$$\triangle ABC$$is equilateral Incircle of $$\triangle ABC$$ touches sides $$AV,AC$$ at $$PQ$$ respectively.A line through $$P$$ and $$Q$$ meets cicumcircle of $$\triangle ABC$$ at $$R$$ and $$S$$. ($$R$$ is nearer to $$P$$) then the value of $$\dfrac {PQ}{PR}$$ equals
A
$$\dfrac {\sqrt {5}-1}{2}$$
B
$$\dfrac {\sqrt {3}+1}{2}$$
C
$$\dfrac {\sqrt {5}+1}{2}$$
D
$$\dfrac {\sqrt {3}-1}{2}$$
Correct option is A. $$\dfrac {\sqrt {5}+1}{2}$$