In the given figure ∠BAP = ∠DCP = 70°, PC = 6 cm and CA = 4 cm, then PD : DB is :

1. 5 : 3

2. 3 : 5

3. 3 : 2

4. 2 : 3

Given,

∠BAP = ∠DCP = 70°

From figure,

∠BAP and ∠DCP are corresponding angles and since they are equal.

∴ AB // CD.

From figure,

PA = PC + AC = 6 + 4 = 10 cm.

In △ PAB and △ PCD,

⇒ ∠PAB = ∠PCD (Both equal to 70°)

⇒ ∠BPA = ∠DPC (Both are equal)

∴ △ PAB ~ △ PCD (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{PB}{PD} = \dfrac{PA}{PC} \\[1em] \Rightarrow \dfrac{PB}{PD} = \dfrac{10}{6} \\[1em] \Rightarrow \dfrac{PB}{PD} = \dfrac{5}{3}.$

Let PB = 5x and PD = 3x.

From figure,

DB = PB - PD = 5x - 3x = 2x.

$\Rightarrow \dfrac{PD}{DB} = \dfrac{3x}{2x} = \dfrac{3}{2}$

⇒ PD : DB = 3 : 2.

Hence, Option 3 is the correct option.