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

It is given that FE divides $$\triangle ABC$$ into two equal parts.

Area of $$\triangle ABC$$ = 2 $$\times \triangle AFE$$

$$\frac{1}{2} \times BC \times AD = \frac{1}{2} \times FE \times AG \times 2$$ 

$$BC \times AD = 2 \times FE \times AG$$

$$\frac{BC}{FE} = \frac{2AG}{AD}$$

Also .

Area of $$\triangle AFE$$ = Area of trapezium BFEC

=>$$ \frac{1}{2} \times FE \times AG = \frac{1}{2} \times (BC + EF) \times DG$$

=>$$ \frac{1}{2} \times FE \times AG = \frac{1}{2} \times BC \times DG + \frac{1}{2 \times EF \times DG}$$

=> $$1 = \frac{BC \times DG}{AG \times FE} + \frac{DG}{AG}$$

=>$$ 1 = \frac{2DG}{AD} + \frac{DG}{AG}$$

=>$$1 - \frac{DG}{AG} =\frac{2DG}{AD}$$

=>$$ \frac{AD}{2DG} = \frac{1}{1 - \frac{DG}{AG}} $$

=>$$\frac{AG + GD}{2DG} =  \frac{1}{1 - \frac{DG}{AG}} $$

=>$$\frac{AG}{DG} + 1 = \frac{2}{1 - \frac{DG}{AG}}$$

Let $$\frac{DG}{AG}}$$ be x.

=>$$\frac{1}{x} + 1 = \frac{2}{1 - x}$$

=>$$\frac{1+x}{x} = \frac{2}{1-x}$$

=>$$x^2 + 2x -1 =0$$

=> x=$$(\sqrt{2} - 1) : 1$$

So , the answer would be option b)$$(\sqrt{2} - 1) : 1$$