In $$\triangle ABC$$, D and E are points on AB and AC respectively such that DE is parallel to BC. If AD = 2 cm, BD = 3 cm, ext-ADE then $$\frac{ar(\triangle ADE)}{ar(\triangle ABC)}$$, is:

A

$$\frac{4}{9}$$

B

$$\frac{16}{81}$$

C

$$\frac{4}{25}$$

D

$$\frac{2}{5}$$