In ΔABC, F and E are the points on sides AB and AC, respectively, such that FE ll BC and FE divides the triangle  in two parts of equal area.   If AD ⊥ BC and AD intersects FE at G, then GD : AG = ?

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

Similar triangle Theorem applied

ΔAGE ∼ ΔADC

(\(\frac{AG}{AD}\))² = \(\frac{ar.\;of\;ΔAGE}{ar.\;of\;ΔADC}\) = \(\frac{1}{2}\)

\(\frac{AG}{AD}\) = \(\frac{1}{\sqrt {2}}\)

[Area of ΔAEF = area of quadrilateral FECD, similarly; [Area of ΔAGE = area of quadrilateral GECD = 1 Unit]

GD = AD - AG = \(\sqrt {2}\) - 1

\(\frac{AG}{GD}\) = \(\frac{1}{\sqrt {2} - 1}\)

GD : AG = ( \(\sqrt {2}\) - 1) : 1