$$x + \sqrt 3 y = \sqrt 3 $$ or $$y = - {1 \over {\sqrt 3 }}x + 1$$

Let $$\theta$$ be the angle which the line makes with the positive x-axis.

$$\therefore$$ $$\tan \theta = - {1 \over {\sqrt 3 }} = \tan \left( {\pi - {\pi \over 6}} \right)$$ or $$\theta = \pi - {\pi \over 6}$$

$$\therefore$$ $$\angle ABC = {\pi \over 6}$$; $$\therefore$$ $$\angle DBE = {\pi \over 6}$$

$$\therefore$$ the equation of the line BD is,

$$y = \tan {\pi \over 6}x + c$$ or $$y = {x \over {\sqrt 3 }} + c$$ ..... (1)

The line $$x + \sqrt 3 y = \sqrt 3 $$ intersects the x-axis at $$B(\sqrt 3 ,0)$$ and, the line (1) passes through $$B(\sqrt 3 ,0)$$.

$$\therefore$$ $$0 = {{\sqrt 3 } \over {\sqrt 3 }} + c$$ or, c = $$-$$1

Hence, the equation of the reflected ray is,

$$y = {x \over {\sqrt 3 }} - 1$$ or $$y\sqrt 3 = x - \sqrt 3 $$