ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP : AB.

Triangle ABC is shown in the figure below:

In ΔAPQ and ΔABC,

∠PAQ = ∠BAC [Common]

∠APQ = ∠ABC [Corresponding angles are equal]

∴ ΔAPQ ~ ΔABC [By AA].

According to question,

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

$\dfrac{\text{Area of ΔAPQ}}{\text{Area of ΔABC}} = \dfrac{1}{2}$

We know that,

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

$\Rightarrow \dfrac{AP^2}{AB^2} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{AP}{AB} = \dfrac{1}{\sqrt{2}}.$

Let AP = x and AB = $\sqrt{2}$x

From figure,

BP = AB - AP = $\sqrt{2}x - x$

$\therefore \dfrac{BP}{AB} = \dfrac{\sqrt{2}x - x}{\sqrt{2}x} \\[1em] = \dfrac{\sqrt{2} - 1}{\sqrt{2}} \\[1em] = \dfrac{\sqrt{2} - 1}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} \\[1em] = \dfrac{2 - \sqrt{2}}{2}.$

Hence, BP : AB = $(\sqrt{2} - 1) : \sqrt{2}$ = $(2 - \sqrt{2}) : 2$.