In the given diagram, PQ // BC and ar∆APQ = ar quad BPQC. The value of AP : PB is:

Given,

PQ // BC and ar ∆APQ = ar quad BPQC.

From figure,

ar ∆ABC = ar ∆APQ + ar quad BPQC = ar ∆APQ + ar ∆APQ = 2 ar ∆APQ.

$\Rightarrow \frac{\text{ar ∆APQ}}{\text{ar ∆ABC}} = \frac{1}{2}$

In ∆APQ and ∆ABC,

⇒ ∠PAQ = ∠BAC [Common]

⇒ ∠APQ = ∠ABC [Corresponding angles as PQ // BC]

∴ ∆APQ ~ ∆ABC (By A.A. axiom)

We know that,

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

$$\begin{aligned}\therefore \frac{\text{ar ∆APQ}}{\text{ar ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{1}{2} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{AP}{AB} = \frac{1}{\sqrt{2}}.\end{aligned}$$

Let AP = $k$, then AB = $\sqrt{2}k$.

$$\begin{aligned}\Rightarrow PB = AB - AP = \sqrt{2}k - k = (\sqrt{2} - 1)k \\[1em] \Rightarrow \frac{AP}{PB} = \frac{k}{(\sqrt{2} - 1)k} = \frac{1}{\sqrt{2} - 1}.\end{aligned}$$

Rationalizing the denominator,

$\Rightarrow \frac{AP}{PB} = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{2 - 1} = \frac{\sqrt{2} + 1}{1}.$

AP : PB = $(\sqrt{2} + 1)$ : 1.

Hence, option 4 is the correct option.