I1.2 In the figure, $A B=A D=D C=4, B D=2 a$. Find $b$, the length of $B C$.
Let $\angle A D B=\theta, \angle C D B=180^{\circ}-\theta$ (adj. $\angle$ s on st. line) In $\triangle A B D, \cos \theta=\frac{a}{4}=\frac{1}{8}$
Apply cosine formula on $\triangle B C D$.
$$
\begin{array}{l}
b^{2}=(2 a)^{2}+4^{2}-2(2 a) \cdot 4 \cdot \cos \left(180^{\circ}-\theta\right) \\
b^{2}=1+16-2 \cdot 4 \cdot(-\cos \theta)=17+8 \times \frac{1}{8}=18 \\
b=3 \sqrt{2}
\end{array}
$$