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

- (a)5 : 3
- (b)3 : 5
- (c)3 : 2
- (d)2 : 3
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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.
$$\begin{aligned}\Rightarrow \frac{PB}{PD} = \frac{PA}{PC} \\[1em] \Rightarrow \frac{PB}{PD} = \frac{10}{6} \\[1em] \Rightarrow \frac{PB}{PD} = \frac{5}{3}.\end{aligned}$$
Let PB = 5x and PD = 3x.
From figure,
DB = PB - PD = 5x - 3x = 2x.
$\Rightarrow \frac{PD}{DB} = \frac{3x}{2x} = \frac{3}{2}$
⇒ PD : DB = 3 : 2.
Hence, Option 3 is the correct option.