AB is a line parallel to the line PQ. If C and D are points on PQ such that the area of $$\triangle ABC$$ is ‘a’ square units, then the area (in sq. units) of $$\triangle ABD$$ is:

A

a

B

$$\frac{a}{2}$$

C

$$\frac{a}{3}$$

D

$$\frac{a}{4}$$