In triangle ABC, $\overline { \mathrm { AB } } = 1$, $\angle \mathrm { A } = \theta$, and $\angle \mathrm { B } = 2 \theta$. Point D on side AB is chosen so that $\angle \mathrm { ACD } = 2 \angle \mathrm { BCD }$. When $\lim _ { \theta \rightarrow + 0 } \frac { \overline { \mathrm { CD } } } { \theta } = a$, find the value of $27 a ^ { 2 }$. (Given that $0 < \theta < \frac { \pi } { 4 }$.) [4 points]