In the figure,the line segment $XY$ is parallel to side $AC$ of $\Delta ABC$ and it divides the triangle into two parts of equal areas. Find the ratio $\frac{AX}{AB}$.

(D) We have $XY \parallel AC$ (Given).
$\angle BXY = \angle A$ and $\angle BYX = \angle C$ (Corresponding angles).
Therefore,$\Delta ABC \sim \Delta XBY$ ($AA$ similarity criterion).
$\frac{\operatorname{ar}(ABC)}{\operatorname{ar}(XBY)} = \left(\frac{AB}{XB}\right)^2$ (Theorem regarding ratio of areas of similar triangles).
Since $XY$ divides the triangle into two parts of equal areas,$\operatorname{ar}(ABC) = 2 \operatorname{ar}(XBY)$.
$\frac{\operatorname{ar}(ABC)}{\operatorname{ar}(XBY)} = \frac{2}{1}$.
Therefore,$\left(\frac{AB}{XB}\right)^2 = \frac{2}{1}$,which implies $\frac{AB}{XB} = \frac{\sqrt{2}}{1}$.
Taking the reciprocal,$\frac{XB}{AB} = \frac{1}{\sqrt{2}}$.
Now,$\frac{AX}{AB} = \frac{AB - XB}{AB} = 1 - \frac{XB}{AB} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$.
Rationalizing the denominator,$\frac{AX}{AB} = \frac{(\sqrt{2} - 1) \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{2 - \sqrt{2}}{2}$.