QQuestion
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
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
(c) 3 : 2
Explanation:
∠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.
\frac{\mathrm{PB}}{\mathrm{PD}}=\frac{\mathrm{PA}}{\mathrm{PC}}=\frac{10}{6}=\frac{5}{3}Let PB = 5x and PD = 3x. Then DB = PB − PD = 2x.
Therefore, PD : DB = 3 : 2.
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