2}\cdot a\cdot b\cdot sin( C )$