In $\triangle \mathrm { ABC }$, $D$ is a point on $BC$, $AD$ bisects $\angle \mathrm { BAC }$, and the area of $\triangle \mathrm { ABD }$ is 2 times the area of $\triangle \mathrm { ADC }$.
(I) Find $\frac { \sin \angle B } { \sin \angle C }$ ;
(II) If $A D = 1 , D C = \frac { \sqrt { 2 } } { 2 }$, find the lengths of $B D$ and $A C$.