In a triangle ABC, a line XY is drawn parallel to BC meeting AB in X and AC in Y. The area of the triangle AXY is half of the area of the triangle ABC. XY divides AB in the ratio of :

Answer: Option C

Note : The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Since, XY || BC, we have :

$$\begin{array}{rl}& \angle AXY=\angle ABC\text{ and }\\ & \angle AYX=\angle ACB\\ & \text{Aslo, }\angle A=\angle A\text{(common)}\\ & \text{So, }△AXY\sim △ABC\\ & \text{Let area (}△ABC\text{)}=x\text{sq. units}\\ & \text{Then,}\\ & \text{Area (}△AXY\text{) = }\frac{x}{2}\text{sq. units}\\ & \frac{{\left(AB\right)}^2}{{\left(AX\right)}^2}=\frac{x}{\left(\frac{x}{2}\right)}\\ & \Rightarrow \frac{AB}{AX}=\sqrt{2}\\ & \Rightarrow \frac{AX+BX}{AX}=\sqrt{2}\\ & \Rightarrow 1+\frac{BX}{AX}=\sqrt{2}\\ & \Rightarrow \frac{BX}{AX}=\left(\sqrt{2}-1\right)\\ & \Rightarrow \frac{AX}{BX}=\frac{1}{\left(\sqrt{2}-1\right)}\text{ Or 1}:\left(\sqrt{2}-1\right)\end{array}$$