In the adjoining figure, XY is parallel to BC. If XY divides the triangle into two equal parts, then $\dfrac{AX}{AB}$ equals:

1. $\dfrac{1}{\sqrt{2}}$

2. $\dfrac{1}{2}$

3. $\dfrac{\sqrt{2} + 1}{\sqrt{2}}$

4. $\dfrac{\sqrt{2} - 1}{\sqrt{2}}$

Given,

In ΔABC and ΔAXY

∠BAC = ∠XAY [Common angle]

∠AXY = ∠ABC [Corresponding angles are equal]

∴ ΔABC ∼ ΔAXY (By A.A. axiom)

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\dfrac{\text{Area of ΔAXY}}{\text{Area of ΔABC}} = \Big(\dfrac{AX}{AB}\Big)^2$ …..(1)

X and Y divides triangle ABC into two equal parts. Therefore,

Area of ΔAXY = $\dfrac{1}{2}$ Area of ΔABC

$\dfrac{\text{Area of ΔAXY}}{\text{Area of ΔABC}} = \dfrac{1}{2}$ …..(2)

Equating Eqn(1) and Eqn(2) :

$\Rightarrow \dfrac{1}{2} = \Big(\dfrac{AX}{AB}\Big)^2 \\[1em] \Rightarrow \dfrac{AX}{AB} = \dfrac{1}{\sqrt{2}}.$

Hence, option 1 is the correct option.